If you go through life forming and sharing opinions, it is a rock-solid certainty that you will be wrong about something. The more opinions you have, the more that will happen. The bigger the issue, the more spectacularly wrong you're likely to be.
In my mid-20s, I stumbled on a brain teaser that, literally, changed how I viewed the world. As melodramatic as it sounds, I haven't been the same since. And, as with so many other things in this world, it's all Jerry Pournelle's fault...
None of us really likes to admit being wrong. One of the most seductive ways to avoid that is to change our opinions retroactively. We say, "No, no, you just misunderstood, you thought I was saying X when I really said Y." Or, even worse, sometimes we just stubbornly refuse to acknowledge the evidence in front of us.
Not that genuine misunderstandings don't happen. But a lot of people, when caught out as wrong, will say it didn't happen. Instead, they conveniently shift their position, but act like they didn't. It's almost as if we rewrite our memories, and by so doing rewrite the history of what we did or said. It's a pathology that's common to the human animal. Opinionated bloviators such as myself are particularly prone to the affliction. I don't claim to be cured, but I think I'm able to recognize the symptoms and, hopefully, manage the disease tolerably.
Silly as it seems, it was a brainteaser that woke me up to it. A brainteaser that not only showed me that I could be completely wrong, but rubbed my nose into my wrongness, repeatedly. Then I watched as some exceptionally intelligent people--mathematicians, engineers, people with various forms of Ph.D., Mensa members, and professional computer programmers also got it wrong, in astounding numbers.
Back in the early 1990s, I used to frequent the Jerry Pournelle RoundTable on the old GEnie network. Pournelle is well-known to old-time computer geeks for his "Computing at Chaos Manor" column in Byte magazine, and by Science Fiction fans for his many novels and short stories. He's also just about the smartest sumbitch I've ever met (I'm not even sure I needed to qualify that by saying "just about"). In his online forum, literally dozens of smart (and some not-so-smart) people would wile away endless hours discussing history, politics, religion, science, computers, books, guns and music. Gosh I loved it.
Anyway, one day one of the members of the RoundTable posted a simple brainteaser, related to a TV game show. It had appeared in Parade magazine, and where it came from before that I do not know. But here it is:
You find yourself on a game show called "Let's Make A Deal." The game is very simple. There are three doors: door #1, door #2, and door #3. Behind one door is a million dollars. The other two doors contain worthless joke prizes. All you have to do is pick which door you want to open, and you get whatever is behind it. But you only get to open one door. By simple math, then, you obviously have a 1 in 3 chance of picking the correct door and becoming an instant millionaire.
You pick a door. As soon as you tell Monty (the gameshow host) what door you want to open, he stops and says, "Okay, you've made your choice. Now, I'm going to do what we always do here on this game. I'm going to open one of the other two doors for you that I know has a booby prize." And he does so. Then he asks, "Okay, now, would you like to stay with your original guess, or would you like to switch to the other door that's still closed? You only get one shot, so do you want to stay with your original choice, or switch?"
Here's the question: is there any compelling reason to switch doors?
To be clear, there is no trickery, and Monty is not cheating. Furthermore, the money has not moved, will not be moved, and if you open the right door, you win the cash. Money is either behind the door you first picked, or behind the remaining unopened door. Should you switch?
Think about that for a moment. To make it easier for you to avoid seeing the answer, I'll put some blank lines below for you to scroll through before seeing the answer:
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Logically speaking, this seems obvious to anyone with a sharp mind: you can switch if you want to, but it makes no difference. Monty has just eliminated one of your choices. Now you're down to two. You didn't know what was behind the doors before, and by opening one of them, you still don't know what's behind the other two. Your odds are 50/50 no matter which door you choose. So, switch or don't, it makes no difference.
Perfectly sensible, right? Absolutely. But it's dead wrong.
The surprising answer is that you should switch doors.
I can tell that some of you are probably reading this and nodding--and then you're thinking to yourself, "No, that's not right. There really is no reason to switch. You can if you want to, or if you think there's cheating going on, but in a pure game, it's 50/50! You've gotten rid of one of the doors is all!"
I know this because it's the argument I made, and most people agreed with me. Dr. Pournelle himself saw it the same way. A stubborn few insisted that no, you really should switch. They were outnumbered and outgunned. It went on for a few days, with quite a few people joining the battle. Many of us were growing impatient with these clueless boobs who could not understand the logic. We explained it over and over: LOOK, you have only two choices! You had three but you eliminated one! You've only got two left! It's 50/50! You're letting yourself get confused by irrelevancies!
Then to my amazement, in the middle of writing a message explaining once again to the boobs why they were wrong, Dr. Pournelle said, "Oh my God, I can't believe I was so stupid!" and switched sides. Yes, you should go for the other door, he said.
I was stunned. As I say, I've always thought he was probably the smartest sumbitch I'd ever met. I'd tangled wits with him a few times, and usually (not always, but usually) came out on the losing side. I've often thought of him as one of my most important teachers--a sensei, a professor, a mensch. That was never formal, and I don't think he was quite aware of how I saw him. It's just that he had a disquieting knack for shaking me out of my comfortable prejudices, making me see things differently than I had before. And there he was, admitting he was wrong.
I and a few others spent two or three more days trying to convince everyone that they were crazy. Then I finally listened to what he and others were saying, and tried it empirically. When I did, I had to face it: I was wrong. It was not a matter of interpretation, and no pretending it didn't happen. I was simply wrong.
The difference between this and other brainteasers wasn't just that I couldn't figure it out. I'd often come across brainteasers I couldn't quite figure out, or that I got wrong. The difference here is that I was certain, damn-well certain that I was right. But I was wrong. Completely.
If you try it empirically, you'll find that if you stay with your original guess, you'll lose two times out of three. If you switch, you'll win two times out of three. By showing you an empty door after your first choice, Monty's given you information. Your original choice had only a 1 in 3 chance of being right. Odds were 2 to 1 that the money was behind one of the other two doors--and he just showed you which of the other two doors was empty.
Here's another way of thinking about it: let's play the same game, but it's got a thousand doors, not just three. Only one has money behind it. You pick the first door, and then Monty opens up 998 of the doors with junk prizes. Do you stay with your original, or do you switch? The odds of your first pick being right are 999-to-1 against, and that didn't change just because he opened up 998 empty doors.
Still don't believe me? I don't blame you. This puzzle popped up all over the online universe, and even made its way into magazines, newspapers, and newsletters. Wherever it went, most people gave the wrong answer. The Mensa monthly journal, for example, published an article claiming that the odds were only 50/50. A month later they printed a retraction. Mathematicians and others wrote to Parade and other publications swearing up and down that the odds were 50/50. It was quite something to watch, if you're enough of a geek to care about this sort of thing.
I also had several experiences where I related the puzzle to friends and co-workers, and they refused to believe you should switch. At several points, I tried placing wagers on it; challenging people to put money on the line usually made them wake up and try it empirically before risking their cash. But even then, I found that some people--smart people, quite often--stubbornly refused to take the bet. They merely insisted they "knew" they were right, and that was that.
It was when I encountered such people that I began to recognize a very human impulse. Sometimes, people will refuse to be change their minds once they've decided what to think. You can be friendly about it, joke about it, be irate or impatient about it, or try to present it any way you want: some people will not be convinced, and will even actively avoid looking at proof that they are wrong.
I also decided I didn't want to be one of those people.
(By the way, if still don't believe I'm right, shoot me an email, and we can set up the same wager. I'm so certain I'm right, I'm willing to gamble $cash$ that I can prove it to you. I'll pay up to $1,000 to anyone who can prove me wrong.)
Besides giving me insights into human behavior, this all emphasized something else to me: the world is not always a matter of opinion. You can't worm out of certain truths. 2+2 never equals 5. When someone you love dies, it's not debateable. If you love someone who doesn't love you back, you can't force the other person's feelings to change. Sexual congress between males and females tends to make babies.
More simply put: the world does not always conform itself to what we want it to be, or think it should be, or even what we are certain it must be. Some things are a matter of opinion or conjecture, but some things are stubborn, immutable facts. And cold, hard, ruthless facts contain the keys to finding whatever provable truths exist in the universe.
Maybe that all sounds obvious. But when you run smack into a circumstance where you are damn well convinced you're right but are wrong anyway, it shakes you up. I believe that whenever this happens, you're faced with three choices: backpeddle and pretend it never happened, stubbornly deny reality, or look the world right in the eye and say, "whoops." This entire experience made me decide to always strive to make that third choice.
It led me to start constantly testing whatever I believed. If I was going to express an opinion, it was going to be something I'd done my best to weigh against ruthless logic and cold, hard, empirically verifiable fact--or to the closest I could find to it. I also resolved to be willing to admit whenever I was not certain about something, or did not have enough information to form an opinion.
When you allow these beliefs to seep into your bones, when you consciously try to keep them in mind no matter where you are or what you're doing, it changes you. You find yourself never trusting 100% of anything you read or see. You wind up saying, "I have no opinion about that" or "that's what I think but I may be wrong" an awful lot. You wind up constantly facing the fact that the amount of things you know for certain are finite, but the things you don't know are infinite. If you keep it up, prejudices start to melt away. Sometimes you find yourself feeling just a little scared: "I have no idea what's going on" is a scary thing to admit.
On the other hand, it can be exhilirating. Because when you do pin down a fact, when you can verify something empirically, when you can demonstrate that 2+2=4 and there really is no arguing about it, my goodness it's powerful. It allows you to look someone right in the eye and say, "I believe that you are mistaken," even if you're talking to a doctor, a lawyer, a President, a politician, a priest, or any other expert who knows more than you about almost any subject. Because they can be wrong too, even on a matter where they generally know much more than you.
Facts are a marvelous levelling device, once you've got hold of one. Theories and logic be damned, a fact is a fact!
Frequently this sort of attitude will be mistaken for arrogance. This is because people commonly mistake confidence and certainty with arrogance. But confidence and certainty are what allowed us to do things like design bridges, put men on the moon, cure diseases, and create things like the internet. As long as your confidence and certainty is always tempered with, "Well, maybe I'm wrong, just show me where," you'll rarely go wrong.
All this forms the linchpin of my personal philosophy of life: the universe is knowable. Facts do exist and do not always conform themselves to our preconceptions or our wishes. When confronted with a fact that challenges your world view, it's your job to change your world view--at least, if you want to be an intellectually honest person. Living this way, you can be very certain about some things, not at all certain about others. Knowing which is which is the constant challenge.
But it's an infinitely rewarding challenge, if you take it seriously.
(And by the way, I still blame Pournelle for everything.)
I dunno, Dean; I spent a whole evening trying to untangle my mind from this puzzle, and finally decided to solve it by working through all possible outcomes and tallying the results.
Here are the 24 possible outcomes:
1. Prize is behind #1. #1 is my initial choice. Monty opens #2. I change to #3. I lose.
2. Prize is behind #1. #1 is my initial choice. Monty opens #3. I change to #2. I lose.
3. Prize is behind #2. #2 is my initial choice. Monty opens #3. I change to #1. I lose.
4. Prize is behind #2. #2 is my initial choice. Monty opens #1. I change to #3. I lose.
5. Prize is behind #3. #3 is my initial choice. Monty opens #1. I change to #2. I lose.
6. Prize is behind #3. #3 is my initial choice. Monty opens #2. I change to #1. I lose.
7. Prize is behind #1. #2 is my initial choice. Monty opens #3. I change to #1. I win.
8. Prize is behind #1. #3 is my initial choice. Monty opens #2. I change to #1. I win.
9. Prize is behind #2. #3 is my initial choice. Monty opens #1. I change to #2. I win.
10. Prize is behind #2. #1 is my initial choice. Monty opens #3. I change to #2. I win.
11. Prize is behind #3. #1 is my initial choice. Monty opens #2. I change to #3. I win.
12. Prize is behind #3. #2 is my initial choice. Monty opens #1. I change to #3. I win.
13. Prize is behind #1. #2 is my initial choice. Monty opens #3. I stay with #2. I lose.
14. Prize is behind #1. #3 is my initial choice. Monty opens #2. I stay with #3. I lose.
15. Prize is behind #2. #3 is my initial choice. Monty opens #1. I stay with #3. I lose.
16. Prize is behind #2. #1 is my initial choice. Monty opens #3. I stay with #1. I lose.
17. Prize is behind #3. #1 is my initial choice. Monty opens #2. I stay with #1. I lose.
18. Prize is behind #3. #2 is my initial choice. Monty opens #1. I stay with #2. I lose.
19. Prize is behind #1. #1 is my initial choice. Monty opens #2. I stay with #1. I win.
20. Prize is behind #1. #1 is my initial choice. Monty opens #3. I stay with #1. I win.
21. Prize is behind #2. #2 is my initial choice. Monty opens #3. I stay with #2. I win.
22. Prize is behind #2. #2 is my initial choice. Monty opens #1. I stay with #2. I win.
23. Prize is behind #3. #3 is my initial choice. Monty opens #1. I stay with #3. I win.
24. Prize is behind #3. #3 is my initial choice. Monty opens #2. I stay with #3. I win.
Half of the outcomes are winners and half are losers, and it doesn't appear to make a damn bit of difference whether you stay with your original choice or abandon it after Monty shows you one of the boobies.
Am I missing some possible outcomes here? If so, I'd appreciate you point out one or more of them.
Merry Christmas and Happy New Year. Hope you feel better soon.
Wow. You sure worked hard on that. Let me print it out and get back to you.
Dean
Okay, you basically did this: you posited where the money would be, and then counted potential outcomes. Like this:
Prize behind #1. You guess 1. Monty shows 2. You switch. You lose.
Prize behind #1. You guess 1. Monty shows 3. You switch. You lose.
Prize behind #1. You guess 1. Monty shows 3. You stay. You win.
Prize behind #1. You guess 1. Monty shows 2. You stay. You win.
Logical. BUT. You've got a case here where two of the options you're showing cannot both happen in the real world--only one or the other may happen.
In practical execution, it really looks like this:
Money behind #1. You guess #3. Host must show you #2--he has no choice. You switch. You win.
Money behind #1. You guess #2. Host shows you #3--he has no choice. You switch. You win.
Money behind #1. You guess #1. Host shows you either #2 or #3, at his option. You switch. You lose. (*)
Money behind #2. You guess #3. Host must show you #1. You switch, you win.
Money behind #2. You guess #2. Host may show you #1 or #3, at his option. You switch, you lose. (*)
Money behind #2. You guess #1. Host must show you #3. You switch, you win.
Money behind #3. You guess #3. Host may show you either #1 or #2, at his option. You switch, you lose. (*)
Money behind #3. You guess #2. Host must show you #1. You switch, you win.
Money behind #3. You guess #1. Host must show you #2. You switch, you win.
In reality, these are the only possible games. In each case I marked with a (*), your way of counting counts that one outcome multiple times.
Remember, Monty's choices are constrained when your initial guess is wrong. When you guess right, his choices open up, but there's still only going to be one game.
Another way of looking at this is that your initial guess was only right 1 in 3 times. Why would showing you a door change the odds?
What if we played the same game with one thousand doors. Money behind 1 door, and boobies behind the other 999 (yay boobies! Ahem...). You guess the first door--and now Monty opens up 998 empty doors, and asks if you want to switch.
Does that make the light go on?
(This is a tough damn riddle, no question about it. Ask me some time why I think it shows that Objectivists are sometimes on the wrong track.)
Ah, OK. You're right. I see it now: in my canonical list of possible outcomes I was treating the "host's choice" of which "boobie" door to show me, in those instances where my initial guess was correct, as if it were somehow *MY* choice.
I didn't have any trouble seeing the principle when the number of doors is 1,000 (or any other large number, for that matter); I just didn't take it on faith that it also applied in the case of only three doors.
I'm still skeptical, which REALLY irks me; now I'll be compelled to spend the rest of my Christmas holiday re-learning MS Visual Basic so I can write a simulator and verify the results empirically. Damn.
So, Dean, (ahem!) tell us why you think this riddle shows that Objectivists are sometimes foolish people?
When taking statistics, one of the hardest problems I had was the following:
A man and a woman have 3 male children. What are the odds of their 4th child being male? The answer is, of course, 50% (well, in the real world the answer tends toward 49%, but never mind!)...
Our instinct is to look at the problem and say: “What are the odds of having 4 male children?” This is gives a very different, and wrong, answer. Once I wrapped my brain around that, I started to do well in the class...
The real answer to your brain teaser is not “yes, change”, we actually don’t have any compelling reason to believe that we chose wrongly the first time. What we do need to do is _ignore the first choice_ and reselect; in eliminating a choice, we are presented with a whole new problem. The new problem must be treated as such and not as an offshoot of the original...
I'm sorry to say it, Andrew, but although I think you're spot on with most of that, you're completely incorrect that when Monty asks you to switch, you've got a new problem.
In point of fact, the original choice is still completely relevent. Fact of the matter is that your odds were 1 in 3 when you made your first pick, and that never changes. From the beginning, odds were that the money was behind one of the other two doors--and by opening one of them, he's just told you which door is most likely to have the money.
The biggest logical mistake people make is, in fact, to assume that you've got a whole new problem once Monty opens the door. This is what leads most people into the ditch.
I really am willing to be wrong on this one. Shown empirical evidence, I will admit my error of reasoning. Has anyone done the experiment? It seems to me that If we have doors A B and C (Door A always representing the door chosen, not necessarily the first door), one of the doors being “true” [P] and the other two being “false” [Q], in the first case the odds of A being true are: [p=1/3]. The odds of A being false are [Q=2/3]. After we learn that (for example) B is not true, the odds of A being true are [p=1/2]...
Why is the second problem not a new problem? Has anyone done the experiment empirically and shown that the Probability of A goes up or down as either B or C are eliminated? Or is this all a thought problem? I remember reading once that gamblers used to believe 8 was the most common number to roll on a D6 (er, 6 sided die) until someone actually sat down and rolled a lot of them to discover that it was actually 7. I am more than a bit leery of empirically provable problems that no one has done more than think about...
Having said all that, I submit that I am incapable of doing the experiment myself: it should take about 1000 trials or so to get meaningful results; that many would require a computer to do easily. My programming skills are non-existent...
Oh, hell! I’d just like to point out that I posted that last, _then_ read your “ The Problem With Pure Reason”, though I believe Kant had a different title...
Andrew: yes, of course I've worked it out empirically. Didn't you read the article? ;-)
As it happens, you don't need anything like 1000 iterations. You can generally do it with 20 or fewer. I've never seen a case that required more than 20-30 iterations to prove the point, anyway. Because the fact is that your odds are 1 in 3 if you don't switch. While you could theoretically guess right 50% of the time by not switching, that would be not at all unlike throwing 7 or 11 on craps ten times in a row.
As I've said, I'm willing to gamble money on this. Do you want to play? It's very simple. I'll set up 30 random iterations of the game, and mail the results in advance to one or two disinterested third parties we both agree to trust. Then we'll play by phone. You take your guess, and I'll tell you about an empty door, and you stay with your initial guess every time.
I'll pay you $2 every time you win. You pay me $1.50 every time you lose. This is very generous of me, right? If it's really 50/50, then if we play 30 games, you should walk away with close to $7.50 (or $75. Or $750). Even if you have a run of bad luck, it would be hard for you not to at least break even, right?
I'll up the ante to $20 to you for each win, $15 to me for each loss, if you like. Or even $200 to you for each win, and $150 to me every time you lose. All you have to do is never switch doors. That's your only rule. You guess, and you never switch after I tell you about an empty door. Oh yeah, and you have to play at least 30 iterations.
I'll even go further. If after 30 iterations you're behind and want to keep going, I'll set up another 50, or 100, or as many more as you wish, and we'll keep going.
I will play the game as long as you want to.
Your only rule is that you must always stick with your first guess, and you have to play the agreed-upon number of times--no backing out once we begin.
Think hard about it before taking my bet. Because I'm telling you, quite openly and with complete confidence, that if you take this bet you will owe me money.
By the way--computer programs aren't the solution to this, since a computer program can be written to get it wrong. You have to work it out empirically.
And if we play this empirical game, you're gonna lose buddy. Wanna put some money on it? [grin]
Sir,
If you say that you have worked it out empirically, I am _quite_ willing to believe you! I am really not so wedded to my own notion that it will take loosing money for me to know that I am wrong. What I still don’t quite understand is why I am wrong...
Having said that, would you tell me what is the [P(A)] after the second try [A=original guess]. What does the math look like?
Eh, never mind, you _told me_ what P(a) was. I just read right past that part. It happens after 8 hours of work...
I would still like to know _why_ the math equals out that way. Why does the problem not change?
Heres the way I look at it:
There are 3 doors, A B C.
If you pick A, and stick with it, there is a 1/3 chance you are right, correct?
That means there is a 2/3 chance that the prize is in either B or C.
So if you switch without any extra info, you have to pick between B or C. That means a 1/2 chance of getting it right multiplied by the 2/3 chance that it is in B or C: 1/2 * 2/3 = 1/3. So there is no benefit of switiching without extra info.
If you get the extra info - such as eliminating one of the doors, you have the full 2/3 advantage; you dont have to pick between two anymore. That is why switching is better.
Hope this helps!
I have never seen a mathematical equation to explain this formula. I confess I don't even know how you'd structure such a formula.
I can only repeat what's been said before. Sushant Prakash describes it rightly, as do other efforts. There are several ways to explain it.
The best seems to be to repeat this. Watch carefully:
Money behind #1. You guess #3. Host must show you #2--he has no choice. You switch. You win.
Money behind #1. You guess #2. Host shows you #3--he has no choice. You switch. You win.
Money behind #1. You guess #1. Host shows you either #2 or #3, at his option. You switch. You lose.
Money behind #2. You guess #3. Host must show you #1. You switch, you win.
Money behind #2. You guess #2. Host may show you #1 or #3, at his option. You switch, you lose.
Money behind #2. You guess #1. Host must show you #3. You switch, you win.
Money behind #3. You guess #3. Host may show you either #1 or #2, at his option. You switch, you lose.
Money behind #3. You guess #2. Host must show you #1. You switch, you win.
Money behind #3. You guess #1. Host must show you #2. You switch, you win.
Follow that sequence carefully, and note the simple plain fact: no matter how you slice it, if you switch, you win 2 times out of 3.
Actually Dean, as a probability Ph.D. student, I'll tell you that there is a caveat. You are making one subtle assumption-- you assume that Monty randomly selects between doors #2 and #3 if he has the choice to show you either. That's probably a good assumption.
However, if Monty *always* shows you door #2 if he can show you either (that is, when you're right), then if he shows you door #2 it doesn't matter whether you switch or not-- although with that assumption if he shows you door #3 then you're 100% if you switch.
Here's the way I think of the original puzzle:
Assume, without loss of generality, that you pick door #1. There are three possible events:
Money behind #1 1/3 chance
Money behind #2 1/3 chance
Money behind #3 1/3 chance
We all agree on this, right? Furthermore, assume that Monty chooses evenly between #2 and #3 if he can show you both. We have the following events:
A: money #1, shows #2 - (1/3)*1/2 = 1/6 chance
B: money #1, shows #3 - (1/3)*1/2 = 1/6 chance
C: money #2, shows #3 - (1/3)*1 = 1/3 chance (no choice)
D: money #3, shows #2 - (1/3)*1 = 1/3 chance (no choice)
These are all the possible events. From this it's easy to see that given Monty showing #3, it's twice as likely that C occurs as B. (A and D can't occure.) Similiarly, given Monty showing #2, it's twice as likely that D occurs as A. Either way, you should switch.
As another note, if Monty *randomly* opens a door and that reveals no prize, then, yes, it makes no difference if you switch. That's assuming that Monty has no idea which ones contain the prize, so he just opens up one, and it happens not to have a prize. Verification of this is left to the reader. :)
Lastly, this puzzle should be well-known to any good bridge player, although there it's usually called the "principle of restricted choice." (If you're missing two touching honors, say a Q and a J, and you see one opponent play one, you should play the other opponent to have the other-- assuming that the first opponent will randomly choose between a Q and a J if both are in his hand.)
The mathematical equation to use is Bayes' formula. Take the set up above, where A, B, C, and D are the only possible events. I write P(A) for the probability that A occurs, P(#2) for the probability that Monty shows you door #2, and so on. P(A|#2) means the probability that A occurs, given that Monty shows you #2. Similarly, P(#2|A) is the probability that #2 was shown, given that A occurs.
Bayes's formula is:
P(A|#2) = [P(A)*P(#2|A)] / [P(#2)]
The same is true with any other events substituted above, like B, C, D, or #3.
We have from my previous post that
P(A)=P(B)=1/6 and P(C)=P(D)=1/3.
Since if events A or D occur, #2 was definitely shown,
P(#2|A) = P(#2|D) = 1.
Similarly, if B or C occur, then #3, not #2, was shown, so
P(#2|B) = P(#2|C) = 0.
Also, it's easy to convince yourself that the setup is symmetric, and thus #2 or #3 can be shown with equal probability. Thus,
P(#2) = P(#3) = 1/2.
(If you're not convinced, it can be shown in a longer proof.) Thus, we have:
P(A|#2) = [(1/6)*1]/(1/2) = (1/6)/(1/2) = 1/3
Similarly,
P(D|#2) = [P(D)*P(#2|D)]/[P(#2)] = [(1/3)*1]/(1/2) = 2/3
P(B|#2) = P(C|#2) = 0.
So, given that door #2 was opened, the probability that A occured, and the money is behind the original door is P(A|#2) = 1/3. The probability that D occured, and the money is behind #3 is P(D|#2) = 2/3. Therefore, you should switch, given that Monty opens #2.
The argument is identical with #3 shown instead. (Or with selecting #2 or #3 initially.)
Finally, for a short, non-mathematical explanation of why your odds can't change of being right at the beginning:
You pick door #1. You have a 1/3 chance of being right. Monty can *always* reveal one of the other two doors, no matter what happens. Since he knows what's behind them, he never fails in revealing a door. Thus, he *can't* change the likelihood that you were right to start with. So you *MUST* remain at a 1/3 chance of winning with your original choice.
Therefore, all Monty is changing is which one of the other doors could be right in the 2/3rds of the time that you were wrong originally. He's eliminating one of them, so if you were wrong in the beginning, which has a 2/3 chance, the correct door is now identified. So, you should switch to it.
Thank you, John. I take only a couple of exceptions:
I was pretty sure in my stating of the problem, I made it pretty clear that Monty was going to show you an empty door.
More importantly, however, you said this:
As another note, if Monty *randomly* opens a door and that reveals no prize, then, yes, it makes no difference if you switch.
Be careful. You are making an assumption here when you say that. ;-)
If Monty is going to randomly open any one of the three doors, including possibly the door I've chosen, you've got a completely different game. If he randomly opens the door you chose, and it's got the money, you obviously do not switch. If he opens the door you chose and it has no money, then obviously you switch, and have a 50/50 chance of being correct.
On the other hand, if Monty is precluded from opening the door you chose--which I think is pretty clear from my description--then it would become irrelevant whether Monty knows where the money is or not. Monty will have to choose at random between the two remaining doors. In that case, if he opens a door with money, obviously you switch to that one. If he opens a door without money, you should switch to the remaining door because the odds of your first guess being right are still only 33%.
Correct?
It has always seemed to me that the key to this being the right answer, and it obviously is, is that Monty's choice is not random. In other words, not only will Monty not show you your choice, Monty is not making a random selection between the other two. Therefore,if your choice was A, Monty, by showing you a goat behind B, has told you something new without changing the original odds; you always knew there was a booby prize behind B or C, but you didn't know which, which offset the 2:1 odds of (B or C) being right, as opposed to A. But now Monty has eliminated the offset, so it remains 2:1 that the prize is behind B or C and it isn't B so you should switch to C.
But what if Monty himself didn't know what was behind B and his choice was random? Now don't the original 2:1 odds that the prize was behind B or C have to be reduced by the 1:2 odds against B containing a booby prize, thus reducing the odds back to 50:50?
Using a more imbalanced scenario, suppose I have two containers, A and B, and 100 marbles, of which 99 are black and 1 is white. I place 99 marbles at random in A and #100 in B. If I then invite you to bet on which contains the white marble, the odds are obviously 99:1 in favor of A. Those odds do not change if I then open container A, and looking at the marbles, remove 98 black marbles from A, leaving one marble in each container. The odds obviously don't now shift to 50:50; they remain 99:1 in favor of white. But if I draw 98 marbles at random from A and all are black, leaving one marble in each, isn't it now 50:50? The original 99:1 probability in favor of A would seem to be offset by the 98:1 probability that the white marble should have been among the 98 selected by me at random from A, if it was in A to begin with, leaving a 1:1, or 50:50 chance of A or B now being correct. (If the white marble was among the 99 in A, and all 99 were extracted at random, there would be a 1 in 99 chance of the white marble being selected first or second or third, etc, or 99th as would now be the case if indeed the white was ever in A).
So it does seem to me that whether the apparent reduction in choices changes the original odds ought to depend on whether the reducing choices are random or not. Or am I missing something?
I think the important bit (and I think I possibly repeat what Edward said) is to ask 'Why has Monty NOT shown me the door he didn't show me?
There are 2 possible reasons why Monty didn't show you 'the door he didn't show you'.
One, he didn't show you it because (as you chose the right door) he has just randomly chosen the other one of the remaining two. This happens 1/3 of the time.
The other possibility is he didn't show you the 'door he didn't show you' because HE COULDN'T SHOW YOU IT as it had the prize behind it. This happens 2/3 of the time.
Here's another little head scratcher demonstrating the joys of Bayesian probability:
We are told that family A has two children; the older child is a boy. What is the probability that his sibling is also a boy? Pretty obviously, 1/2.
Now we are told that family B has two children, one of whom is a boy. What is the probability that the other child is a boy? Answer: 1/3.
Huh? How can knowing whether the boy is older change the probability? Rev. Bayes strikes again!
In the case of family A, the only combinations of {gg,gb,bg,bb} (oldest child first) allowed are {bg,gb}, hence 50% probability that the younger child is a boy.
But without knowing whether the boy we do know about is younger or older, the probability space is {bb,bg,gb}. Thus the chance of two boys is only 1/3.
This might seem like harmless fun, but try explaining Bayes's Theorem to a jury so they can weigh the significance of a DNA fingerprint.
**In the case of family A, the only combinations of {gg,gb,bg,bb} (oldest child first) allowed are {bg,gb}, hence 50% probability that the younger child is a boy.**
I thought the only combinations allowed would be bg and bb, since oldest child is boy. Was it just a mistake or am I not understanding something?
Following up to your reply to John Thacker's comments, what about if the gameshow host doesn't have to open a door for you? (i.e. he can either open a booby-prize door which you haven't selected, or not open any door)
Then, he's only opening the door to try and influence you, and so it comes down to psychology and game theory. If his goal is to stop you winning, and he knows you know about this problem, he will know you will change when asked! So his strategy would be *only* to open a door if you originally selected the correct door. So, the correct strategy would be to stick.
[I reread your description, and Monty clearly doesn't say anything about opening other doors *before* you make your selection]
So, if I trust Monty, and think he's just trying to make the game more exciting, I'd change. However, if I don't trust Monty, and think he's trying to screw me, I'd stick with my original choice.
So maybe the moral of this story is that you should go with your 'gut-feel' after all! What's more important: an ability to judge Monty's motivation, or an ability to prove mathematically this brain-teaser?
Edward: I have to repeat what I said before.
If Monty will never open the original door you picked, but has to randomly pick between the two remaining doors (because he doesn't know what's behind them), then there are two possibilities:
1) He opens the door with the money. You switch and win, obviously.
2) He opens an empty door. In which case, he has just accidentally helped you make a discovery: which one of the unchosen doors definitely doesn't have the money. But the odds of your original guess are not changed by Monty's knowledge or non-knowledge: they were still 33%, and haven't changed simply because Monty's made a random stab in the dark. If his random choice shows you an empty door, you should still switch, and will win 66% of the time.
It is virtually impossible to achieve a 50/50 game here. The only way I can see it is if Monty chooses completely at random, including the possibility that he will open the door you've already chosen. Then, if he opens your door and it's got money, you don't switch, and you win. If it doesn't, then you obviously switch, and then--and only at that moment--do your odds become 50/50.
If the game is truly random, what Monty knows or does not know does not affect your odds. What Monty and you mutually discover when he opens a door, however, does change them.
David: Excellent reasoning!
Although I thought I covered everything, and made it very clear that Monty would only choose an empty door from one of the unchosen two in my original description, I did not state that Monty has the option of refusing to open any doors. Drat it, you caught me. You're the first one to catch that in the nearly one year time span since I first posted this riddle!
If Monty has a choice to not bother opening any door at all, then the odds become densely scrambled, because he could be playing head games. Although I thought I made it clear that there was no trickery here, I apparently didn't do a good enough job of it. Good on yer. ;-)
Dean, you're correct that I should have stated the revised problem better, but here goes:
You select one of the three doors. Monty randomly opens one of the other two, and it does NOT contain the prize. Now your odds are indeed 50-50, and it doesn't matter if you switch. You could just trust me, or run it empirically. But I'm absolutely right on this one. I'll make the same sort of large money bet with you over a large number of tries.
Quick explanation:
Without loss of generality, assume that I pick door A, and Monty shows door B always, and that he shows door B to be empty. Initially, there are three possibilities for the prize, each with probability 1/3: the prize can be behind A, B, or C. However, his action eliminates only the cases where B has the prize. The remaining events are split evenly between A and C: 1/2 chance of each.
To put it another way, note that it's exactly like the previous problem except that Monty always prefers to show you B if he has a choice between B and C, and we assume that he shows you B. Remember that in that case I showed that the odds are 50-50%.
You can play "true" Let's Make a Deal style, where you're not allowed to switch to B if it reveals a prize. Then your overall chances of winning are 1/3. If you're allowed to switch to B, then your OVERALL chances are 2/3 (because you always win if it's in B, but you also get one more door), but the chances of winning once B is revealed to have no prize is 1/2. (Note: (1/3)*1 + (2/3)*(1/2) = 2/3 )
Unlike the original problem, you don't eliminate half the events where A holds the prize, since Monty can't choose between opening doors B and C.
In the original game, if you pick A, switching always effectively allows you to pick B-or-C, since switching always means you win if the prize was behind B OR C to begin with, because of how Monty is forced. In this new game, switching just trades A for C. It's winning on A vs. C if you can't switch to B if it wins, or winning on A-or-B versus winning on B-or-C if you can switch to a winning B.
Yet Another Explanation:
I'm assuming that you pick door A, and then Monty opens B randomly, and it happens to not have the prize. But if you had picked door C, Monty still would open door B, and it would still not have the prize if it didn't before. These cases are obviously exactly the same, and so the same strategy must be right. But it can't possibly make sense to switch, since Monty opens B regardless of if you pick A or C. Your picking A or C doesn't move the prize. So, you could have just told Monty "I'll pick one of A or C," and he would still open B, showing it to not have the prize. Then you pick from A or C after that. Note that Monty's opening of B is thus INDEPENDENT of which one of A or C you actually open. Thus, all these cases still have the same probability. Picking from A or C after he opens B is clearly 50-50.
Sorry Sushant - that was a typo. I did indeed mean bg and bb.
I'd actually been told this riddle a efw years ago, and I hadn't been convinced at that time. Your explanation set me thinking, and now I understand it, thanks a lot!
Dean,
Having read your response to me but also John's rejoinder to you, it still seems clear to me that the randomness of Monty's choice does change the odds. As one of the earlier posts pointed out, if you flip a coin and it comes up heads three times in a row, the odds on the fourth try are still 50/50, regardless of the odds against getting heads four times in a row. The reason is that the original odds against four in a row are reduced by the odds against the three heads in a row which have already come up. Thus the fourth flip is a new game and the odds are 50/50.
The same must hold true for Monty. If his choice is random, then the odds must change, just as the odds against flipping heads four times in a row decrease each time heads is thrown.
When Monty, on the other hand, showed us the goat behind B, there were no "odds against" his choice which could impact the original odds. The probability of Monty finding the goat behind B was 100% because Monty already knew where the goat was. The original 2/3 probability of (B or C) being correct therefore remained unaffected. That's why switching was the right answer.
I agree that it sounds anomalous to say that the odds are affected by Monty's state of knowledge. But in fact Monty's knowledge seems to be the only reason the second round of choice should not be analyzed as a new problem, with 50/50 odds. Otherwise, how do you answer the coin-flip scenario?
Okay, okay, I misstated the matter above. I had part of it right but then wandered into the weeds.
If Monty must choose randomly between the two doors that you did not choose, and it comes up empty, you've gotten to a 50/50 game.
To work this out empirically, you have to exclude all cases where Monty opens up a door at random and shows you the money. By stating that Monty opened an empty door on accident, you've changed the odds--because Monty's odds of guessing an empty door are not 50/50.
Still, in the real world this circumstance could occur, and my description did not preclude it. I'm going to have to rewrite it so it does.
I wrote a Monte (Monty? :) ) Carlo simulation of this last night. After one billion runs with no switching, I got 333314878 wins. With switching every time, the number becomes 666645895.
C code available on request.
Okay, I've edited the original, to close up the holes you clever people have found. The following paragraph:
...has been amended, and now reads as follows:
That should close up the loopholes, yes?
I remember having a bad time of this puzzle at first, when I was taking probability. I found a clear way to see it, eventually. I hope this helps.
The key is to remember Monty's limitations:
-the door he opens must be a loser, and
-he can't open the door you picked
You pick a door. You have a 2/3 chance of picking a loser, and a 1/3 chance of picking the winner. Monty then opens a losing door, but not the one you picked.
If you picked a loser, Monty was just forced to open the other loser so if you switch you're guaranteed to get the winner. The odds of this are 2/3.
If you picked the winner, Monty opens a loser but which one doesn't matter. If you switch you'll just be going to the other loser. Fortunately the odds of this are only 1/3.
You win 2/3 of the time if you switch.
To the people who say it's 50/50- it's only an "honest" 50/50 if Monty is allowed to open the door you picked (if it happens to be a loser).
In that case, you pick and have a 2/3 chance of picking a loser and a 1/3 chance of picking the winner. Monty then open one of the losers, even the one you picked. Then what you have is:
-you picked a loser and Monty opened the other loser. If you switch you win. Happens 1 in 3.
-you picked a loser and Monty opened your door. You have to repick, and it's 50/50. Happens 1 in 3.
-you picked the winner, and Monty opened one of the losers. If you switch you'll get the other loser. Happens 1 in 3.
David Gillies: How many lines long is the code? I'd love to post it here. Why not? :-)
Sure Dean, no problem. I've emailed you the source. On my ancient PIII 500MHz a billion tries takes about ten minutes.
You guys are all far beyond the mental breakthrough promised in the article. And yet still no flying cars. hmm.
I won't pretend I can outsmart you probablists. I learned something here. Thanks. Now I wanna watch the show so I can scream "Switch, Penguin Lady! Switch!"
Update: I just tried this on one of my best friends. His initial reaction was, natch, that it makes no difference if you switch. I pointed out the fact that if you pick the correct door first time and switch, you always lose, but if you don't and switch you win every time, and you do the latter twice as often as the former. He bought it, straight out the door! This is a smart, but mathematically naive guy (no advanced maths beyond grade school algebra).
BTW, for the conundrum about whether a boy or a girl is more likely in a two-child family where one knows that one of the children is a boy, it is pedagogically simpler to substitute a fair coin, viz.: "I toss two coins. The first is a head. What is the probability that the second is a head? Answer (obviously): 1/2" vs. "I toss two coins. One of them is a head. What is the probability that the other is a head? Answer: 1/3"
While you're mentioning that, David, I'll re-post this email you sent me last night:
That is a wonderful story.
It's my observation that a certain minority population accepts the correct answer immediately. My wife--who is gifted in math, and used to tutor calculus to engineering students--pretty much "got it" right away and with little struggle.
Most people blow this at first, yet pretty consistently I've seen that some people grasp the right answer easily. How smart you are seems to have no bearing on the matter.
(The guys with the degrees in probability are probably still laughing at us.)
Erdös hated the Monte Carlo approach (not elegant enough) but even he had to concede the switching strategy was correct after being shown the results (of course, spurred by this he went off to find an analytical solution). Generally even the most recalcitrant of anti-switching proponents will be convinced if you offer to wager them on the outcome of ten or so tries. As you say, empirical results trump everything.
I don't have a degree in 'probability' although statistical analysis and manipulation played a very heavy part in the mathematics portion of the syllabus for my Physics degree. I think a certain, albeit small, amount of mathematical education is essential, if only because the majority of people are so innumerate that the Gambler's Fallacy represents to them an Iron Law of the Universe (go on - try it - ask the average man in the street what the probability is of getting heads on the 11th toss of a coin if they've all come up heads so far.)
I first saw this conundrum about 15 years ago on Usenet news (rec.puzzles) It was generating heated discussion back then. It was amazing how few people said, "right, let's just set up a few (dozen/hundred/million/billion) trials". Kinda like the story about the Greek philosophers arguing about how many teeth a horse has, and Archimedes (or was it Socrates?) stepping up to the plate and saying, "look, guys, why don't you just look in its mouth?"
Anyway, you gonna post my Monte Carlo code, or what <grin>?
1)It's most likely your first choice is wrong.
2)The quizmaster then has to show you the (only) other wrong door;
3)So your best bet is to switch.
One more reason for me to mistrust a jury of my peers.
There is only one prize. You have one chance, Monty has two. Knowing that one of Monty's doors is a dud adds no information since with only one prize at least one of Monty's doors has to be a dud. So, SWITCH!
Gee. Of course Monty can't open your door or not open a door. Duh. You've chosen a door with a 1/3 chance of being the right door. The chances of the other two doors being right are 2/3. Monty shows you a door that isn't right. So your door is still 1/3 likely. The remaining door is, therefore, 2/3 likely to be right. So switch!
The clearest way I can think of to explain it is as follows:
After you make your choice of Door A or B or C, the Host tells you that you can stick with your original choice, or choose BOTH of the other two doors. Pretty easy to see which of those options has the better odds.
By opening one of the other two, the Host has effectively given you the option of picking both.
It took one good explanation to turn this one around for me. I first encountered the problem in my grade 11 chem class, and I actually managed to convince the class, including my teacher who had given the problem, that there's no good reason to switch. That weekend, however, my grandfather, who was a math teacher, explained to me why I was totally wrong by explaining the solution like this:
On your initial guess, you have a 1/3 chance of being right. If we were somehow able to select both of the other doors, and get the money if it was behind either of them, then our odds would increase to the 2/3 level. But how can we do this? We get some help from Monty. When Monty shows us that one of the other two doors doesn't have the money, it is now possible to select the two-door group, and be sure that if the money was ever behind either of them, we have found it. By switching, we are able to gain the 2/3 advantage.
When trying to argue the point, I've found that it helps to put it in terms where it seems insane to disagree. Ask the unbeliever to imagine the problem as having n doors. Which would you prefer: a single guess, or n-1 guesses? Or even more concretely, in the case of 1000000 doors, 1 guess, or 999999 guesses? It usually shows the way.
Thinking on it some more, I've come up with another way of explaining it.
The key fact that seems to lead people astray is interpreting sticking with the original choice as making that choice at the 1 in 2 level. This is not the case. The chances of your selection being right were 1/3 when you made your choice, and they remain at that level no matter what. Since the odds of the inital guess being right are 1/3, and we know that if the inital guess is wrong, then the other unopened door is right, then the other door's chances of being the correct choice must be 1-1/3=2/3.
The problem would be totally different if, say, after revealing the door, Monty then invited your wife onstage, and asked her to choose from one of the two doors, without telling her which one you picked. Her chances of winning are 1/2, since she makes a blind guess at two doors, with only one being correct.
In my previous post, I mentioned a scenario where your wife comes onstage, and chooses a door, oblivious of your answer. Here's a variant of that situation that drives home just how different her situation is: Suppose Monty tells your wife (or partner, I should probably say; women could be playing too...) which selection you made. Now suppose that your partner will agree with you X% of the time, and will choose the other door (100-X)% of the time.
The problem: What does X have to be in order for your partner to have a 50/50 shot? What does X have to be in order for your partner to have the best odds? The worst odds?
This problem is interesting, because the answer also provides a response to those who say that the odds in the initial problem are 50/50; it shows them a way in which they can be correct.
Last post, I promise!
I posted my last comment without providing the answer to my simple, but surprising question.
Following the lines is the simple analysis of the answer
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Suppose your wife agrees with you x% of the time. Therefore, there are two ways she can be right. Either she goes with you, and is correct, or changes her answer, and is correct. The first occurs x*(1/3) of the time, while the second occurs (1-x)*(2/3) of the time. Adding these, we get x/3 + 2/3 - 2x/3 = 2/3 - x/3 = (2-x)/3.
Therefore, if we want your partner to be correct 50%, or 1/2 of the time, we just solve for x:
(2-x)/3 = 1/2
4-2x = 3
-2x = -1
x = -1/-2 = 1/2.
Therefore, in order for your partner to be correct 50% of the time, s/he should agree with you 50% of time. Though this sounds odd at first, it makes sense: By going with you 50% of the time, and going the other way 50% of the time, the choice is back to a random selection between the two choices, with a probability of 1/2.
To find you partner's greatest odds, we can either apply common sense, or calculus. We'll go with calculus, just for fun. Let f(x) = 2/3-x/3. Therefore f'(x) = -1/3, so f'(x) is never zero. Therefore, we know that the max and min points on the interval of [0,1] (the only possible values for x) must occur at the end points. By inspection, we find that f(0) = 2/3, whereas f(1) = 1/3. Therefore, your partner's best odds happen when they never agree with you, whereas their worst odds happen when they always agree with you.
It should be noted that never agreeing with you is identical to yourself always switching, and always agreeing with you is identical to never switching. This shows, among other things, that not only should you switch, but always switching is actually the best way to win, while always sticking is the worst.
I know just enough probability theory to have thought the poser of this problem was wrong. I even went so far as to exchange emails with him (there's a website out there somewhere; google-search "Monty Hall Problem" or "Monty Hall Dilemma" and you'll see) that got increasingly agitated until the guy on the other end finally got through to me.
Where this problem is hard is not in the math. Where this problem is hard is certainly in the way that it's worded. It looks like an easy, obvious problem, but it's really a slightly less easy, slightly less obvious problem in a very clever disguise.
I only stumbled upon this puzzle yesterday. My initial reaction, even halfway thru the piece was "It HAS to be 50/50". The kicker for me was the "1000 door" analogy.
For the record, I've tried this (using pieces of paper for the doors with "lose" and win" on the hidden side of the paper) with my colleagues and my family. My wife stuck to her guns and lost 10 of 12 games. She switched on the last two. My youngest son immediately embraced the idea of switching to win and won 9 times out of ten.
I'm seriously considering giving up the day job and becoming a street hustler. "OK, buddy, you think THAT one is the ace? Right...let me improve your odds by showing you that this one" - flip over a card - "is not the ace....want to change your first guess?" I could retire at 45.
Thanks, Dean.
Whoa! That took forever to read through! Interesting problem, but I'm afraid it only took me about four posts to agree with Dean. I'm probably repeating everybody else here, but the simplest way to put it is this:
You have a greater probability of being wrong (2/3) than you do of being right (1/3).
If your guess is wrong, then the remaining closed door is the prize. If your guess is right then the remaining door is nothing.
Since there is a greater chance of being wrong, there is therefore a greater chance of the remaining door being the money.
i.e.
Choose A -> Open B -> C is money -> WIN
Choose A -> Open C -> B is money -> WIN
Choose A -> Open either -> A is money -> LOSE
Assuming you always switch, you win 2/3 times, regardless of which door you originally choose.
Well, so many comments about the mathematical problem, that we almost forgot what Dean's article was really about: being so sure about something that you cannot accept the possibility of being wrong...
Let me tell you how I learned about this quiz. Two weeks ago my wife and I went to the movie with two other couples. One of the guys brought the problem, and I almost immediately gave him the 50-50 answer. He had seen the problem in a article, which had the mathematical prove for the 1/3-2/3 solution (the same Sushant brought on Dec-30). The third guy (who later confessed he had seen the problem and the solution before) agreed with the article's solution. I called him an idiot, I said he didn't honour the Economics degree he holds, I couldn't understand how someone smart could be agreeing to that bullshit...
The following day I asked my friend to e-mail me the article, and saw the demonstration by myself... Then I thought: wow, now we have 3 idiots: our friend, and the 2 analysts who signed the article (my point was that they should've broken one of Baye's Theorem assumptions)! A google-search later and I found out that I was the real idiot. It still took me some hours to understand why the 1/3-2/3 answer was correct...
Since then, I've told the problem to almost every friend and co-worker I have. Most of them don't buy the 1/3-2/3 approach.... The guy who brought me the problem originally, said "I know that Math and Statistics say I should switch doors, but I stay with my original choice!"
Weird how we all tend not to accept that this "small-truth" of ours might be wrong. I have to admit that I thought Dean's original article a bit too dramatical (sorry, Dean...), but it made me consider how it was when someone brought the idea of the Earth not being like a Pizza, or the Darwin's evolutionary concept.... People were defying God's truth, their whole world breaking in front of their eyes! If we spent so much brain-time on a probability quiz, just think on how painful it was to rebuild their whole concept of truth....
Well, now let me give my contribution to the quiz itself: I think the best way to explain it to skeptical people is this: consider you play this game 300 times, and all of them you chose door#1 and DON'T switch doors after Monty opens an empty door. How many times do you expect to win? (People will agree that this number is 100). How many times do you expect the door that Monty has opened to be the one with the prize (People will agree that, by definition, this number is zero). Then, since you are playing 300 times, the third door must be the one with the prize the remaining 200 times.... By the way, this approach also shows that if Monty's choice is not always wrong, then changing doors might not be worth!
Wow. My brain hurts.
Consider this scenario: you get picked out of the audience, and as you're walking to the stage you decide to choose door #1. Then, before you say anything, Monty opens one of the empty doors to make the game easier.
From his perspective you now have a 50/50 shot at the prize. But if he opens door #2 or #3, you've still got better odds if you change your mind, right?
So is Monty wrong if he announces your 50/50 odds to the audience? And if he is, how could you prove it to him?
Dean, if you were so sure you were right the first way of thinking, then you realized you were wrong, is it not possible that you are wrong in the way you think about it now?
I think you mentioned that people who think the new choice is 50-50 are flawed because the probablities don't change from one round to the next. That is exactly the point though. Even after monty opens a door, the other door that you didnt pick, and he didnt open, still has a 1/3 chance of being right. and since the door you originally picked also has a 1/3 chance of being right, 1/3 = 1/3 and therefore each door is equally likely to have the money behind it. When you say the door you should switch to has a 2/3 chance, you are mistaken because the 2/3 chance is ONLY the chance that its not behind the door you picked. As soon as Monty opens a door, it doesnt change the probability that its behind the other door (the one you didnt pick, and the one he didnt open). That door still has a 1/3 chance of having the money, not 2/3. I have not studied your 'empirical' tests carefully, but I feel confident I can find a flaw. And no, I do not want to bet money, not because I think I am wrong, but simply because I am not a 'gambling money' kinda guy
Also, in your response to Dave D's demonstration of cases, I dont believe your response is correct. You say this:
Money behind #1. You guess #3. Host must show you #2--he has no choice. You switch. You win.
Money behind #1. You guess #2. Host shows you #3--he has no choice. You switch. You win.
Money behind #1. You guess #1. Host shows you either #2 or #3, at his option. You switch. You lose.
Money behind #2. You guess #3. Host must show you #1. You switch, you win.
Money behind #2. You guess #2. Host may show you #1 or #3, at his option. You switch, you lose.
Money behind #2. You guess #1. Host must show you #3. You switch, you win.
Money behind #3. You guess #3. Host may show you either #1 or #2, at his option. You switch, you lose.
Money behind #3. You guess #2. Host must show you #1. You switch, you win.
Money behind #3. You guess #1. Host must show you #2. You switch, you win.
Which seems all well and good and proves your point, but if you initially pick the door with money, Monty can show you either door, so the probability that he'll show you door 2 is the same that he'll show you door 3 (assuming the money is behind 1 and you picked 1 and that there is not some other governing rule that says he always picks a certain door or something) So I think Dave D is more correct when he shows 4 cases, not your 3. Yes, Monty will in actuality only show you one door, but they both can happen, which is what probability is all about.
Oh well, I withdraw! =)
Okay I ran a bunch of tests, and yes you do win 2/3 of the time more if you switch. Empirical evidence proves true.
The rub of course is that on the gameshow, you only get one chance to play, not 30, or 1000! so in the long run its better to switch, but for your one time playing, you can obviously be the unlucky 1/3 who picked right originally, switches and loses!
Ok Check this...
I pick a door.
There is a one in three chance I am right.
The host does his hting... it makes no difference... and opens a door... who cares!!
Now the process starts all over again, just as if I had been asked to pick a door, and there was a one in two chance of me being right.
I still have the same probability of winning at this point, which is 50%.
It doesnt matter at all what happened in the beginning. If you are doing Empirical Evidence tests it does. But not if youre there on that game show. At the one moment that counts... the moment when you decide between 2 doors... you have an equal chance of winning and losing.
You have a 50/50 chance at the one moment that matters.
And thats all that matters.
Boom. Your head just exploded, didnt it?
Ok, I've read this problem several times, and, quite frankly, I get upset at the reasoning given.
Unless I've been taught incorrectly, probability isn't serial- if someone flips 20 coins at the same time, 50% will probably be heads, but if they flip 1 coin 20 times, the 20th flip has a 50/50 chance even if the first 19 came out as tails.
This question basically sets up a serial problem and the justification is based on the probability pre-revelation.
I see this as flawed logic.
If I'm wrong about probability not being serial, please feel free to let me know that I've been short changed on my public education- it won't be the first time.
As the story is told and as I've been taught, however, the story is not the same as choosing one door and then being told to choose between that and the other 2 doors. One of the doors is determined, just as much as those previous 19 flips. The probability is calculated after the determination.
-Drew
I read the original puzzler, then was interrupted by a co-worker. I then saw that there are only three possibilities, the money is behind door number 1, door number 2, or door number 3. I choose number three and the host opens 1 or 2. In the case where the money is behind my door, it is best that I stay put. But in the two cases that the money is behind the remaining door, it is best that I switch. It makes sense, so I answer that there is a compelling reason to switch. I scroll down and am simply amazed at how difficult this can be for (*some!*)people to grasp, even after reading everything.