Wow, this brings back memories. We debated this one for *weeks* when I was in grad school! There was one guy who was a Ph.D. chemist who *never* came around to be convinced by this. Marilyn vos Savant once posted this in one of her columns and I have never seen such outrage. I recall one angry letter from a statistician (again, a Ph.D.) vowing never to read her column again. "There is enough mathematical illiteracy out there without your contributing to it with this nonsense!" or some such. It just goes to show you that even Ph.D. scientists are not immune to the set of prejudices we like to refer to as "common sense."
Now remind me some day to post the OTHER big argument from my grad school days -- the island with the blue-eyed and brown-eyed people. ;-)
I remember discussing the problem in an undergrad math class... Dr. Barefoot, now he's one of the smartest people I've ever known and imposing as hell too, 6"4' ish with broad shoulders, but the most softspoken person I've ever known. (Got a little swept away here, he died of cancer last summer at 53 which is too damn young!)
I remember this problem, and I'm sure I got it wrong the first time I tried to answer--much like I did when I read this article. I made a snap judgement...without thinking...or writing it out. But it doesn't take too much convincing to make me believe that you should switch. Monty has exerted control over the indepedent probabilities, and thus throws the problem into a logic nightmare.
I'm always a fan of the coin throwing problem. If I flip a coin 10 times and it comes up heads 9 times in a row, what is the probability that it will come up heads on the 10th throw? Um...1/2. The same as it was the first 9 tosses.
When you first pick door C, your odds of winning are 1/3, right? That means there's a 2/3 chance the prize will be behind door A or B. If Monty Hall shows you B and it's empty, then there's a 2/3 chance the prize is behind door A.
Here's the trick. Initially your odds of winning are 1 in 3 because their is completely symmetry among the 3 doors. Naively, we tend to think there is also symmetry between the remaining 2 doors *after* we have been shown the empty one... but that's not true.
Monty Hall has *broken* the symmetry by showing you the other door. Why? He did not choose a door randomly! He intentionally showed you a crap door just to keep the game more exciting! His decision was not random, but informed. In doing so, he has provided you (indirectly) with information.
If you know any computer programming languages, it can be helpful to run a simulation of this with a random number generator and (say) a thousand trials. As you write the logic for this experiment, you will begin to see how/why it works.
When you first pick door C, your odds of winning are 1/3, right? That means there's a 2/3 chance the prize will be behind door A or B. If Monty Hall shows you B and it's empty, then there's a 2/3 chance the prize is behind door A.
Isn't there also then a 2/3 chance that the prize is behind door C?
I don't get it, and I'm a pretty stunningly brilliant guy you know!! :-)
What a beautiful problem!
It also stresses the importance of information in decisionmaking.
We never have such levels of certainty in daily life, so we should be all the more humble about our opinions.
This has been for years one of the two questions I ask any graduate student foolish enough to want to study with me. It is also used as the basis of the probability section of the new NSF's special teaching program for secondary mathematic's teachers that was written by Prof. Daniel Fendel at SF State (who also wrote a review of Oncogenes, Aneuploidy and AIDS on Amazon).
This question I used to test their quantitative thinking even in the face of apparent contradictions with common sense.
The other question I asked them was one that a 9-ball hustler stumped me for several hours with in Oakland back in the 60s. I used this to test the ability of the prospective student to think non-linearily as it is now called.
Two folks decide to have a race to determine whose car is SLOWER. How do they do this?
Unlike the first question, this is a standard brainteaser in that once you get the answer it is really obvious.
Thank you so much for the opportunity to write this.
This may be another way of looking at it that lets you grok it.
When you have your first choice, you have a 1 in 3 chance. That isn't going to change if you don't change your answer. Start to finish, that door is 1 in 3.
When Monty opens the second door, he is essentially letting you change your answer and pick both doors. Not only is he letting you pick both doors (2:3) he is letting your throw out the wrong one.
That means that you can keep your first choice (which is locked in at 1:3) or you can change your answer and get two for the price of one (2:3).
The contestant makes a choice. His odds of selecting the winner is 1/3. Monty opens one door, asks do you wish to change your choice. If you say no, your odds of being winner remains at 1/3. If you decide to change doors, you're odds of winning increase to 1/2. Right ?
I think the reason most people (including myself a while ago) get this wrong is a failure to take into account what Cynical Nation pointed out, That Monty Knows where the prize is. Without that assumption I don't think the math works ( right?), and that assumption never seems to be stated clearly, (Dean, you didn't, nor did the person who told me the thing, who never claimed to know why the answer is what it is).
There is still a linguisic error which threw me initially"
"Now, I'm going to open one of the other two doors for you that has a booby prize."
Doesn't this say that both doors conceal a booby prize? Shouldn't it read one of the other two doors that MAY have a booby prize? I'll bet this is where you think you've already fixed the error you mentioned, but it still reads wrong to me.
Rot-13'd or not you are both quite correct, and got it much faster than dim-witted me sotted out in a poolroom at 2 in the am, and thinking about angular momentum when I shouldn't have been thinking at all. Good pool hustlers, with or without formal education, are expert at both psychology and practical physics.
You guys should read the thread that gave birth to this reposting. I think you might enjoy it...at least some of it.
About the trading cars... I can see how that would work.
I was thinking something different: if you raced and gave $100 to the car that WON the race, you'd have incentive to make them drive as fast as possible too.
Heh. This was on my probability final (Math 2C, a class that no longer exists) at Caltech. I got it wrong then. I got it right today :) You don't forget this one :)b
I hope nobody will post an answer to your really old brainless teaser because once more in a completely predictable fashion (unlike say the prediction, even with 2 years of when a person infected by HIV will develope an AIDS-defining disease -- unless medicated of course..then it is clear.. 2 years is the maximal life expectancy of a completely asymptomatic HIV AB+ after 10-15 on mM AZT) you have jumped in "full feet and no pajamas" as Gurdjieff might have put it.
This is NOT as far as I can tell a blog for playing brain teaser games...there are hundreds of sites devoted to exactly that I believe.
Dean had a point in reposting his essay that was quite different from your own.
Changing the subject is a favorite tactic of .......? (fill in the blank...no brains required)
Catch 22 - Here's an answer to Catch 22 It's hard to follow (sorry)
1.You put on the scale AAAA –BBBB
2. If equal put CCC-AAA
3. If equal weigh the remaining C against a neutral ball
4.If step 2 has the C's going, say Down, your step 3 is weighing C(a) against C(b) and the one that goes down is the one or it's C(c) and its heavier
5. iF original step one looks like sends the A's down and the B"S up (for example)
step 2 would be weighing AAAB against ACCC, if left side goes down you have three A's that could be lighter (weigh one A against another and the lighter one or the remaining one is it)
6.if the right goes down it is either the A form the left that's lighter or the B from the right that's heavier, (just weigh one against a neutral) if the scale is even your left with three B's that must be heavier (which you can figure out like step 5)
Cynical Nation:
"Marilyn vos Savant once posted this in one of her columns and I have never seen such outrage. I recall one angry letter from a statistician (again, a Ph.D.) vowing never to read her column again. 'There is enough mathematical illiteracy out there without your contributing to it with this nonsense!' or some such. It just goes to show you that even Ph.D. scientists are not immune to the set of prejudices we like to refer to as 'common sense.'"
She did, indeed. The letters went on for weeks. I think there was one MIT professor who agreed with her, and that was about it. She got a lot of flak for the two-dogs problem ("You have two dogs and determine that the sex of one is male. What's the probability that the other is also male?"), too. But then, probability tends to get weird reactions from people.
Can someone confirm my slightly alternative premise. It doesn't matter if you switch doors, so long as your door selection is truly among the final two. For example, if you flipped a coin to decide among the last two doors, do you not now have a 1 in 2 chance of being correct, even when choosing the original door?
Very interesting. This confirms what I have been learning here every day, i.e., that Dean is a brilliant man, one of the most intelligent people I've ever known. I would never have got this. Funny thing is that I was reading this post just a few days ago while browsing through Dean's old archives. It's like rummaging around in an attic, finding some long-forgotten old heirloom or something, and then suddenly seeing that very thing on the breakfast table the next morning.
Dean wrote:
"Theories and logic be damned, a fact is a fact!"
Funny about that, too. I'm reading a book on the history of Western philosophy since Kant, leading up to "postmodernism", which says exactly the opposite. If you want to find a "reality-based community", Dean is the King of it.
Actually, CN's explanation is sort of incorrect, I think. When confronted with three doors, you have a 1/3 chance of being right with your pick. When Monty takes away a door, you have a second independent pick, at 1/2 chance, not 2/3 chance. The loss of a door doesn't alter your chance in the second pick. You still don't know which is the good door, and there are two, so it's one-in-two.
However, you usually have to be a student of statictics to remeber that, given two events that cannot occur simultaneously, the probability of one specific combined event happening is simply the sum of the individual probabilities. Otherwise, it is the multiplication of probabilities.
If event A and event B are both coin flips, then two coin flips could occur simultaneously. Therefore, the probability of two flips being head is 1/2 x 1/2 = 1/4.
But, in the door example, event B (pick from two doors where one door is removed) cannot happen without event A (pick from three doors). So, you add probablities. Taking another pick gives you an overall 1/3 + 1/2 = 5/6 chance. The kicker is that not taking the second pick leaves you with your original single chance of being right, at 1/3.
And, to not piss off Harvey Bialy, I'll add a more apropos comment:
Dean tries to show the difference between opinion and fact. He adds: "the world is not always a matter of opinion." and that, sometimes, what is immutable fact goes against our opinions. It's fine to go from fact to opinion. However, to many people go from opinion-made-into-fact.
I'll add that, ideally, policymakers use facts to determine policy. It helps immensely to "get your facts straight" before making policy, however, it is no guarantee of the validity of the generated policy. First, you must be willing to look at facts, then find the right facts, then understand them fully and, even then, when you make policy from it, you still only make opinion, not another fact.
Isn't any wonder that most policy-making results in failure? :) :)
If time is on my side, I will write a program tonight that actually walks through several million iterations of this keeping track of win/losses based on changing / not changing.
I am not sold on the logic here as I think there's a semantic difference that is what causes the arguments. But we'll see if I get ~5,000,000 wins when I change and ~5,000,000 loses when I don't ;)
My problem is this. Let's say the contestant is Lenny from that movie 'Memento' (Leonard suffered short term memory loss). At the point Monty opens the boobie door, Lenny forgets what he chose and that there was even a third door. Now he chooses between two doors and it may just happen that he chooses the same door again. The 2nd time, his choice is 50/50 or 1 in 2. Not 1 in 3 if he happens to pick the same door he did the first time, right?
Hopefully my application will prove me wrong and cause me to lose a lot of sleep ;)
alphadog...i hope you were using "pissed off" really loosely because i have not been angered or annoyed by any of this. quite the contrary.
your explanation btw is quite lucid. i don't know if anybody else has used the one i like here or not because i have honestly not carefully read all the wordss on this scroll. but it is brief and maybe useful to some who still have trouble getting their minds around this apparent and very instructive contradiction between what we think we know and what is really so:
just increase the number of doors to 100. after 98 are opened you are given the same choice. keep or switch. whay do you do now?
Dean's post---is discussing thought process. My comment above addresses a question whose
solution involves thought processes. So how is presenting a thought process problem
changing the subject ? Seems another poster, asked a “slower car” problem.
Unless things have changed, Dean’s World is still an open forum and not a closed panel.
I learned this lesson long ago, when I first fully understood both quantum mechanics and statistical mechanics (the underlying theory behind thermodynamics). It appeared strange to me at the time, but I've learned that the lessons in fact and indeterminancy apply to more than relatively simply many particle problems.
That is why I get very frustrated when people do not listen to me. I do not cherry-pick the data I use to form my opinions, and I try to get the best data available.
More often than not, however, I run into people who have "set" their opinion, and no data will ever convince them that their opinion is not in alignment with the facts.
This was the post that first drew me into Dean's World (it was linked by Chaos Manor at the time), and the attitude behind it has kept me here ever since.
There are some tricky things in this that make it harder to analyze. Monte will never open the door with the car. So, if you pick door A and it has the car, he will open door B or C 50% of the time. But, if you pick door B while A has the car, he will ALWAYS open door C.
Let's have A always have the car. We thus have
Your pick opens switch no switch
A 33% B 16% C 16% A 16%
C 16% B 16% A 16%
B 33% C 33% A 33% B 33%
C 33% B 33% A 33% C 33%
-----------------------------------------
67% win 33% win
lookee here my blogobud....dr. "slower car" bialy's points in his initial post were clearly the following:
1. to express his amazement that this "highly instructive contradiction between what we think we know and what is so" was so important to him and dean and for the same reason. isn't blogging wonderful?
2. i thought it was pretty clear that my two questions were to test both sides of the brain so to speak...the quantitative without which no scientist can work correctly and the sudden flip of thought (intuition) required for the other question to see if the spark of a possible eureka lay in the student. q2 did not ask any reader to solve it, although i imagined a few would and even as quickly as two did.
it was not posted under a rubric of "a discussion thread abt thought processes" nor as an invitation to submit a bunch of brain teasers or puzzle sites, although of course i am in no position to censor any contributions from anyone.
Two folks decide to have a race to determine whose car is SLOWER. How do they do this?
You caught me in another moment with this one, the guy that introduced my parents and a teacher I once had actually attempted this back in the 40's. Both of them were on the fender of their car adjusting the carborator as the cars idled down the main drag in Flagstaff, Arizona.
Crazy Huh! When you get the same story from different sources you tend to believe it.
like i just wrote isn't blogging a terrrifc way to experience the small world nature of our connectivities....'like' a neural net caltechgal if you are still there..right? or what i absolutely know to be a fact -- like the genetic/metabolic maps in organisms as simple as bacteria and as complicated as humans.
alphadog,
Great explanation except that your winning chances go up to 2/3, not 1/2.
I think the best way to say it is this:
Say you plan to change doors.
2/3 of the time, your first pick is wrong. When that happens, Monty HAS to show you the right door, as there is only one empty door for him to open, he opens it, and you choose the other door, which will necessarily have the prize.
1/3 of the time, your first pick is right, you change doors, and you're screwed.
So 2/3 of the time, you win.
Actually, there is no need for me to write my app because I proved this myself on the way home beyond a shadow of a doubt. Simply use a bigger number like 10,000 doors.
You pick one. 9,998 others are shown to be not winning doors. The question isn't if you pick 1 out of 2, it's is my first 1 in 10,000 shot correct? Knowing that Monty knows which of the two is correct, you'd probably be safer choosind his door since that one has a 99,999 in 10,000 chance of being correct. Whittle it back down to 3 doors and Monty has a 2/3 chance of pointing to the correct door.
Interesting! Although not life changing for me. I've always lived with the horrific knowledge that I know nothing at all and even the things I'm right on are statistical abberations...
Not participating in the HIV discussion as I am woefully ill equipped to say anything of value (except that I would not be surprised in the slightest if HIV wasn't the cause of AIDS - the earth ain't flat either).
As to the brain teaser and the underlying revelation, like Owen, I also happened upon Dean's World back when he first posted this, and it took me a good hour or two to unconvince myself that Dean was wrong. A good number of people I know refuse to accept it even after empirical testing proved them wrong. Reminds me of Heinlein's famous quote (paraphrasing from memory):
"Man is not a rational being, but a rationalizing one."
On a lighter note, I've been fortunate enough to get on Who Wants To Be A Millionaire once, and if I ever get on there again, and hitting the 64 000 question haven't the FAINTEST idea what the right answer is and I have a 50/50 left, I'm going to do the following:
get out a coin, flip it - heads means AB, tails CD. then heads A/C, tails B/C.
I will then ask for 50/50. If whatever I randomly picked is one of the remaining options, I will switch, as it will be 75% probable that the one I picked before is wrong. If it doesn't remain, I will flip the damn coin again.
Oh, and last time I went I was out of lifelines by the 32K question and guessed that the first soccer World Cup winner was Brazil. My brother, who was sitting in the audience behind me could have told the answer to me in his sleep, but the dolt didn't even have the common decency to have a coughing fit :)
Oh, and it's Uruguay. When I've forgotten my own name, I'll still remember that one.
Dean, you're the first person I've met who admitted being on GEnie at the same time I was, although I spent a lot more time around the Zymurgy RT; homebrewing is an addictive hobby.
This reminds me of another brain teaser: if a couple has two children, what are the odds that they will have two of the same sex or one of each?(I'm probably wording it poorly, but you get the idea)
The answer: your are twice as likely to have two children of the same sex. It's a question of combinations, of which there are only 3: 2 boys, 2 girls, or 1 boy and 1 girl. Two of the choices result in two same sex children while only only leads to the alternative.
I've had people argue this example with me many times; some of them are still convinced I'm crazy. Okay, they have a point, but they're still wrong.
Key here seems to be to focus on the first choice. Those who say that after Monty opens the door the choice is 50/50 are saying that, when presented with three doors, you can choose at random one that, after an event about which you have no information, has a 50/50 chance of being the only right one out of three. That's impossible.
i don't really get the two dogs problem. it's something crazy like the odds are two to one that the other dog is female. but i don't understand the math behind it.
oh well.
as far as dean's larger point: well, some of the universe is knowable. my mathematical proof would be as follows:
say the odds of winning at powerball is 88 million to one. once the payoff is above $88 million, mathematically, powerball should be one of the best bets ever seen in gambling. Is it?
No it's not, since anyone with $88 million bucks lying around would buy 88 million tickets with all possible combinations covered, and they'd end up with $X/N where x is whatever the amount above 88 million is, and N is the amount of people with $88 million lying around, X/N = quite probably < $88M.
Wolfeyes,
The missing dollar is simple: the order by which the story is presented causes you to add the wrong direction.
Here's a simpler example of the same distortion technique: "I have 11 fingers. See? [holds up all 10 fingers and counts] 1,2,3,4,5 on the first hand, and6,7,8,9,10 on the second. Wait, that isn't right. I coulda sworn...okay, let's count the other direction [counts backward]: 10,9,8,7,6 on the second hand...ah, yes: we already know the first hand has 5 fingers, so 6 + 5 = 11. Simple."
If that analogy doesn't help, there's a pretty good explanation of that puzzle here.
I believe it's impossible to buy $88 million lottery tickets in a cost effective manner. That's why they make you go buy them, and you can't just order them and do it electronically.
I believe rich people have offered the state to cover every number in order to win, but the state always says no.
What I was getting at is that it's a good bet to buy a lottery ticket if the payoff is over $88 million, in principle. But it ignores the fact that we don't have a lifespan that comes anywhere close to justifying 88 million tries.
Similarly, we can understand some stuff about the universe. But we're still pretty small in comparison, even all of us.
Does this blog rock, or what? The politics of the election was really holding you back, Dean. Between this and the HIV/AIDS discussion, I'm lovin' every minute of it!
j
Powerball >$88M would make sense if you were the only person playing. If you buy a full set of tickets, you're then gambling that nobody ELSE will win, because that would divide the pot. The California Lottery has actually sold at least one full set of tickets. IIRC, they actually lost money (~$15M) because someone else also won.
You pick door #1 with 1/3 odds. Monty takes door #2 out of the equation. You now have a brand new choice. You can choose door #1 or door #3. It doesn't make sense to me to look at it as "staying" with door #1 or switching to door #3; you have a new choice, with a 50/50 chance of getting it right.
I must be missing something here, because I used to see it Dean's way but now I don't understand it. Why does your memory of choosing door #1 affect the odds? If your memory was wiped clean after your original choice, and you now had a choice between door #1 and #3, how is that any different?
Moe: I believe I addressed your grammatical concern when I had Montie say, "Now, I'm going to open one of the other two doors for you that has a booby prize."
This should tip the reader off that Montie knows what's behind the doors, shouldn't it?
David in New York: Hmm. I thought my phrasing was clear enough. How about this for a rewrite?
"Now I'm going to do what we always do in this game. Now that you've made your choice, I'm going to open one of the other two doors for you, one with a booby prize."
That should clear up any linguistic ambiguities, yes?
The fastest way is to see that your original odds of winning do not go up just because Montie opened up a door that he knew had no prize. It was 2/3rds likely to be one of those other two doors, and Montie has just eliminated...
Argh. Take Mark Shaw's advice. Just click here and read. It has explanations from all sorts of angles.
Or just do what I did: work it out empirically. Instead of doing it in your head (because your head is fooling you man, take my word for it), get a deck of cards and a friend.
Take out a king and two dueces. Shuffle those three cards, then lay them out. Peek at the cards but don't show your friend. Have him pick a card. Then you take one of the deuces (remember, you know where they both are because you looked) and flip it over. But remember you can't flip the one he picked. So if he picked the King, you flip either of the remaining cards you want. If he didn't pick the King, then you have only one choice--the remaining deuce.
Play this for 10 or 20 iterations and you'll see it--likely you'll see it even before that. But for faster results, ask your friend to either always stick (in which case he'll win 33% of the time) or always switch (in which case he'll win 67% of the time).
Forgive me if this was already said... there are a lot of comments here. The way that makes the most sense to me (and I got a C in Statistics, so I'm no stats whiz) is this:
Each door initially has a probability of 1/3.
By initially picking door A, you have 1/3 probability, and a 2/3 probability that it is in door B or C which I will now call door B&C.
Why do I call it door B&C? Because you get to open both of them! The only difference is that the opening of both the doors only counts as one guess, as the host does it for you. So the choice comes down to this: open door A and get 1/3 probability, or open doors B and C (or door B&C) and get 2/3 probability. Because doors B and C collectively have 2/3 probability by addition, eliminating one of them makes that door have 0 probability and makes the other door take on ALL of that 2/3 probability.
And now, a moneymaking opportunity:
Now that we know that keeping your original guess gives you 1/3 and changing gives you 2/3, what would happen if you performed the game many times with a person who randomly either stayed or changed. When he stays, he gets 1/3, and when he changes he gets 2/3. Since he randomly chooses, his total prize attainment percentage will be the average of 1/3 and 2/3.
[(1/3)+(2/3)]/2 = 1/2
So if you set up a booth at a county fair and took turns letting people bet against you and you betting against them, and assuming that half of your opponents changed and half stayed, and you of course changed every time, you would win 2/3 of the time, and they would win 1/2 of the time, giving you a 1/6 advantage. You could make a lot of money doing this until someone caught on. :-D Even if they switched more than half of the time, you'd still make money. Only when they change as often as you change (which is all the time) would you stop making money.
Mark: If you've ever seen a gambling game popular at small church fundraisers and such, I believe what you describe is not far from what they call "chuck-a-luck" dice.
Okay, but what if you choose the same door? Isn't that statistically the same as choosing the other door? You now have a 1/2 chance of either door being right so you are effectively making a new choice. The first time you had a 1/3 chance, the second time you have a 1/2 chance no matter which door you chose the first time. The first choice doesn't matter all that much, it's the second choice that matters. Whichever door you choose first Monty will show you a booby-prize door. In any case, there's a 1/3 chance of any door being right in the first case and 1/2 chance of any door being right. As somebody above said, if you flipped a coin 9 times and it's heads each time, the odds of the next one being heads is still 1/2. The previous choices don't affect the current choice.
I try to admit when I'm wrong, but I still don't see this. I think I understand it, but I still don't agree.
On an aside, I learned the same lesson from reading Heinlein at a young age. He knew people.
The fact that one of the doors you didn't initially choose doesn't have the prize isn't something you didn't know, but what it does is tell you which of the other two could be the door with the prize. By eliminating that 1/3 probability from the opened door, it transfers that probability to the other door. (one way of thinking of it)
Okay, but what if you choose the same door? Isn't that statistically the same as choosing the other door? You now have a 1/2 chance of either door being right so you are effectively making a new choice.
It's not statistically the same, and here is why: the door you initially pick (but do not open) is not fair game for elimination by the host. If it were your odds for the remaining two doors would be 50/50, because your choice wouldn't have any bearing on the door he eliminates.
If the first door you choose is wrong (2/3 probability that it will be), there is also a 2/3 probability that you'll get the right door if you switch, because you get to open both the other doors, one of which will have the prize 2/3 of the time!
Now thing about this: if you switch, you are opening two doors of your choosing (which is obviously 2/3 odds). Your first choice merely indicates which one you think (or guess) is NOT the one with the prize. Statistics say that the first door you choose will be wrong 2/3 of the time (because it will only be right 1/3 of the time). If the first door you choose is wrong, you get to open the other two doors and win... and that happens 2/3 of the time, if you switch.
If you stay, you're gambling that your first choice was right, a 1/3 chance.
Look at the 1000 doors puzzle. I'm sure you see that by not changing your original choice, your chances are 1 in a thousand.
Back to 3 doors: If you are going to stick with your original choice, your chance is one in three. The opening of the 2nd door by the host is helpful information, but only if you use it. The useful information he is giving you is this: "If the door you first picked is wrong (2/3 chance), then the prize is between the one of the remaining doors that I did not open."
Basically, the game works because you're banking on the fact that 2/3 of the time your first choice will be wrong. So you get to "protect" that wrong door, 2/3 of the time and force the host to eliminate the other door.
1/3 of the time you'll choose the door with the prize the first time, and switching will hurt you. But 2/3 of the time you will be right if you switch.
The Monty Hall puzzle had me totally stumped until the 100-98 door soln. showed me that switching must be best. Previous to that all I could figue out was that if I flipped a coin as my second chance to choose, I would get an increase over my first, but still only 50-50.
This is just an anecdote:I solved a triangle problem once which a guy put up saying it was virtually impossible if you hadn't ever seen it before, that 500 math graduates in a bunch hadn't been able to solve it, blah blah. I got what had to be the right ans. after about 3 hrs. of trying to recreate my past, but did not post it until no one else did for a few days, because I figured since I was so rusty, I was gonna sit back and observe in peaceful anonomity. But then I realized to my horror I had thrown away my method ! [using equations] and I couldn't recreate it, except for pulling one fortuitious move which would only work in lucky cases - assume a=b. Naturally, I couldn't prove I got the ans. in any other way, and I couldn't prove it to myself, either. My honor was impugned by the guy who put the problem up. But what could I say? When the guy who posted the problem showed his ans., it was pretty easy, as usual, no equations necessary. One other person got it using a program, after I put mine up. There were probably about 5 of us trying, or who would at least show themselves, the rest of the millions being nothing better than crass cowards. Since that time of my ignominy, I have limited myself to no more than 20 posts/day. It still hurts. [I don't know what the hell will be my fate after Bialy gets through with me later, due to my last post on hiv. So I will simply say, Goodbye, now before it's too late.]
Weeks later it dawned on me out of nowhere that a 5th grader would have been able to do the problem with graph paper, easily getting as close as the programer, who didn't get the exact ans.. Duh! Except that they don't bother with graph paper anymore, do they? 500 math majors/graduates can't be wrong.
This makes sense and almost makes me believe
Basically, the game works because you're banking on the fact that 2/3 of the time your first choice will be wrong.
My main objection is that prior choices shouldn't affect current choices. (As in the coin-flipping example from Jeremy way above) Therefore, no matter which door you chose first you have a 50/50 chance of picking the correct door the second time.
Robert Speirs seems to say this doesn't make sense ("it's impossible"), I'm not sure why he says that but I've seen his comments too many times to just dismiss them out of hand.
I will have to think more about this.
As an aside, I avoided statistics in college because I took a related, business management, course so I'm not all that conversant with sadistics. I did tutor somebody in sadistics so I have some knowledge, but I generally worked on only certain aspects and didn't get a feeling for the math as a whole.
Dishman: I see your point. I don't understand the math behind two or more people betting to split a prize, though I'm sure i can if someone would be kind enough to explain it. (I know I'm a bit off topic here, sorry.)
I'm not saying you're wrong but ALL of you are doing a piss poor job of explaining this. The coin analogy is bad because that's always a 1 in 2 situation.
The doors are a 1 in 3 situation until one is opened and then becomes a 1 in 2 situation. Now, if that's not a fifty/fifty proposition please point out how for us dumbfolk.
"It's not statistically the same, and here is why: the door you initially pick (but do not open) is not fair game for elimination by the host. If it were your odds for the remaining two doors would be 50/50, because your choice wouldn't have any bearing on the door he eliminates."
Before that I was stuck on the coin analogy where the probability stays at 50% even after 9 straight instances of heads. But this is more akin to my real view of the coin analogy, which is that tails is more likely in that 10th flip, because there were in fact 9 flips, and tails should occur 5 times. It's an gambling mentality. The probability of heads coming up 10 in 10 is less than the probability of it coming up 9 times in 10. Bet tails. Change your door. The second choice is 50-50, when viewed in isolation, but viewing it in isolation will cost you money, or the car.
My main objection is that prior choices shouldn't affect current choices. (As in the coin-flipping example from Jeremy way above) Therefore, no matter which door you chose first you have a 50/50 chance of picking the correct door the second time.
The reason that the prior choice affects the current choice is that you're more likely to choose the wrong door the first time, forcing the host to eliminate the other wrong door, which gives you the answer. In order for it to be like a coin toss, the two events have to be possibly simultaneous. You could flip two coins at the same time... they wouldn't affect each other.
But you cannot make your first door choice at the same time that you chose between staying or changing your choice, because the door eliminated by the host depends on the choice you make. He may have been wanting to eliminate door A, but if you pick door A, he has to eliminate door B, which tells you that 2/3 of the time, the prize will be behind door C. They are simply not independent events, so you cannot approach them as such.
When you chose the first door (which will be WRONG 2/3 of the time), you are squatting on that door. The host can't eliminate it. If he could, your choice between the remaining two doors would be a tossup... 50/50, because your choice didn't affect which doors the host eliminated. But if you "reserve" that first door, your force him to eliminate a different door, which 2/3 of the time will put the prize behind the third door.
again:
- 2/3 of the time your first choice will be wrong
- if your first choice is wrong, the host is forced to eliminate the other wrong choice
I don't think anyone will disagree with those two statements. The first is obvious, and the second stems from the first.
- every time your first guess is wrong, it is better to switch
Why? Because if you're sitting on one crap door, and the host just eliminated the other one, it should be obvious where the prize is.
- because switching will give you the prize 2/3 of the time when compared to 1/3, you should do it every time... it doubles your odds
A friend sent me this website that provides empirical evidence for any skeptics that remain out there. It runs automatically for 1000 tries, or can be stepped manually through the selection process, showing you first hand how the results turn out. Pretty cool. And it saves me the trouble of writing the code to prove it to some others.
What if the host has no idea which door the prize is behind, and eliminates one of the other two doors randomly? What are your chances of getting the prize if you stay? What if you switch? These events aren't independent, because the host can't eliminate the one you first choose, but it is possible that he will eliminate the door with the prize, if your first choice is wrong, leaving the prize door vulnerable.
And I have to agree with someone several comments back... this is way more fun than before the election!
I think Marilyn's dog problem went something like this, IIRC.
A family has two dogs. One of the dogs is male. What is the probability that the other dog is also a male? The answer, counterintuitively, was 1/3.
Here was the way it worked. Across the universe of two-dog families, there are 4 equally probable possible pairings:
MM
MF
FM
FF
By telling us that one of the dogs is male, she has excluded the last possibility. That leaves us with 3 equally probable possibilities, but in only one of them is the other dog male. Hence the answer, 1/3.
Note that the problem doesn't work if the male dog is identified in any other way: if, for example, the problem read "The older dog is male," the puzzle wouldn't work, and the answer would be 50%, as you would expect. The result comes from the fact that the male dog in the puzzle could be either the "A" dog or the "B" dog.
All right, I'm not making sense anymore. I'll stop.
Just a thought: this would NOT work if Monty can open the door you picked. That's why it is better to switch. We are assuming that for the sake of excitement, Monty will make you choose between your original door and a door you don't know about. Should Monty open your door, the probability is 50/50, because you have no knowlege about doors other than the one you picked. However, you HAVE knowledge if he opens a junk door that is not your own. Thus, it is better to switch.
I kept a fairly open mind and was convinced fairly quickly, but still, sometimes my "better sense" keeps telling me I'm wrong.
Veeshir, one of the solutions given depends on the fact that wherever the prize is, it has to be behind one of the two doors left, to a probability of 1. Since the door you initially choose still has a probability of 1/3 or being correct, the other door must have a probability of 2/3. I don't think it matters what the host knows about your first choice. He must eliminate one bad choice anyway.
I didn't get it either, due to thinking I could only increase my liklihood to 50% by switching from the 1/3. I didn't even get that until I was told the 50-50 chance was wrong. Then I still couldn't get above 50%.
Looking at the 100 - 98 choices is helpful. If you remove one choice as your first choice, it has a 1/100 chance of being correct. But if you make the host remove one negative choice from the now remaining 99, each choice within the 99 now has a 1/99 chance of being correct [the 98 remaining in play + your 1, making each remaining choice in play 1/99 likely to be correct]. And so on, until after 98 negative choices, the "other" of your choices, the only other one remaining, is nearly assured of being where the prize is. So you must switch to it.
You can use this as a rule, regardless of how many choices are left in play for you to switch to. If any negative ones have been removed from the remaining choices after you have made yours earlier, you must switch to one of the remaining apart from your first choice. The chance of your first choice being correct has stayed the same, while that of any of the rest has increased.
For those who are still having trouble understanding it, here's another way to think about it. All the odds, be they 1 in 3, 1 in 2, or whatever, are based on the assumption that the selection was made at random. Your choice is random, but Monty's isn't, as he systematically excludes both your door and whichever door holds the prize. When you select Door A, there is a one in three chance your original selection was correct, in which case Monty will open either B or C at random. There is also a one in three chance that B is correct, and Monty will have no choice but to select C, and there is a one in three chance that Monty will open C because B has the prize. Between those three possibilities, there is a 2/3 chance that of the two doors you didn't originally select will hold the prize, and that Monty will identify the other one for you.
Try this counter-example. Rather than hand-pick a door with a booby prize, Monty selects one of the three doors the same way you did: at random. As luck would have it, the door he opened just happens to be one of the two you did not select, and it also just happens not to be the one with the cash prize. Then he asks you if you want to change your selection. This time, the odds really are 1 in 2 either way.
Wrong. The odds still favor your making the switch.
If he has eliminated one of the two doors that you didn't pick, it doesn't matter whether he did it with or without knowledge. Your best bet is to switch. If you don't understand this, then you don't understand the problem.
XRLQ is close, very close, but, Patterico still has it right. Even if Monty doesn't know and he purely lucks into his guess of an empty door, you should switch. Because the odds of your original choice being right were still only one in 3.
The odds only get tricker if Monty's making as wild a guess as you are in that he might accidentally open the door with the money. But then it's a very different question.
Which I guess is why the quibbling over semantics doesn't really matter. If he lucked into his empty door, it's still 2/3rds likely to be the remaining door. Instead of giving you information, he's stumbled on information to share with you. Either way, you should switch unless you have some evidence that he's cheating and intentionally trying to make you switch. (But again, we already said, he ALWAYS does this trick with the empty door, which means he knows, and which means his hand is forced and he can't be playing to psych you out.)
"If he lucked into his empty door, it's still 2/3rds likely to be the remaining door. Instead of giving you information, he's stumbled on information to share with you."
"The odds only get tricker if Monty's making as wild a guess as you are in that he might accidentally open the door with the money."
What's the difference between these two scenarios? In one, you imply Xrlq is right (which I think is true), in the other you say he's wrong. But they seem like the same case.
Xlrq, I disagree with your counter example, wracking my brain to see why. What I see is that you only gain new information about your choice if Monty picks your door. Then you either have the prize or you know you don't, in the latter case making each other door worth 1/2, so 50-50 is what the odds are. But if Monty picks an empty door not yours, then you still don't know any more about your door which originally had a 1/3 chance. If you assume this, then the remaining door has a 2/3 chance of hiding the prize. I'm using the rule that it does not matter how the empty door is chosen, your door's chance does not change. This could be wrong, and I can't do it mathematically yet.
I also don't see how Monty choosing one of the other two doors could alter their original chance of having the prize 2/3 of the time.
Think of it this way. When Monty opens a door, he gives you a lot of new information about that door, and some information about the other doors he could have opened, but didn't. In the original selection, only the two doors you did not select were candidates for being opened. Thus, the fact that Monty didn't open your door tells you nothing about what was behind it. The fact that he didn't open the other door tells you plenty.
That's why my counterexample doesn't work. If Monty got to pick all three at random - or if he got to pick any of the three that didn't have the prize - then his choice of his door would tell you just as much about your own selection as it would about the third door.
Bottom line: if your original answer to Dean's teaser was 50-50, your basic idea was right; you just forgot to take into account that Monty was playing with a stacked deck.
Maor: I was unclear in my phrasing. If Monty is opening a random door, the entire premises of the game are different simply because he might accidentally open the door with the money. That makes the odds of winning trickier at the moment the game begins.
Still, either way, if Monty is making a wild guess and stumbles upon an empty door, you should still always switch because in that moment after he makes his bad guess, it's 2/3rds likely the money will be behind the remaining door.
By stating that Monty ALWAYS opens an empty door, we sort of lessen the confusion and allow you to concentrate on the odds of switch/no-switch.
Still: If you make a pick, then Monty opens an empty door, in that moment it's 2/3rds likely the money will be behind the remaining door, whether he guessed or not.
If Monty opens the door with the money, that's a different scenario than the one we've been discussing.
But if he opens a door that you didn't pick, and it doesn't have the money, you should switch to the other door you didn't pick. It doesn't make a damn bit of difference whether Monty *accidentally* picked the door with no money, or whether he *knew* he was picking that door. His knowledge is 100% irrelevant to the probabilities.
This is where Xrlq was wrong. His counter-example has it exactly wrong:
Try this counter-example. Rather than hand-pick a door with a booby prize, Monty selects one of the three doors the same way you did: at random. As luck would have it, the door he opened just happens to be one of the two you did not select, and it also just happens not to be the one with the cash prize. Then he asks you if you want to change your selection. This time, the odds really are 1 in 2 either way.
No, they aren't. This is the common fallacy that the brain teaser is designed to address. In that situation, your odds are 1 in 3 if you stick with your original choice, and 2 in 3 if you switch.
Regardless of Monty's knowledge.
If X disagrees, I'll play him for money. We live close enough to each other to do it. I'll do my best to make it an expensive lesson for him.
If we play long enough, I'll soon own Patterico's house. They key is not just that Monty avoided the door with the prize, but also that he excluded your door from consideration among the doors that would or would not be opened. If he simply said "I will open one of the three doors which does not have the prize," and your door stood an equal chance of being opened to that of the other door that wasn't opened, then there is no reason to treat "your" door any differently from the other door that was left unopened. After all, the mere act of selecting a door at random did not influence the chances of anything.
If my previous counterexample didn't make things obvious enough, let's try this one. The real game won't work with two players since two players' choices may leave him with only one door to choose from, but my game, where Random Monty has all three to choose from, will work with any number. Suppose I choose Door A at random, Patterico chooses Door B at random, and Random Monty - who gets to choose between all three doors - ends up with Door C, with no prize (either by luck, or because he randomly chose between the two doors that didn't have the prize). If Patterico is right, I have a one in three chance of winning by sticking with my original choice, and a 2 in 3 chance of winning if I switch to his. Conversely, *he* has a 1 in 3 chance of winning of *he* sticks to his original choice (the same one that now has a 2 in 3 chance of winning if I take it), and a 2 in 3 chance if he switches to mine (the one that is favored to lose if I keep it myself). And if that wasn't absurd enough, two impossible scenarios now become possible: we run a 4 in 9 chance of *both* winning if we exchange doors, and a 1 in 9 chance of both losing if we don't.
Also, saying he chooses it totally at random is a different example -- the rule has to be that it doesn't count when he chooses the door you chose.
But it doesn't matter whether *he* knew where the prize was. It just matters that he doesn't get to pick the door you picked -- regardless of whether he *knows* what door you picked, or where the prize is.
If that's the rule, then knowledge is indeed irrelvant. My point was that the distinction to draw is not between what I chose and what I didn't choose, but between the doors that were subject to a particular vetting process and those that weren't. If I picked A, the fact that Monty didn't open A tells me nothing; I knew he wasn't going to pick that door (or, in Patterico's version, I knew that it wouldn't count if he did). But the fact that he didn't choose B (or C), which he could have chosen if the conditions were right, tells me a great deal about what he knows about B (or C).
Assume the contestant picked door C. Then Monty faces 1 of 3 situations: coded as set (00 01 10).
(i.e. doors A and B are empty, door A is empty door B the win, or door A the win and B is empty).
What does Monty do? If:
Situation 1 he just opens either A or B.
Situation 2 he opens door A.
Situation 3 he opens door B.
I think everyone will agree that Monte will act as above.
How often will Monte open A?
In Situation 1 he opens it 1/2 but Situation 1 occurs in only 1/3 of the games. So 1/3 x 1/2 = 1/6.
In Situation 2 he opens Door A but Situation 2 occurs only 1/3. So 1/3.
In Situation 3 he never opens A. So 0/3.
Therefore Monty will open A in
1/6 + 1/3 + 0/3 = 1/2 (of the time when) a contestant chooses C. This makes sense because we expect him to open B as often as A in that case.
Wrap - look (back at) the coding set of A and B: (00 01 10). You can clearly see that 2/3 of the time Monty MUST ACT to avoid the winning door. And he always does; so 2/3 of the time the door he avoids is the win.
Sorry this had to be long. But the problem is tough and has been missed by the best.
Introduction: (borrowed from a previous post by Cynical Nation)
" I think Marilyn's dog problem went something like this, IIRC.
A family has two dogs. One of the dogs is male. What is the probability that the other dog is also a male? The answer, counterintuitively, was 1/3.
Here was the way it worked. Across the universe of two-dog families, there are 4 equally probable possible pairings:
MM
MF
FM
FF "
=================================================
rksmann comments: Without much analysis I think a there are 3, not 4, combinations - two males, two females, or one male with one female.
Nothing was said about the dogs being ordered, so how is FM different than MF? Drop FM.
So MM, MF, and FF remain. If we are told at least one dog is male then FF is eliminated, leaving MM and MF. But since already we know there is a male just strip that letter from the notation and MM becomes M and MF becomes F.
So our choice is reduced to M or F. It looks like 50-50 to me. The chance the other dog is male is 1/2.
If Monty eliminates one of the two doors that you did not originally pick AT RANDOM, it does not benefit you to switch. Why?
Well, 2/3 of the time your original pick will be wrong. 1/3 of the time it will be right. That 1/3 odds for the original door isn't going to change. But if Monty eliminates one of the other doors at random, the 2/3 chance that they share is halved. Meaning, every time that your first choice is incorrect, you have a 50/50 chance of being right if you switch. But that's only 50/50 of 2/3, which is, in fact, 1/3, which is the same odds as staying.
"Meaning, every time that your first choice is incorrect, you have a 50/50 chance of being right if you switch. But that's only 50/50 of 2/3, which is, in fact, 1/3, which is the same odds as staying."
+++++++++++++++++++++++++++++++++++++++++++++++++++++
It seems to me that
"every time your first choice is wrong you have a 100% chance of being right if you switch."
"Meaning, every time that your first choice is incorrect, you have a 50/50 chance of being right if you switch. But that's only 50/50 of 2/3, which is, in fact, 1/3, which is the same odds as staying."
+++++++++++++++++++++++++++++++++++++++++++++++++++++
It seems to me that
"every time your first choice is wrong you have a 100% chance of being right if you switch."
+++++++++++++++++++++++++++++++++++++++++++++++++++++
Amendment: I assumed the game ended when Monty, acting randomly, opened the winning door. But maybe you meant that Monty did not end the game and did not show the contestant the result of opening that door.
If so, 1/3 of the chance to win has already evaporated and switching would not help. (as you concluded).
This puzzle is very deep and confuses because Monty does act randomly 1/3 of the time but is constrained 2/3 of the time.
Here's the problem with the dog example. It's not ordering; you can always find a way to do that. Call the alpha dog #1 and the omega #2, or number them according to their size, the order in which the family got them, or any other criteria you like. As long as one dog can be clearly identified as Dog #1 and the other as Dog #2, you're fine.
Now, to Ken's error: assuming that the probabilities of MM, MF and FM are equal. They're not. There are four equally likely possibilities, two of which result in MM, but only one of which leads to MF, and one to FM. Here are the possibilities:
The dog we knew to be male is Dog #1. Dog #2 is male also.
The dog we knew to be male is Dog #1. Dog #2 is female.
The dog we knew to be male is Dog #2. Dog #1 is male also.
The dog we knew to be male is Dog #2. Dog #1 is female.
Summary: it still looks 50-50 to me. but i'm always cautious about these problems.
unless I missed some problem condition, i just don't see that there are more than 3 ways to own 2 dogs: MM MF FF.
i agree there are other orders in which you can inspect pairs (plural) of dogs and determine their sex: inspect 4 pairs and you can encounter MM MF FM FF, or FF MF FM MM, or MM FF FM MF or, you get the idea.
once you assign a dog tag such as First (#1) and Second (#2) then then you are saying something not in the original problem. the problem just says SOME dog is male.
the next 2 paragraphs try to illustrate my view.
suppose the owner walks the two dogs. he passes a stranger who sees some dog is male. does that indicate anything about the other dog? if so, how?
or, you meet an old friend and ask about his family. he says things are fine and they recently had a second child. you remember they had one child, a little boy, when you last met. at this point have you learned anything about the sex of the second child?
RYI RKSMANN = KENMANN. i just registered and somehow seem to have two ids. bad Ken!
Ken: I agree the odds are 50/50 on the dog problem. The information we're given is 100% probability that either D-1 or D-2 is male, and therefore, a 50/50 chance that both are. It works whether you assign the dogs numbers that were not discussed in the original problem (the four combinations I listed above), or whether you arbitrarily assign the number 1 to "some dog" and 2 to "the other" (two combinations, where D-1 is definitely male, and D-2 is either male or female).
They only way it comes out 1/3 is when you mix the two strategies and treat FM and MF as two separate possibilities out of four (the four I outlined above), on the one hand, while treating MM as though it were only one out of two (MM and MF), on the other, then forgetting that you've done that and assigning equal odds to all three.
Word problems are tough. I appreciate your thoughts and agreement at 50-50.
++++++++++++++++++++++++++++++++++++++++++++++++++
Comment. The reason I gave two examples (dog-walking and second-child) is that the first says nothing about order but comes out 50-50.
The second-child example does order the children (#1 is a boy) but still comes out 50-50.
Each dog's attributes are simply independent of the other dog. As are each childs.
++++++++++++++++++++++++++++++++++++++++++++++++++
Second comment. I don't understand why anyone says I assigned a 1/3 probability equally to each combination. What I said was there are 3 ways to own 2 dogs.
The assumption that 1/2 of the dogs on earth are male - regardless of whether you have inspected them or not - drives the problem.
I suppose I will hear from someone about sampling one dog with and without replacement soon.
You are wrong with "two children" question. You have a child. That child has a sex. You have a second child. There is a ~50% chance that the second child's sex will be the same as the first child's sex.
Therefore there is a roughly 50% chance your two children will have the same sex. (Not exactly 50% since more than 50% of all children are boys.)
As for Powerball, more than one person can select the winning numbers, in which case the pot is split among the winners. So it's still a lousy bet. :-)
As for the two dogs problem. It is not solveable given the information stated. You need to know if "you checked one dog, and he was male", or if "you checked both dogs, and at least one was male."
If it's the first case, then there's a 50% chance the other dog is also male (the probabilities are independent). If it's the second, then there's a 2/3 chance the other dog is female. (There are 4 equally probable cases: MM MF FM FF. You've eliminated the last case. So there are three possible outcomes. 1 is it's another male, two are it's a female.)
The dogs have to be "ordered". Because the dogs are two separate objects, each with it's own sex. (The oldest being male and the youngest being female is a different situation than the oldest being male and the youngest being female.)
Interesting variation on the Monty Hall paradox. Let me see if I understand your version correctly:
1: You pick a door.
2: Monty randomly picks a door.
3: If Monty picks your door, the game is over. You win or lose based on what is shown (to let you re-pick if there's nothing there, and keep your winnings if there's money there, would be unfair).
4: If Monty picks a door other than yours, and that door has the money behind it, game over, you lose.
5: If Monty picks a door not yours, with no money behind it, you have to decide whether or not to switch.
In this game there are 9 different ways things can come out. Of those 9:
1 time: You make correct choice, Monty makes it too. Game Over.
2 times: You make correct choice, Monty doesn't.
2 times: You make wrong choice, Monty shows correct one. Game Over.
2 times: You make wrong choice, Monty makes same wrong choice. Game Over.
2 times: You make wrong choice, Monty makes different wrong choice.
So under the rules I listed above, it's 50 / 50.
Now, let's try slightly different rules: You pick, then Monty randomly picks one of the two other doors. If there's money there, the game's over, otherwise you can switch if you want to.
This is still a 50 / 50 proposition. It's that way because 1/3 of the time Monty reveals the money, and you lose.
This is just a phrasing of the many comments and solutions posted.
++++++++++++++++++++++++++++++++++++++++++++++++++
In the original game Monty always opens 1 of 2 doors. He can choose randomly in 1 of 3 games - when the player has chosen correctly.
But in 2 of 3 games the player has chosen a losing door. When this happens Monty MUST obey the rules and leave the winning door closed.
Therefore. Monty is leaving the winning door closed in 2 of 3 games. And the player should always switch to it.
+++++++++++++++++++++++++++++++++++++++++++++++++++++ Variations:
In rule variations where Monty always opens some door at random (from 2 doors or all 3 doesn't matter) either the game concludes when Monty opens the door or it continues. Then the player can switch or keep his original door.
But since Monty has obeyed no rule in opening a door his action tells the player nothing. The player now knows the door Monty opened was a loser but he still knows nothing about the other two and the odds are 50-50, switch or not.
I wouldn't switch; you feel worse if you lose by switching doors than losing because you did not.
+++++++++++++++++++++++++++++++++++++++++++++++++ I notice Mark has a solution to the 12 Cue Ball Problem. He posted at 12:33. I am close* and want to try again before reading his.
*Close counts with dynamite, but not logic. And the missing rungs on a ladder are the ones that stop you.
If Monty opens one of the remaining doors at random, and gets an empty door, the money is still most likely behind the remaining door. You should switch. The only change is he's discovered information rather than known it.
Greg: you're right, I should have built a truth table before opining. Basically, a rule that says you've checked at least one dog, and at least one of them is male, has these four possibilities:
D1 is male and D2 is male --> OK.
D1 is male and D2 is female --> OK.
D1 is female and D2 is male --> OK
D1 is female and D1 is female --> NOT OK
After forcing myself to see the correct answer, I now understand what's going on. If you had told me that one specific dog was male (which only had a 50-50 probability before), that would have given me more new information about the dogs than telling me that at least one of the two is male (something that had a 3/4 chance from the beginning). Since you've given me less information, you've still increased the likelihood of two males (it was originally 1 in 4) but by a lesser margin (now 1 in 3 rather than 1 in 2).
The dog problem becomes more obvious when you add more dogs to the equation. Suppose I told you Joe Schmoe has sixteen dogs, and the first one that greets you (or the tallest one, or the oldest one, or the friendliest one, or some other characteristic that refers to one specific dog) will be male. That tells you something you didn't already know. But if I tell you he has sixteen dogs and at least one of them is male, you'll look at me funny and say "well, duh."
Dean, Ken's right. If Monty chooses at random among the doors you haven't chosen, the fact that he chooses Door 2 won't communicate any information that is specific to Door 3. The original trick only works because for all you know, Monty could have chosen Door 3, and didn't.
When in doubt, ignore your intuitions, and build a table of all possible outcomes. For anyone door you might pick (let's call it Door 1), there are six possibile random outcomes, all equally probable:
1. Door 1 has the prize, Monty picks 2, so you switch to 3 and lose.
2. Door 1 has the prize, Monty picks 3, so you switch to 2 and lose.
3. Door 2 has the prize, Monty picks 2 and voids the game (or forfeits, or you forfeit, whatever).
4. Door 2 has the prize, Monty picks 3, so you switch to 2 and win.
5. Door 3 has the prize, Monty picks 2, so you switch to 3 and win.
6. Door 3 has the prize, Monty picks 3 and voids the game.
Once you throw out 3 and 6, you're left with 4 guessing games, two of which you can win by sticking with your original choice, the other two by switching. 50-50.
The trick is that while options 1-6 are all equally likely to occur under the Random Monty scenario, they're not equally likely when Monty plays by the traditional rules. By systematically avoiding the prize, Monty doesn't just throw out the 3s and the 6s, as the "Random Monty, Plus Correction" model dictates. Rather, he turns every potential 3 into a 4, and every potential 6 into a 5, thereby doubling the odds of scenarios 4 and 5 (the two scenarios where you win by switching your bet). Meanwhile, the odds of 1 and 2 (the two scenarios where you win by sticking with your original bet) are left untouched.
Maybe. I never get too adamant about these things. My reply will consider that Monty randomly opens 1 of the 2 remaining doors and skip the 3 door variation.
You believe that because there was a 2/3 chance that the player did not choose correctly there must be a 2/3 chance the prize is behind another door. And I agree. However there is not a 2/3 chance of the prize being behind a specific door.
Examine the contest as 2 games not 1.
Game #1 ends when Monty randomly opens a door. If the prize is behind it the player loses. Otherwise Game #2 begins.
In Game #2 the player looks at 2 doors and sticks or switches. He knows that at first there were 2 chances he was wrong and only 1 that he was right. But the Game #1 has eliminated 1 chance he wa wrong. So now there is only 1 chance he can lose and 1 that he can win. 50-50 at this point.
It is a little like a survival show. The longer you stay in the better your odds. Simply by reaching Game #2 the player is safely past 1 chance of losing.
The doggy just keeps barking. Some answer 1/2, as I do. Some say 2/3, and maybe other fractions are advocated by other factions. Does my problem statement differ from than that used by others?
I am working from this problem statement "a family owns two dogs. at least 1 is male. what is the chance the other is female?."
Further, I must assume 1/2 of dogs are male (not to mean that each dog is 1/2 male). And the psychology of selecting pets does not matter.
Yet another angle on why knowledge matters, and after-the-fact corrections aren't enough. Instead of thinking of Monty as "selecting" the door he opens, think of him as "selecting" the door he doesn't open. If he knows where the prize is and intentionally "selects" it every chance he gets (and he gets that chance two times out of three), then you'd be a fool not to follow his lead. But if he's shooting in the dark just like you, there's no more reason for you to follow his lead than there is for him to follow yours.
Okay. I see what you're saying now. But there's more subtlety here. You're pulling a fast one by saying the game is "void" if Monty finds the money. Is it void? What does "void" mean? Does it mean:
1) I automatically win?
2) I automatically lose?
3) We play a new game?
Let's leave out the "play a new game" option because that just leads us to too many possible iterations. Let's declare it: if Monty finds the money, I automatically lose.
To simplify, let's say money's always in door 2 and I always switch.
Game 1: I choose door 1. Monty chooses door 2. I lose.
Game 2 I choose door 1. Monty chooses door 3. I switch and win.
Game 3: I choose door 2. Monty chooses door 1. I switch and lose.
Game 4: I choose door 2. Monty chooses door 3. I switch and lose.
Game 5: I choose door 3. Monty chooses door 2. I lose.
Game 6 I choose door 3. Monty chooses door 1. I switch. I win.
I only win one time in three in this scenario. However, you're right I guess: If Monty can accidentally find the money, and I find myself having to choose between switching or not, my odds are now 50/50 in that precise moment. Because that moment only happens to me four out of six times--the other 2 out of 6, it's an automatic loss.
Overall chance of winning still remain at 1 in 3 when the game begins. Odds go up to 50/50 if Monty doesn't find the money.
By the way, when I say "pulling a fast one" I didn't mean it to come across that way. You were simplifying for example.
Still: you can't really toss those "monty finds it" chances out. They affect the overall game odds. Basically, the game's stacked against you if Monty's guessing. (Or they're stacked in your favor if you win when he hits the money. Where you go if the game is reset, my head doesn't want to go there yet.)
Depends on what you mean by "toss them out." In terms of winning vs. losing, I guess it depends on whether you technically have a right to switch your bet to the door Monty just opened, instead of the one he left closed. In the real game, you'd never do that because the door he just opened always revealed a booby prize. In the randomized version, you could switch to his choice every time he stumbled across the prize. Or you can assume, as I did, that this is supposed to be a guessing game on the second round, so you throw out the result completely when the prize is exposed and there is nothing left to guess about.
Regardless of how/if you count it in terms of winning vs. losing, you have to exclude the cases where Monty finds the prize from this analysis. The whole do I switch / do I keep my original choice conundrum only makes sense when you don't know where the prize is. If you do know where the prize is, switch to that.
I can't consider a "void" game as anything except a loss for the Player. If he leaves without the Prize he has lost.
To assume he might get a "consolation" i.e. a chance to play again, seems a distraction.
Never-the-less:
+++++++++++++++++++++++++++++++++++++++++++++++++++++
If a 'void' does lead to a replay his chance of a win is NOW what it was at the start.
Avoid this logic: a second game (or a third, or) will improve his chances. It arises from the idea that he has some chance to win in each game; therefore when he plays more games those chances must accumulate.*
They don't! Or this would be valid too: "he has some chance to lose in each game, so, as he plays more games, the chance to lose rises to 100%.
+++++++++++++++++++++++++++++++++++++++++++++++++++
Notes:
* The consolation replay scheme is quite distinct from just playing the game repeatedly. When you do that sometimes the Player wins but he NEVER loses because he just plays again.
Repeatedly would be like a night of Poker with one Player allowed to bet play money while the others use real $$$.
++++++++++++++++++++++++++++++++++++++++++++++++++
I sense that we agree now. Perhaps! Many of the posts are difficult to parse.
I don't really care if you call Games 2 and 5 a win, a loss or a draw. The important thing is that when these two games arise, there is nothing you can do to improve or worsen your odds of winning. So while deciding whether or not it is generally a better strategy to keep your original bet or to switch, you have to act as though Games 2 and 5 don't exist, and focus on the four others, where your decision to stay or switch will ultimately determine whether you win or lose the round.
Once you look at it that way, you see that there's no point switching your bet for Random Monty. If you do, you'll win Games 2 and 6, and lose the rest. If you don't, you'll win Games 3 and 4 and lose the rest. In other words, a wash.
Yes, XRLQ is right. It's quibbling whether you say you "throw out" or declare a "loss" if Monty finds the money. The fact is that if you find yourself in a situation where Monty is PURELY GUESSING and he HAPPENS to open an empty door, in THAT MOMENT (not before), your odds are 50/50.
This is, I confess, even more counter-intuitive than the original problem.
But I can explain it like the original problem in a way that might make sense.
Make it ten doors. You pick one. Monty opens one--empty. Opens another--empty. Opens another--empty. By pure random chance, he's totally and completely guessing, he enver opens the door with the money, until you're finally left with two doors. Your first pick, and the one left.
Get your mind around this: every time he opens a door and it's empty, the odds go up that your door has the money. 1 in 10... 1 in 9... 1 in 8... 1 in 7... 1 in 6.... 5... 4...3... down to 1 in 2.
Note that at the start of the game, your odds were 1 in 10. It's a lousy game and no fun to play because you'll almost always lose. But it illustrates (once again counterintuitive) point.
I like to think that the odds on the randomized example are counter-counterintuitive, i.e., they are very intuitive to someone who hasn't dealt with the original problem. Two unopened doors, one prize, no obvious signs of manipulation - of course the odds are 50-50! I mean, what else could they be?
The trick is, in the original problem, Monty is manipulating the game; he's just doing it in a very subtle way most readers miss. So they ignore his actions, and place their bets as though they were in the randomized version of the game.
Now remind me some day to post the OTHER big argument from my grad school days -- the island with the blue-eyed and brown-eyed people. ;-)
I'm always a fan of the coin throwing problem. If I flip a coin 10 times and it comes up heads 9 times in a row, what is the probability that it will come up heads on the 10th throw? Um...1/2. The same as it was the first 9 tosses.
Isn't probability fun?
Here's the trick. Initially your odds of winning are 1 in 3 because their is completely symmetry among the 3 doors. Naively, we tend to think there is also symmetry between the remaining 2 doors *after* we have been shown the empty one... but that's not true.
Monty Hall has *broken* the symmetry by showing you the other door. Why? He did not choose a door randomly! He intentionally showed you a crap door just to keep the game more exciting! His decision was not random, but informed. In doing so, he has provided you (indirectly) with information.
If you know any computer programming languages, it can be helpful to run a simulation of this with a random number generator and (say) a thousand trials. As you write the logic for this experiment, you will begin to see how/why it works.
I don't get it, and I'm a pretty stunningly brilliant guy you know!! :-)
It also stresses the importance of information in decisionmaking.
We never have such levels of certainty in daily life, so we should be all the more humble about our opinions.
By the way, I see the logic behind what you are saying, I just don't understand the logic behind what you are saying.
Why would your 1/3 probability of being right change?
This has been for years one of the two questions I ask any graduate student foolish enough to want to study with me. It is also used as the basis of the probability section of the new NSF's special teaching program for secondary mathematic's teachers that was written by Prof. Daniel Fendel at SF State (who also wrote a review of Oncogenes, Aneuploidy and AIDS on Amazon).
This question I used to test their quantitative thinking even in the face of apparent contradictions with common sense.
The other question I asked them was one that a 9-ball hustler stumped me for several hours with in Oakland back in the 60s. I used this to test the ability of the prospective student to think non-linearily as it is now called.
Two folks decide to have a race to determine whose car is SLOWER. How do they do this?
Unlike the first question, this is a standard brainteaser in that once you get the answer it is really obvious.
Thank you so much for the opportunity to write this.
When you have your first choice, you have a 1 in 3 chance. That isn't going to change if you don't change your answer. Start to finish, that door is 1 in 3.
When Monty opens the second door, he is essentially letting you change your answer and pick both doors. Not only is he letting you pick both doors (2:3) he is letting your throw out the wrong one.
That means that you can keep your first choice (which is locked in at 1:3) or you can change your answer and get two for the price of one (2:3).
Gurl qevir rnpu bgure'f pne. Orpnhfr gurl jnag gur bgure'f pne gb or snfgre, gurl jvyy rnpu qevir gurz nf snfg nf cbffvoyr.
Translate it here.
Let each person drive the others car.
(And let the first person cut the car in half and the other select which part he wants.)
Sorry.
"Now, I'm going to open one of the other two doors for you that has a booby prize."
Doesn't this say that both doors conceal a booby prize? Shouldn't it read one of the other two doors that MAY have a booby prize? I'll bet this is where you think you've already fixed the error you mentioned, but it still reads wrong to me.
Rot-13'd or not you are both quite correct, and got it much faster than dim-witted me sotted out in a poolroom at 2 in the am, and thinking about angular momentum when I shouldn't have been thinking at all. Good pool hustlers, with or without formal education, are expert at both psychology and practical physics.
You guys should read the thread that gave birth to this reposting. I think you might enjoy it...at least some of it.
You have 12 white cue balls and a balance scale.
All the balls weigh exactly the same except one. The one ball not weighing the same might be heavier or
lighter than the others.
The question is:
How do you find the odd ball and how do you tell if it is heavier or lighter ?
You may use the scale only three times. Yes, there is a solution.
I was thinking something different: if you raced and gave $100 to the car that WON the race, you'd have incentive to make them drive as fast as possible too.
I hope nobody will post an answer to your really old brainless teaser because once more in a completely predictable fashion (unlike say the prediction, even with 2 years of when a person infected by HIV will develope an AIDS-defining disease -- unless medicated of course..then it is clear.. 2 years is the maximal life expectancy of a completely asymptomatic HIV AB+ after 10-15 on mM AZT) you have jumped in "full feet and no pajamas" as Gurdjieff might have put it.
This is NOT as far as I can tell a blog for playing brain teaser games...there are hundreds of sites devoted to exactly that I believe.
Dean had a point in reposting his essay that was quite different from your own.
Changing the subject is a favorite tactic of .......? (fill in the blank...no brains required)
how come you can't think so good when it comes to hiv and aids?
1.You put on the scale AAAA –BBBB
2. If equal put CCC-AAA
3. If equal weigh the remaining C against a neutral ball
4.If step 2 has the C's going, say Down, your step 3 is weighing C(a) against C(b) and the one that goes down is the one or it's C(c) and its heavier
5. iF original step one looks like sends the A's down and the B"S up (for example)
step 2 would be weighing AAAB against ACCC, if left side goes down you have three A's that could be lighter (weigh one A against another and the lighter one or the remaining one is it)
6.if the right goes down it is either the A form the left that's lighter or the B from the right that's heavier, (just weigh one against a neutral) if the scale is even your left with three B's that must be heavier (which you can figure out like step 5)
"Marilyn vos Savant once posted this in one of her columns and I have never seen such outrage. I recall one angry letter from a statistician (again, a Ph.D.) vowing never to read her column again. 'There is enough mathematical illiteracy out there without your contributing to it with this nonsense!' or some such. It just goes to show you that even Ph.D. scientists are not immune to the set of prejudices we like to refer to as 'common sense.'"
She did, indeed. The letters went on for weeks. I think there was one MIT professor who agreed with her, and that was about it. She got a lot of flak for the two-dogs problem ("You have two dogs and determine that the sex of one is male. What's the probability that the other is also male?"), too. But then, probability tends to get weird reactions from people.
Without that fact we wouldn't have Las Vegas, Powerball or Atlantic City.
Dean wrote:
"Theories and logic be damned, a fact is a fact!"
Funny about that, too. I'm reading a book on the history of Western philosophy since Kant, leading up to "postmodernism", which says exactly the opposite. If you want to find a "reality-based community", Dean is the King of it.
However, you usually have to be a student of statictics to remeber that, given two events that cannot occur simultaneously, the probability of one specific combined event happening is simply the sum of the individual probabilities. Otherwise, it is the multiplication of probabilities.
If event A and event B are both coin flips, then two coin flips could occur simultaneously. Therefore, the probability of two flips being head is 1/2 x 1/2 = 1/4.
But, in the door example, event B (pick from two doors where one door is removed) cannot happen without event A (pick from three doors). So, you add probablities. Taking another pick gives you an overall 1/3 + 1/2 = 5/6 chance. The kicker is that not taking the second pick leaves you with your original single chance of being right, at 1/3.
So, what's better, 1/3 or 5/6? :)
-alphadog
Dean tries to show the difference between opinion and fact. He adds: "the world is not always a matter of opinion." and that, sometimes, what is immutable fact goes against our opinions. It's fine to go from fact to opinion. However, to many people go from opinion-made-into-fact.
I'll add that, ideally, policymakers use facts to determine policy. It helps immensely to "get your facts straight" before making policy, however, it is no guarantee of the validity of the generated policy. First, you must be willing to look at facts, then find the right facts, then understand them fully and, even then, when you make policy from it, you still only make opinion, not another fact.
Isn't any wonder that most policy-making results in failure? :) :)
Great essay, Dean.
I am not sold on the logic here as I think there's a semantic difference that is what causes the arguments. But we'll see if I get ~5,000,000 wins when I change and ~5,000,000 loses when I don't ;)
My problem is this. Let's say the contestant is Lenny from that movie 'Memento' (Leonard suffered short term memory loss). At the point Monty opens the boobie door, Lenny forgets what he chose and that there was even a third door. Now he chooses between two doors and it may just happen that he chooses the same door again. The 2nd time, his choice is 50/50 or 1 in 2. Not 1 in 3 if he happens to pick the same door he did the first time, right?
Hopefully my application will prove me wrong and cause me to lose a lot of sleep ;)
your explanation btw is quite lucid. i don't know if anybody else has used the one i like here or not because i have honestly not carefully read all the wordss on this scroll. but it is brief and maybe useful to some who still have trouble getting their minds around this apparent and very instructive contradiction between what we think we know and what is really so:
just increase the number of doors to 100. after 98 are opened you are given the same choice. keep or switch. whay do you do now?
solution involves thought processes. So how is presenting a thought process problem
changing the subject ? Seems another poster, asked a “slower car” problem.
Unless things have changed, Dean’s World is still an open forum and not a closed panel.
That's why I like to comment.
That is why I get very frustrated when people do not listen to me. I do not cherry-pick the data I use to form my opinions, and I try to get the best data available.
More often than not, however, I run into people who have "set" their opinion, and no data will ever convince them that their opinion is not in alignment with the facts.
beat this:
http://www.panix.com/~mshaw/3door.txt
Mark Shaw
Thanks for expanding my mind, Dean!
Let's have A always have the car. We thus have
Your pick opens switch no switch
A 33% B 16% C 16% A 16%
C 16% B 16% A 16%
B 33% C 33% A 33% B 33%
C 33% B 33% A 33% C 33%
-----------------------------------------
67% win 33% win
The second row should be shifted one column to the right. The total line should have the 67% under the "switch" column and 33% under "no switch."
lookee here my blogobud....dr. "slower car" bialy's points in his initial post were clearly the following:
1. to express his amazement that this "highly instructive contradiction between what we think we know and what is so" was so important to him and dean and for the same reason. isn't blogging wonderful?
2. i thought it was pretty clear that my two questions were to test both sides of the brain so to speak...the quantitative without which no scientist can work correctly and the sudden flip of thought (intuition) required for the other question to see if the spark of a possible eureka lay in the student. q2 did not ask any reader to solve it, although i imagined a few would and even as quickly as two did.
it was not posted under a rubric of "a discussion thread abt thought processes" nor as an invitation to submit a bunch of brain teasers or puzzle sites, although of course i am in no position to censor any contributions from anyone.
You caught me in another moment with this one, the guy that introduced my parents and a teacher I once had actually attempted this back in the 40's. Both of them were on the fender of their car adjusting the carborator as the cars idled down the main drag in Flagstaff, Arizona.
Crazy Huh! When you get the same story from different sources you tend to believe it.
like i just wrote isn't blogging a terrrifc way to experience the small world nature of our connectivities....'like' a neural net caltechgal if you are still there..right? or what i absolutely know to be a fact -- like the genetic/metabolic maps in organisms as simple as bacteria and as complicated as humans.
thanks.
harvey
Great explanation except that your winning chances go up to 2/3, not 1/2.
I think the best way to say it is this:
Say you plan to change doors.
2/3 of the time, your first pick is wrong. When that happens, Monty HAS to show you the right door, as there is only one empty door for him to open, he opens it, and you choose the other door, which will necessarily have the prize.
1/3 of the time, your first pick is right, you change doors, and you're screwed.
So 2/3 of the time, you win.
You pick one. 9,998 others are shown to be not winning doors. The question isn't if you pick 1 out of 2, it's is my first 1 in 10,000 shot correct? Knowing that Monty knows which of the two is correct, you'd probably be safer choosind his door since that one has a 99,999 in 10,000 chance of being correct. Whittle it back down to 3 doors and Monty has a 2/3 chance of pointing to the correct door.
Interesting! Although not life changing for me. I've always lived with the horrific knowledge that I know nothing at all and even the things I'm right on are statistical abberations...
You just explained it so this dumb blond finally understood the odds involved.
Now, if only someone could explain the missing dollar from the $30 hotel bill question, I'd know all the answers of the universe.
Where's my dad when I need him?
As to the brain teaser and the underlying revelation, like Owen, I also happened upon Dean's World back when he first posted this, and it took me a good hour or two to unconvince myself that Dean was wrong. A good number of people I know refuse to accept it even after empirical testing proved them wrong. Reminds me of Heinlein's famous quote (paraphrasing from memory):
"Man is not a rational being, but a rationalizing one."
On a lighter note, I've been fortunate enough to get on Who Wants To Be A Millionaire once, and if I ever get on there again, and hitting the 64 000 question haven't the FAINTEST idea what the right answer is and I have a 50/50 left, I'm going to do the following:
get out a coin, flip it - heads means AB, tails CD. then heads A/C, tails B/C.
I will then ask for 50/50. If whatever I randomly picked is one of the remaining options, I will switch, as it will be 75% probable that the one I picked before is wrong. If it doesn't remain, I will flip the damn coin again.
Oh, and last time I went I was out of lifelines by the 32K question and guessed that the first soccer World Cup winner was Brazil. My brother, who was sitting in the audience behind me could have told the answer to me in his sleep, but the dolt didn't even have the common decency to have a coughing fit :)
Oh, and it's Uruguay. When I've forgotten my own name, I'll still remember that one.
Dean, you're the first person I've met who admitted being on GEnie at the same time I was, although I spent a lot more time around the Zymurgy RT; homebrewing is an addictive hobby.
This reminds me of another brain teaser: if a couple has two children, what are the odds that they will have two of the same sex or one of each?(I'm probably wording it poorly, but you get the idea)
The answer: your are twice as likely to have two children of the same sex. It's a question of combinations, of which there are only 3: 2 boys, 2 girls, or 1 boy and 1 girl. Two of the choices result in two same sex children while only only leads to the alternative.
I've had people argue this example with me many times; some of them are still convinced I'm crazy. Okay, they have a point, but they're still wrong.
i don't really get the two dogs problem. it's something crazy like the odds are two to one that the other dog is female. but i don't understand the math behind it.
oh well.
as far as dean's larger point: well, some of the universe is knowable. my mathematical proof would be as follows:
say the odds of winning at powerball is 88 million to one. once the payoff is above $88 million, mathematically, powerball should be one of the best bets ever seen in gambling. Is it?
No it's not, since anyone with $88 million bucks lying around would buy 88 million tickets with all possible combinations covered, and they'd end up with $X/N where x is whatever the amount above 88 million is, and N is the amount of people with $88 million lying around, X/N = quite probably < $88M.
The missing dollar is simple: the order by which the story is presented causes you to add the wrong direction.
Here's a simpler example of the same distortion technique: "I have 11 fingers. See? [holds up all 10 fingers and counts] 1,2,3,4,5 on the first hand, and6,7,8,9,10 on the second. Wait, that isn't right. I coulda sworn...okay, let's count the other direction [counts backward]: 10,9,8,7,6 on the second hand...ah, yes: we already know the first hand has 5 fingers, so 6 + 5 = 11. Simple."
If that analogy doesn't help, there's a pretty good explanation of that puzzle here.
I believe rich people have offered the state to cover every number in order to win, but the state always says no.
What I was getting at is that it's a good bet to buy a lottery ticket if the payoff is over $88 million, in principle. But it ignores the fact that we don't have a lifespan that comes anywhere close to justifying 88 million tries.
Similarly, we can understand some stuff about the universe. But we're still pretty small in comparison, even all of us.
j
I must be missing something here, because I used to see it Dean's way but now I don't understand it. Why does your memory of choosing door #1 affect the odds? If your memory was wiped clean after your original choice, and you now had a choice between door #1 and #3, how is that any different?
This should tip the reader off that Montie knows what's behind the doors, shouldn't it?
David in New York: Hmm. I thought my phrasing was clear enough. How about this for a rewrite?
"Now I'm going to do what we always do in this game. Now that you've made your choice, I'm going to open one of the other two doors for you, one with a booby prize."
That should clear up any linguistic ambiguities, yes?
There are dozens of ways to explain this.
The fastest way is to see that your original odds of winning do not go up just because Montie opened up a door that he knew had no prize. It was 2/3rds likely to be one of those other two doors, and Montie has just eliminated...
Argh. Take Mark Shaw's advice. Just click here and read. It has explanations from all sorts of angles.
Or just do what I did: work it out empirically. Instead of doing it in your head (because your head is fooling you man, take my word for it), get a deck of cards and a friend.
Take out a king and two dueces. Shuffle those three cards, then lay them out. Peek at the cards but don't show your friend. Have him pick a card. Then you take one of the deuces (remember, you know where they both are because you looked) and flip it over. But remember you can't flip the one he picked. So if he picked the King, you flip either of the remaining cards you want. If he didn't pick the King, then you have only one choice--the remaining deuce.
Play this for 10 or 20 iterations and you'll see it--likely you'll see it even before that. But for faster results, ask your friend to either always stick (in which case he'll win 33% of the time) or always switch (in which case he'll win 67% of the time).
Each door initially has a probability of 1/3.
By initially picking door A, you have 1/3 probability, and a 2/3 probability that it is in door B or C which I will now call door B&C.
Why do I call it door B&C? Because you get to open both of them! The only difference is that the opening of both the doors only counts as one guess, as the host does it for you. So the choice comes down to this: open door A and get 1/3 probability, or open doors B and C (or door B&C) and get 2/3 probability. Because doors B and C collectively have 2/3 probability by addition, eliminating one of them makes that door have 0 probability and makes the other door take on ALL of that 2/3 probability.
And now, a moneymaking opportunity:
Now that we know that keeping your original guess gives you 1/3 and changing gives you 2/3, what would happen if you performed the game many times with a person who randomly either stayed or changed. When he stays, he gets 1/3, and when he changes he gets 2/3. Since he randomly chooses, his total prize attainment percentage will be the average of 1/3 and 2/3.
[(1/3)+(2/3)]/2 = 1/2
So if you set up a booth at a county fair and took turns letting people bet against you and you betting against them, and assuming that half of your opponents changed and half stayed, and you of course changed every time, you would win 2/3 of the time, and they would win 1/2 of the time, giving you a 1/6 advantage. You could make a lot of money doing this until someone caught on. :-D Even if they switched more than half of the time, you'd still make money. Only when they change as often as you change (which is all the time) would you stop making money.
I try to admit when I'm wrong, but I still don't see this. I think I understand it, but I still don't agree.
On an aside, I learned the same lesson from reading Heinlein at a young age. He knew people.
It's not statistically the same, and here is why: the door you initially pick (but do not open) is not fair game for elimination by the host. If it were your odds for the remaining two doors would be 50/50, because your choice wouldn't have any bearing on the door he eliminates.
If the first door you choose is wrong (2/3 probability that it will be), there is also a 2/3 probability that you'll get the right door if you switch, because you get to open both the other doors, one of which will have the prize 2/3 of the time!
Now thing about this: if you switch, you are opening two doors of your choosing (which is obviously 2/3 odds). Your first choice merely indicates which one you think (or guess) is NOT the one with the prize. Statistics say that the first door you choose will be wrong 2/3 of the time (because it will only be right 1/3 of the time). If the first door you choose is wrong, you get to open the other two doors and win... and that happens 2/3 of the time, if you switch.
If you stay, you're gambling that your first choice was right, a 1/3 chance.
Look at the 1000 doors puzzle. I'm sure you see that by not changing your original choice, your chances are 1 in a thousand.
Back to 3 doors: If you are going to stick with your original choice, your chance is one in three. The opening of the 2nd door by the host is helpful information, but only if you use it. The useful information he is giving you is this: "If the door you first picked is wrong (2/3 chance), then the prize is between the one of the remaining doors that I did not open."
Basically, the game works because you're banking on the fact that 2/3 of the time your first choice will be wrong. So you get to "protect" that wrong door, 2/3 of the time and force the host to eliminate the other door.
1/3 of the time you'll choose the door with the prize the first time, and switching will hurt you. But 2/3 of the time you will be right if you switch.
This is just an anecdote:I solved a triangle problem once which a guy put up saying it was virtually impossible if you hadn't ever seen it before, that 500 math graduates in a bunch hadn't been able to solve it, blah blah. I got what had to be the right ans. after about 3 hrs. of trying to recreate my past, but did not post it until no one else did for a few days, because I figured since I was so rusty, I was gonna sit back and observe in peaceful anonomity. But then I realized to my horror I had thrown away my method ! [using equations] and I couldn't recreate it, except for pulling one fortuitious move which would only work in lucky cases - assume a=b. Naturally, I couldn't prove I got the ans. in any other way, and I couldn't prove it to myself, either. My honor was impugned by the guy who put the problem up. But what could I say? When the guy who posted the problem showed his ans., it was pretty easy, as usual, no equations necessary. One other person got it using a program, after I put mine up. There were probably about 5 of us trying, or who would at least show themselves, the rest of the millions being nothing better than crass cowards. Since that time of my ignominy, I have limited myself to no more than 20 posts/day. It still hurts. [I don't know what the hell will be my fate after Bialy gets through with me later, due to my last post on hiv. So I will simply say, Goodbye, now before it's too late.]
Weeks later it dawned on me out of nowhere that a 5th grader would have been able to do the problem with graph paper, easily getting as close as the programer, who didn't get the exact ans.. Duh! Except that they don't bother with graph paper anymore, do they? 500 math majors/graduates can't be wrong.
Basically, the game works because you're banking on the fact that 2/3 of the time your first choice will be wrong.
My main objection is that prior choices shouldn't affect current choices. (As in the coin-flipping example from Jeremy way above) Therefore, no matter which door you chose first you have a 50/50 chance of picking the correct door the second time.
Robert Speirs seems to say this doesn't make sense ("it's impossible"), I'm not sure why he says that but I've seen his comments too many times to just dismiss them out of hand.
I will have to think more about this.
As an aside, I avoided statistics in college because I took a related, business management, course so I'm not all that conversant with sadistics. I did tutor somebody in sadistics so I have some knowledge, but I generally worked on only certain aspects and didn't get a feeling for the math as a whole.
(I didn't read the numerous comments so I don't know if someone else suggested this, sorry :) )
The doors are a 1 in 3 situation until one is opened and then becomes a 1 in 2 situation. Now, if that's not a fifty/fifty proposition please point out how for us dumbfolk.
"It's not statistically the same, and here is why: the door you initially pick (but do not open) is not fair game for elimination by the host. If it were your odds for the remaining two doors would be 50/50, because your choice wouldn't have any bearing on the door he eliminates."
Before that I was stuck on the coin analogy where the probability stays at 50% even after 9 straight instances of heads. But this is more akin to my real view of the coin analogy, which is that tails is more likely in that 10th flip, because there were in fact 9 flips, and tails should occur 5 times. It's an gambling mentality. The probability of heads coming up 10 in 10 is less than the probability of it coming up 9 times in 10. Bet tails. Change your door. The second choice is 50-50, when viewed in isolation, but viewing it in isolation will cost you money, or the car.
The reason that the prior choice affects the current choice is that you're more likely to choose the wrong door the first time, forcing the host to eliminate the other wrong door, which gives you the answer. In order for it to be like a coin toss, the two events have to be possibly simultaneous. You could flip two coins at the same time... they wouldn't affect each other.
But you cannot make your first door choice at the same time that you chose between staying or changing your choice, because the door eliminated by the host depends on the choice you make. He may have been wanting to eliminate door A, but if you pick door A, he has to eliminate door B, which tells you that 2/3 of the time, the prize will be behind door C. They are simply not independent events, so you cannot approach them as such.
When you chose the first door (which will be WRONG 2/3 of the time), you are squatting on that door. The host can't eliminate it. If he could, your choice between the remaining two doors would be a tossup... 50/50, because your choice didn't affect which doors the host eliminated. But if you "reserve" that first door, your force him to eliminate a different door, which 2/3 of the time will put the prize behind the third door.
again:
- 2/3 of the time your first choice will be wrong
- if your first choice is wrong, the host is forced to eliminate the other wrong choice
I don't think anyone will disagree with those two statements. The first is obvious, and the second stems from the first.
- every time your first guess is wrong, it is better to switch
Why? Because if you're sitting on one crap door, and the host just eliminated the other one, it should be obvious where the prize is.
- because switching will give you the prize 2/3 of the time when compared to 1/3, you should do it every time... it doubles your odds
A friend sent me this website that provides empirical evidence for any skeptics that remain out there. It runs automatically for 1000 tries, or can be stepped manually through the selection process, showing you first hand how the results turn out. Pretty cool. And it saves me the trouble of writing the code to prove it to some others.
What if the host has no idea which door the prize is behind, and eliminates one of the other two doors randomly? What are your chances of getting the prize if you stay? What if you switch? These events aren't independent, because the host can't eliminate the one you first choose, but it is possible that he will eliminate the door with the prize, if your first choice is wrong, leaving the prize door vulnerable.
And I have to agree with someone several comments back... this is way more fun than before the election!
A family has two dogs. One of the dogs is male. What is the probability that the other dog is also a male? The answer, counterintuitively, was 1/3.
Here was the way it worked. Across the universe of two-dog families, there are 4 equally probable possible pairings:
MM
MF
FM
FF
By telling us that one of the dogs is male, she has excluded the last possibility. That leaves us with 3 equally probable possibilities, but in only one of them is the other dog male. Hence the answer, 1/3.
Note that the problem doesn't work if the male dog is identified in any other way: if, for example, the problem read "The older dog is male," the puzzle wouldn't work, and the answer would be 50%, as you would expect. The result comes from the fact that the male dog in the puzzle could be either the "A" dog or the "B" dog.
All right, I'm not making sense anymore. I'll stop.
:-)
I kept a fairly open mind and was convinced fairly quickly, but still, sometimes my "better sense" keeps telling me I'm wrong.
So it is a 1/3 chance against a 2/3 chance. I am a little depressed that I've spent this much time and other people's effort figuring it out. Oh well.
I didn't get it either, due to thinking I could only increase my liklihood to 50% by switching from the 1/3. I didn't even get that until I was told the 50-50 chance was wrong. Then I still couldn't get above 50%.
Looking at the 100 - 98 choices is helpful. If you remove one choice as your first choice, it has a 1/100 chance of being correct. But if you make the host remove one negative choice from the now remaining 99, each choice within the 99 now has a 1/99 chance of being correct [the 98 remaining in play + your 1, making each remaining choice in play 1/99 likely to be correct]. And so on, until after 98 negative choices, the "other" of your choices, the only other one remaining, is nearly assured of being where the prize is. So you must switch to it.
You can use this as a rule, regardless of how many choices are left in play for you to switch to. If any negative ones have been removed from the remaining choices after you have made yours earlier, you must switch to one of the remaining apart from your first choice. The chance of your first choice being correct has stayed the same, while that of any of the rest has increased.
Try this counter-example. Rather than hand-pick a door with a booby prize, Monty selects one of the three doors the same way you did: at random. As luck would have it, the door he opened just happens to be one of the two you did not select, and it also just happens not to be the one with the cash prize. Then he asks you if you want to change your selection. This time, the odds really are 1 in 2 either way.
Wrong. The odds still favor your making the switch.
If he has eliminated one of the two doors that you didn't pick, it doesn't matter whether he did it with or without knowledge. Your best bet is to switch. If you don't understand this, then you don't understand the problem.
The odds only get tricker if Monty's making as wild a guess as you are in that he might accidentally open the door with the money. But then it's a very different question.
Which I guess is why the quibbling over semantics doesn't really matter. If he lucked into his empty door, it's still 2/3rds likely to be the remaining door. Instead of giving you information, he's stumbled on information to share with you. Either way, you should switch unless you have some evidence that he's cheating and intentionally trying to make you switch. (But again, we already said, he ALWAYS does this trick with the empty door, which means he knows, and which means his hand is forced and he can't be playing to psych you out.)
"The odds only get tricker if Monty's making as wild a guess as you are in that he might accidentally open the door with the money."
What's the difference between these two scenarios? In one, you imply Xrlq is right (which I think is true), in the other you say he's wrong. But they seem like the same case.
I also don't see how Monty choosing one of the other two doors could alter their original chance of having the prize 2/3 of the time.
That's why my counterexample doesn't work. If Monty got to pick all three at random - or if he got to pick any of the three that didn't have the prize - then his choice of his door would tell you just as much about your own selection as it would about the third door.
Bottom line: if your original answer to Dean's teaser was 50-50, your basic idea was right; you just forgot to take into account that Monty was playing with a stacked deck.
Still, either way, if Monty is making a wild guess and stumbles upon an empty door, you should still always switch because in that moment after he makes his bad guess, it's 2/3rds likely the money will be behind the remaining door.
By stating that Monty ALWAYS opens an empty door, we sort of lessen the confusion and allow you to concentrate on the odds of switch/no-switch.
Still: If you make a pick, then Monty opens an empty door, in that moment it's 2/3rds likely the money will be behind the remaining door, whether he guessed or not.
But if he opens a door that you didn't pick, and it doesn't have the money, you should switch to the other door you didn't pick. It doesn't make a damn bit of difference whether Monty *accidentally* picked the door with no money, or whether he *knew* he was picking that door. His knowledge is 100% irrelevant to the probabilities.
This is where Xrlq was wrong. His counter-example has it exactly wrong:
No, they aren't. This is the common fallacy that the brain teaser is designed to address. In that situation, your odds are 1 in 3 if you stick with your original choice, and 2 in 3 if you switch.
Regardless of Monty's knowledge.
If X disagrees, I'll play him for money. We live close enough to each other to do it. I'll do my best to make it an expensive lesson for him.
If my previous counterexample didn't make things obvious enough, let's try this one. The real game won't work with two players since two players' choices may leave him with only one door to choose from, but my game, where Random Monty has all three to choose from, will work with any number. Suppose I choose Door A at random, Patterico chooses Door B at random, and Random Monty - who gets to choose between all three doors - ends up with Door C, with no prize (either by luck, or because he randomly chose between the two doors that didn't have the prize). If Patterico is right, I have a one in three chance of winning by sticking with my original choice, and a 2 in 3 chance of winning if I switch to his. Conversely, *he* has a 1 in 3 chance of winning of *he* sticks to his original choice (the same one that now has a 2 in 3 chance of winning if I take it), and a 2 in 3 chance if he switches to mine (the one that is favored to lose if I keep it myself). And if that wasn't absurd enough, two impossible scenarios now become possible: we run a 4 in 9 chance of *both* winning if we exchange doors, and a 1 in 9 chance of both losing if we don't.
Also, saying he chooses it totally at random is a different example -- the rule has to be that it doesn't count when he chooses the door you chose.
But it doesn't matter whether *he* knew where the prize was. It just matters that he doesn't get to pick the door you picked -- regardless of whether he *knows* what door you picked, or where the prize is.
His knowledge is irrelevant.
Assume the contestant picked door C. Then Monty faces 1 of 3 situations: coded as set (00 01 10).
(i.e. doors A and B are empty, door A is empty door B the win, or door A the win and B is empty).
What does Monty do? If:
Situation 1 he just opens either A or B.
Situation 2 he opens door A.
Situation 3 he opens door B.
I think everyone will agree that Monte will act as above.
How often will Monte open A?
In Situation 1 he opens it 1/2 but Situation 1 occurs in only 1/3 of the games. So 1/3 x 1/2 = 1/6.
In Situation 2 he opens Door A but Situation 2 occurs only 1/3. So 1/3.
In Situation 3 he never opens A. So 0/3.
Therefore Monty will open A in
1/6 + 1/3 + 0/3 = 1/2 (of the time when) a contestant chooses C. This makes sense because we expect him to open B as often as A in that case.
Wrap - look (back at) the coding set of A and B: (00 01 10). You can clearly see that 2/3 of the time Monty MUST ACT to avoid the winning door. And he always does; so 2/3 of the time the door he avoids is the win.
Sorry this had to be long. But the problem is tough and has been missed by the best.
" I think Marilyn's dog problem went something like this, IIRC.
A family has two dogs. One of the dogs is male. What is the probability that the other dog is also a male? The answer, counterintuitively, was 1/3.
Here was the way it worked. Across the universe of two-dog families, there are 4 equally probable possible pairings:
MM
MF
FM
FF "
=================================================
rksmann comments: Without much analysis I think a there are 3, not 4, combinations - two males, two females, or one male with one female.
Nothing was said about the dogs being ordered, so how is FM different than MF? Drop FM.
So MM, MF, and FF remain. If we are told at least one dog is male then FF is eliminated, leaving MM and MF. But since already we know there is a male just strip that letter from the notation and MM becomes M and MF becomes F.
So our choice is reduced to M or F. It looks like 50-50 to me. The chance the other dog is male is 1/2.
Stayed up to late - hope my answer is correct.
Well, 2/3 of the time your original pick will be wrong. 1/3 of the time it will be right. That 1/3 odds for the original door isn't going to change. But if Monty eliminates one of the other doors at random, the 2/3 chance that they share is halved. Meaning, every time that your first choice is incorrect, you have a 50/50 chance of being right if you switch. But that's only 50/50 of 2/3, which is, in fact, 1/3, which is the same odds as staying.
"Meaning, every time that your first choice is incorrect, you have a 50/50 chance of being right if you switch. But that's only 50/50 of 2/3, which is, in fact, 1/3, which is the same odds as staying."
+++++++++++++++++++++++++++++++++++++++++++++++++++++
It seems to me that
"every time your first choice is wrong you have a 100% chance of being right if you switch."
"Meaning, every time that your first choice is incorrect, you have a 50/50 chance of being right if you switch. But that's only 50/50 of 2/3, which is, in fact, 1/3, which is the same odds as staying."
+++++++++++++++++++++++++++++++++++++++++++++++++++++
It seems to me that
"every time your first choice is wrong you have a 100% chance of being right if you switch."
+++++++++++++++++++++++++++++++++++++++++++++++++++++
Amendment: I assumed the game ended when Monty, acting randomly, opened the winning door. But maybe you meant that Monty did not end the game and did not show the contestant the result of opening that door.
If so, 1/3 of the chance to win has already evaporated and switching would not help. (as you concluded).
This puzzle is very deep and confuses because Monty does act randomly 1/3 of the time but is constrained 2/3 of the time.
Now, to Ken's error: assuming that the probabilities of MM, MF and FM are equal. They're not. There are four equally likely possibilities, two of which result in MM, but only one of which leads to MF, and one to FM. Here are the possibilities:
Summary: it still looks 50-50 to me. but i'm always cautious about these problems.
unless I missed some problem condition, i just don't see that there are more than 3 ways to own 2 dogs: MM MF FF.
i agree there are other orders in which you can inspect pairs (plural) of dogs and determine their sex: inspect 4 pairs and you can encounter MM MF FM FF, or FF MF FM MM, or MM FF FM MF or, you get the idea.
once you assign a dog tag such as First (#1) and Second (#2) then then you are saying something not in the original problem. the problem just says SOME dog is male.
the next 2 paragraphs try to illustrate my view.
suppose the owner walks the two dogs. he passes a stranger who sees some dog is male. does that indicate anything about the other dog? if so, how?
or, you meet an old friend and ask about his family. he says things are fine and they recently had a second child. you remember they had one child, a little boy, when you last met. at this point have you learned anything about the sex of the second child?
RYI RKSMANN = KENMANN. i just registered and somehow seem to have two ids. bad Ken!
The CBP looks tough. And I have nothing so far.
They only way it comes out 1/3 is when you mix the two strategies and treat FM and MF as two separate possibilities out of four (the four I outlined above), on the one hand, while treating MM as though it were only one out of two (MM and MF), on the other, then forgetting that you've done that and assigning equal odds to all three.
Word problems are tough. I appreciate your thoughts and agreement at 50-50.
++++++++++++++++++++++++++++++++++++++++++++++++++
Comment. The reason I gave two examples (dog-walking and second-child) is that the first says nothing about order but comes out 50-50.
The second-child example does order the children (#1 is a boy) but still comes out 50-50.
Each dog's attributes are simply independent of the other dog. As are each childs.
++++++++++++++++++++++++++++++++++++++++++++++++++
Second comment. I don't understand why anyone says I assigned a 1/3 probability equally to each combination. What I said was there are 3 ways to own 2 dogs.
The assumption that 1/2 of the dogs on earth are male - regardless of whether you have inspected them or not - drives the problem.
I suppose I will hear from someone about sampling one dog with and without replacement soon.
:)
You are wrong with "two children" question. You have a child. That child has a sex. You have a second child. There is a ~50% chance that the second child's sex will be the same as the first child's sex.
Therefore there is a roughly 50% chance your two children will have the same sex. (Not exactly 50% since more than 50% of all children are boys.)
As for Powerball, more than one person can select the winning numbers, in which case the pot is split among the winners. So it's still a lousy bet. :-)
As for the two dogs problem. It is not solveable given the information stated. You need to know if "you checked one dog, and he was male", or if "you checked both dogs, and at least one was male."
If it's the first case, then there's a 50% chance the other dog is also male (the probabilities are independent). If it's the second, then there's a 2/3 chance the other dog is female. (There are 4 equally probable cases: MM MF FM FF. You've eliminated the last case. So there are three possible outcomes. 1 is it's another male, two are it's a female.)
The dogs have to be "ordered". Because the dogs are two separate objects, each with it's own sex. (The oldest being male and the youngest being female is a different situation than the oldest being male and the youngest being female.)
Interesting variation on the Monty Hall paradox. Let me see if I understand your version correctly:
1: You pick a door.
2: Monty randomly picks a door.
3: If Monty picks your door, the game is over. You win or lose based on what is shown (to let you re-pick if there's nothing there, and keep your winnings if there's money there, would be unfair).
4: If Monty picks a door other than yours, and that door has the money behind it, game over, you lose.
5: If Monty picks a door not yours, with no money behind it, you have to decide whether or not to switch.
In this game there are 9 different ways things can come out. Of those 9:
1 time: You make correct choice, Monty makes it too. Game Over.
2 times: You make correct choice, Monty doesn't.
2 times: You make wrong choice, Monty shows correct one. Game Over.
2 times: You make wrong choice, Monty makes same wrong choice. Game Over.
2 times: You make wrong choice, Monty makes different wrong choice.
So under the rules I listed above, it's 50 / 50.
Now, let's try slightly different rules: You pick, then Monty randomly picks one of the two other doors. If there's money there, the game's over, otherwise you can switch if you want to.
This is still a 50 / 50 proposition. It's that way because 1/3 of the time Monty reveals the money, and you lose.
This is just a phrasing of the many comments and solutions posted.
++++++++++++++++++++++++++++++++++++++++++++++++++
In the original game Monty always opens 1 of 2 doors. He can choose randomly in 1 of 3 games - when the player has chosen correctly.
But in 2 of 3 games the player has chosen a losing door. When this happens Monty MUST obey the rules and leave the winning door closed.
Therefore. Monty is leaving the winning door closed in 2 of 3 games. And the player should always switch to it.
+++++++++++++++++++++++++++++++++++++++++++++++++++++ Variations:
In rule variations where Monty always opens some door at random (from 2 doors or all 3 doesn't matter) either the game concludes when Monty opens the door or it continues. Then the player can switch or keep his original door.
But since Monty has obeyed no rule in opening a door his action tells the player nothing. The player now knows the door Monty opened was a loser but he still knows nothing about the other two and the odds are 50-50, switch or not.
I wouldn't switch; you feel worse if you lose by switching doors than losing because you did not.
+++++++++++++++++++++++++++++++++++++++++++++++++ I notice Mark has a solution to the 12 Cue Ball Problem. He posted at 12:33. I am close* and want to try again before reading his.
*Close counts with dynamite, but not logic. And the missing rungs on a ladder are the ones that stop you.
If Monty opens one of the remaining doors at random, and gets an empty door, the money is still most likely behind the remaining door. You should switch. The only change is he's discovered information rather than known it.
D1 is male and D2 is male --> OK.
D1 is male and D2 is female --> OK.
D1 is female and D2 is male --> OK
D1 is female and D1 is female --> NOT OK
After forcing myself to see the correct answer, I now understand what's going on. If you had told me that one specific dog was male (which only had a 50-50 probability before), that would have given me more new information about the dogs than telling me that at least one of the two is male (something that had a 3/4 chance from the beginning). Since you've given me less information, you've still increased the likelihood of two males (it was originally 1 in 4) but by a lesser margin (now 1 in 3 rather than 1 in 2).
The dog problem becomes more obvious when you add more dogs to the equation. Suppose I told you Joe Schmoe has sixteen dogs, and the first one that greets you (or the tallest one, or the oldest one, or the friendliest one, or some other characteristic that refers to one specific dog) will be male. That tells you something you didn't already know. But if I tell you he has sixteen dogs and at least one of them is male, you'll look at me funny and say "well, duh."
When in doubt, ignore your intuitions, and build a table of all possible outcomes. For anyone door you might pick (let's call it Door 1), there are six possibile random outcomes, all equally probable:
1. Door 1 has the prize, Monty picks 2, so you switch to 3 and lose.
2. Door 1 has the prize, Monty picks 3, so you switch to 2 and lose.
3. Door 2 has the prize, Monty picks 2 and voids the game (or forfeits, or you forfeit, whatever).
4. Door 2 has the prize, Monty picks 3, so you switch to 2 and win.
5. Door 3 has the prize, Monty picks 2, so you switch to 3 and win.
6. Door 3 has the prize, Monty picks 3 and voids the game.
Once you throw out 3 and 6, you're left with 4 guessing games, two of which you can win by sticking with your original choice, the other two by switching. 50-50.
Maybe. I never get too adamant about these things. My reply will consider that Monty randomly opens 1 of the 2 remaining doors and skip the 3 door variation.
You believe that because there was a 2/3 chance that the player did not choose correctly there must be a 2/3 chance the prize is behind another door. And I agree. However there is not a 2/3 chance of the prize being behind a specific door.
Examine the contest as 2 games not 1.
Game #1 ends when Monty randomly opens a door. If the prize is behind it the player loses. Otherwise Game #2 begins.
In Game #2 the player looks at 2 doors and sticks or switches. He knows that at first there were 2 chances he was wrong and only 1 that he was right. But the Game #1 has eliminated 1 chance he wa wrong. So now there is only 1 chance he can lose and 1 that he can win. 50-50 at this point.
It is a little like a survival show. The longer you stay in the better your odds. Simply by reaching Game #2 the player is safely past 1 chance of losing.
The doggy just keeps barking. Some answer 1/2, as I do. Some say 2/3, and maybe other fractions are advocated by other factions. Does my problem statement differ from than that used by others?
I am working from this problem statement "a family owns two dogs. at least 1 is male. what is the chance the other is female?."
Further, I must assume 1/2 of dogs are male (not to mean that each dog is 1/2 male). And the psychology of selecting pets does not matter.
What problem statements are others using?
Very well said.
If Monty doesn't know, his action cannot guide the player. He cannot even tell the Player the answer because he doesn't know it.
Gotta stop!!!! Sure I will!!!!
1) I automatically win?
2) I automatically lose?
3) We play a new game?
Let's leave out the "play a new game" option because that just leads us to too many possible iterations. Let's declare it: if Monty finds the money, I automatically lose.
To simplify, let's say money's always in door 2 and I always switch.
Game 1: I choose door 1. Monty chooses door 2. I lose.
Game 2 I choose door 1. Monty chooses door 3. I switch and win.
Game 3: I choose door 2. Monty chooses door 1. I switch and lose.
Game 4: I choose door 2. Monty chooses door 3. I switch and lose.
Game 5: I choose door 3. Monty chooses door 2. I lose.
Game 6 I choose door 3. Monty chooses door 1. I switch. I win.
I only win one time in three in this scenario. However, you're right I guess: If Monty can accidentally find the money, and I find myself having to choose between switching or not, my odds are now 50/50 in that precise moment. Because that moment only happens to me four out of six times--the other 2 out of 6, it's an automatic loss.
Overall chance of winning still remain at 1 in 3 when the game begins. Odds go up to 50/50 if Monty doesn't find the money.
Still: you can't really toss those "monty finds it" chances out. They affect the overall game odds. Basically, the game's stacked against you if Monty's guessing. (Or they're stacked in your favor if you win when he hits the money. Where you go if the game is reset, my head doesn't want to go there yet.)
Regardless of how/if you count it in terms of winning vs. losing, you have to exclude the cases where Monty finds the prize from this analysis. The whole do I switch / do I keep my original choice conundrum only makes sense when you don't know where the prize is. If you do know where the prize is, switch to that.
Monty Problem and Void games.
I can't consider a "void" game as anything except a loss for the Player. If he leaves without the Prize he has lost.
To assume he might get a "consolation" i.e. a chance to play again, seems a distraction.
Never-the-less:
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If a 'void' does lead to a replay his chance of a win is NOW what it was at the start.
Avoid this logic: a second game (or a third, or) will improve his chances. It arises from the idea that he has some chance to win in each game; therefore when he plays more games those chances must accumulate.*
They don't! Or this would be valid too: "he has some chance to lose in each game, so, as he plays more games, the chance to lose rises to 100%.
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Notes:
* The consolation replay scheme is quite distinct from just playing the game repeatedly. When you do that sometimes the Player wins but he NEVER loses because he just plays again.
Repeatedly would be like a night of Poker with one Player allowed to bet play money while the others use real $$$.
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I sense that we agree now. Perhaps! Many of the posts are difficult to parse.
Once you look at it that way, you see that there's no point switching your bet for Random Monty. If you do, you'll win Games 2 and 6, and lose the rest. If you don't, you'll win Games 3 and 4 and lose the rest. In other words, a wash.
This is, I confess, even more counter-intuitive than the original problem.
But I can explain it like the original problem in a way that might make sense.
Make it ten doors. You pick one. Monty opens one--empty. Opens another--empty. Opens another--empty. By pure random chance, he's totally and completely guessing, he enver opens the door with the money, until you're finally left with two doors. Your first pick, and the one left.
Get your mind around this: every time he opens a door and it's empty, the odds go up that your door has the money. 1 in 10... 1 in 9... 1 in 8... 1 in 7... 1 in 6.... 5... 4...3... down to 1 in 2.
Note that at the start of the game, your odds were 1 in 10. It's a lousy game and no fun to play because you'll almost always lose. But it illustrates (once again counterintuitive) point.
The trick is, in the original problem, Monty is manipulating the game; he's just doing it in a very subtle way most readers miss. So they ignore his actions, and place their bets as though they were in the randomized version of the game.