this is simple

you are changing your chance of winning from 1 in 3 to 1 in 2 if yu change. If you don't then your chance is still one in 2 as by choosing that particular door you are asuming that the other 2 are wrong. When one door is opened your choice is now of 2 doors if you change you are saying that there is only one of 2 doors it isnt so a 1in2 chance therefore increasing your chances.

Increase chances??? Is this answer correct? I'm tempted to say it makes no difference as Monty has automatically increased chances to 1 in 2 by revealing a booby prize door.

Changing now will not increase my chances. Effectively this game was never 1 in 3. It's always 1 in 2 as long as Monty always reveals one booby prize.

I greatly miss the old Pournelle RT. Damn IDT for ruining the great thing that GEnie was.

I do not miss this particular argument, however, because I didn't care one way or the other. I knew I was never going to be on "Let's Make a Deal."

Indeed!

Thanks for the essay, I thoroughly enjoyed reading it.

== Mal

Thanks for the puzzle. I got it right, but that's because I've run across similar puzzles before, and thus recognized the problem class.

It just goes to show you that probability is hard for human beings to do right. But then, we already had an existence proof of that fact, called Las Vegas.

On the philosophical issues, here's a thought.

There's more than one kind of 'reality.' In some things, what's real is determined by objective, scientific facts (water is made up of two parts hydrogen to one part oxygen, even if we don't like it; further, there's nothing we can do about that fact). Call this class of facts W1.

But in some other things, 'reality' is determined by opinion and desire. Our laws are not made by the British Parliment, and that's true because we want it to be that way. If we didn't want it to be that way, we could change it by an act of collective will. Call this class of facts W2.

In still others, 'reality' is determined by logic and mathematics, but said logic may or may not have application to 'realities' of type W1 or W2. For example, there used to be logical 'proofs' that the universe is utterly deterministic, but the proofs were based (among other assumptions) on the idea that every sub-atomic particle has an exact position and velocity. It is a fact of class W1 that the physical universe doesn't work that way.

People tend to mix up these areas. The former Soviet Union (OOH! It's been ten years, and I still love to type 'former Soviet Union.') foundered on a massive confusion of W2 events (such as "What do the Party bosses like?") and W1 events (e.g., "What's the best way to advance technology?").

And of course, 'best' is ambiguous. Is the 'best' way to advance new technology the one that gets the most technical results for the least economic inputs, or the one that makes the rulers happiest?

Note the areas interpenetrate and overlap. People who accuse Mark Twain of racism because HUCKLEBERRY FINN uses the word 'nigger' are arguing about facts of class W1, and they are wrong. People who say the book offends them because it uses the word 'nigger' are reporting facts of class W2. It's pointless to argue they shouldn't be offended. They are.

So it's important to keep in mind whether you're discussing a logical argument, and if so, what the assumptions are, and what they are relevent to; or a physical argument, where we need scientific data; or a psychological argument, in which case we'd better know just what mental states are under discussion.

Another example: recently, over at "abuzz", I've been in a long dispute about Pearl Harbor, whether FDR knew the attack would take place there on Dec. 7th, whether he pushed us into war with Japan because he had guessed that Hitler would declare war on the U.S., and similar subjects. This is a W1 argument about a W2 situation: what was going on in FDR's head in 1941. It's remarkable how many people present as arguments on the issue reports of their own psychology ("I think FDR knew." "I can't believe any President would do that,") and how few present any evidence of what President Roosevelt believed and wanted at the time.

Hi Tim,

Maybe you already know, but maybe you don't. Jerry still has a very active discussion forum at:
http://www.jerrypournelle.com

Hey, Dean! Long time, no see. I've occasionally wondered what you were up to, and then I find the link on Jerry's site here. I, too, miss the days of the Pournelle RT.

As for the puzzle: Yeah, the whole RT was like that. Even though I was in charge of Games, Chat, Sports and some other stuff for GEnie, being the product manager for that RT made the whole time worth while for me.

I sure wish I had kept archives...

-Jess

Let me see if I have this right.

Before the other door is opened, you have three alternate futures, all equally likely. Opening one door closes off one, but it equally consistent with the other two. That is why people thing, "hmmm, 50/50".

BUT, the door that stayed closed ONLY stays closed in half of the future where you have the right door. And the door that stayed closed stays closed in ALL of the futures where IT is the right door. SO:

All the future started out equally likely
ALL the futures with the prize behind the openned door are foreclosed
HALF the futures with the prize behind your door are foreclosed
ALL the futures with the prize behind the remaining closed door are open.

On a count, that is 1:3 for sticking, 2:3 for switching.

I've even had some vague exposure to the Bayesian approach, and I've only "always knew this" for the last four minutes.

Maybe this is the key to seeing how taking "hardline" approaches often fails in the long run ... softer lines are more likely to let people pretend that they were "previously misunderstood" when they decide to switch sides, while harder lines more often push people into taking unambiguous stands, even when they aren't positive which side of the line they wish to be on when the demand to "take a side" is made.

I am French, and not very sure of my English.
A friend send me your puzzle, and I try to demonstrate (if not arrogant, as you said!!)the résult.

The thing is, when you choose the first door, you have 2 chances of 3 to be wrong. Then, of the 2 doors left, one is good, one is bad. And Monty opens the bad one. Now, you are sure that if you had chosen a bad door the first time (2 chances of 3), the last door is good. (And if you had chosen the good one (1 chance of 3), the last door is bad).
You see now that then you switch, you have 2 chances of 3 to have the good door. It is the odd to have the bad one the first time.

I like your sentence "It leds me to test whatever I believed". It is the start of Descartes's philosophy.

I thoroughly enjoyed reading your article. I was convinced (99% sure) that it was 50/50. The thing that swayed it for me was the argument about the 1000 doors.
One person (Steven Onge)has made a mistake on assumptions immediately after reading the article. My laws are in fact made by the British Government...this is the internet.

Suppose Monty doesn't open any doors. Instead he says "I can open a door with a booby prize behind it." Everyone agrees that this is possible.

Now he says "Would you like to switch?" Now you can see the you have two chances if you switch, only one if you don't.

Monty does NOT open one door at random and find it empty. In that case, the odds are indeed 50/50 for the remaining door. Then the prize could have been behind the door he opened.

When Monty opens the door with no prize, he is not adding information, since you already knew he could do that.

Where we are misled in this problem is counting Monty's door opening as a guess. It's not.

This is my explanation of the mystery.

You have a 1 in 3 chance if you stick with your original choice, and a 2 in 3 chance if you switch. This is a fact.

If Monty opened a door at random and found it empty, each remaining door would have a 1 in 2 chance of being correct. 1 in 3 times, Monty would win the prize. 1 in 3 times you would win the prize, and 1 in 3 times no one would win the prize.

Monty CHOOSES to open a door with no prize. This has no effect on the probabilities.

Suppose Monty draws a line so that your door is on one side of the line, and the other two are on the other. The offer is made, stay on your side or choose the the two doors on the right. Of course you will choose to go with the two doors. Does it matter if Monty opens up a losing door? You knew one of them was a loser already.

Nothing Monty does can change the fact that there are two chances on his side of the line.

It appears that Monty is giving information by opening the door, but that's an illusion.

Recently I discovered that this article and the attendant discussion got chopped off and lost, probably some time ago during a conversion we did. I have restored it, and as many of the comments as I could, although the dates on most of them will be wrong.

I will add one thing though: Michael is mostly right, except Monty is giving you information: he's identified one of the three doors it could not possibly be behind. Given that you know he will only choose a door it won't be behind, that's information. It allows you to say for certainty that it's one of the two remaining doors, without changing the odds that your first guess was 1 in 3.

It's funny how semantics can mess with your interpretation of things like this!

Holy Crap! i am a millionaire!

I still don't get it, and it's not because I don't want to. I love counterintuitive things like this.

Only it seems to me that once you get a chance to change your answer, it's a completely new event. You have now qualified for the final event - the prize is behind one of two doors. You can pick either door. Your previous choice has nothing to do with it.

Think of it as two separate events - the first one with three doors, has probability 1 of succeeding, since even if you did choose the booby prize, Monty just opens the other one, and you're still in the game. The second game has probability 1/2.

You should neither always switch nor always stay. You should switch half the time, and stay the other half. In other words, retract your guess, and then make a new one, which may be the same as the original one. Am I wrong?

I'm too lazy to read all the responses (and am racing out the door to go to work), but essentially, I think your lack of belief stems from the way you're phrasing the question/teaser. I think this because until I rephrased it, it made no sense.

Look at it this way: You have three doors. One of them has a prize behind it. You want the prize. Consider that after you've made your choice, the game is over. You don't have the option to switch. However, the evil Monty is going to force you to switch at the last minute. Now, what is your goal to win?

Your goal now is to pick a losing door. By doing so, you'll win. When there are three doors, what are the chances you'll win? 2 out of 3. With 10 doors? 9 out of 10. etc.

I know you could say that on Let's Make a Deal Monty doesn't always offer you the choice. However, in the various examples of this game that I've seen online (you know, the ones offering "empirical" evidence), that factor has been eliminated.

I understand that essentially I'm playing a semantic game, but it's one that certainly assists in understanding the odds/processes involved.

You're wrong, because in this game scenario, the deck is *always* stacked. There's only one significant gated choice to consider, only one choice that is enforced -- and the odds at the instance of that choice are what counts. You could start with 1,000,000 doors and the odds of selecting the correct door would still only be 2:1 if Monty eliminates all but two doors (the one you pick and another one). The implication that the odds being at 1,000,000:1 is illusory because that particular ratio is never enforced: there is 100% certainty that you will be given 2:1 odds when the actual prize door is revealed. Therefore, discounting certainties, your odds will always be 2:1 in the way the game above is set out.

I'll take check or cash. :D

If you're so certain, do you want to take my wager?

I don't, and here's why... the odds are not deterministic; they depend on the way in which you make your choices. If you agree to flip a coin to decide whether to stick or switch, I'd take your wager. Or if I'm feeling adventurous I might take your wager for just one instance of the game. It is however true that your odds will be what you say they are if you switch *every* time, as a matter of policy.

I stand corrected!

Part of what's interesting about the problem is the way in which it's presented. It offers certain deterministic assumptions, chief among them that Monty exercises knowledge of the prize location in his choices. If it were a purely random system it'd be 50/50, but it's not.

Well, yes, but in a truly random system, Monty might well accidentally open the door containing the money. Which would be a strange thing for Monty to do.

But you have now grasped the problem. And Monty does say he always does this, and that he's going to open a door he knows is open. That's your tipoff.

Initially when I heard of this years ago, I thought it was 50:50, and then a great many intelligent people managed to convince me that no, you are better to switch doors, although I never felt 100% comfortable.

But Dean's philosophy inspired me, so I reasoned it through as simply and thoroughly as possible (ie by brute force):

There are only four ways I can see that the game could play out, each of equal probability:

1) I pick the prize door, and Monty opens garbage door 1. If I switch, it can only be to garbage door 2, and I lose.

2) I pick the prize door, and Monty opens garbage door 2. If I switch, it can only be to garbage door 1, and I lose.

3) I pick garbage door 1, Monty then has no choice, and can only show me garbage door 2. If I switch, it must be to the prize door, and I win.

4) I pick garbage door 2, Monty then has no choice but to show me garbage door 1. If I switch, it must be to the prize door, and I win.

Am I missing something here? As far as I can see there are only these four scenarios, and half are for switching and half against. That looks like 50:50 on whether to switch or not.

What is perhaps causing the confusion is that when people think about this they either conflate scenarios 1) and 2) into only one scenario-- "Monty opens a garbage door", or assume that in scenarios 3) and 4) Monty actually has some kind of choice, which he does not.

To make it more interesting, if you do the same exercise with 4 doors, you get 9 possible scenarios, 6 of which with a 50:50 chance of you winning if you switch, and 3 of which switching always makes you lose (the three where you picked right the first time). With four doors, you are definately better off *not* to switch after Monty shows you a loser, and how's that for counterintuitive?

oops, I stepped away from the computer, took maybe two steps, and realized that I was dead wrong.

You DO in fact have to conflate scenarios 1) and 2). The fact of the matter is that I pick the right door only 1/3 of the time, even if Monty can play it out two different ways. I pick the wrong door the other two times, and if I've picked the wrong door, switching always lets me win.

Similarly with 4 doors-- don't switch and you win 1 out of 4; switch and you win 1.5 out of 4.

I stand corrected.

I had to write a Perl program to prove it to myself:

http://ungwe.org/blog/2003/04/08/12:02/

(it's at the bottom of that page).

Yes, it works out as you say, that changing doors gives you an overall 66% chance of winning, but it's too much for my brain to see how it works.

It's weird, as a programmer I instinctively wanted to make the bit where Monty opens a wrong door not happen if I wasn't going to switch, because it can't make a difference, but I had my program do it anyway just to be sure...

Rob Bray's two posts finally made this click for me. Like Geoff, I had to write a script to back up my newfound insight with cold, hard numbers. Here's the fruit of that work:

http://efficacy.net/experiments/deal/

Basically, the way I understand it now is this: the player has a 1 in 3 chance of picking the winning door and a 2 in 3 chance of picking the losing door in the first step of the game. Once a losing door is eliminated, you either stay with your original odds by staying with your original pick, or reverse the odds by choosing the other door. Strange, but true!

I have it now! After looking at Rob's way of listing all the possibilities, it becomes easy...

If you switch after Monty eliminates one wrong door, then you always swap your choice (if your first choice would have won, you change to a loosing one, and vice verse). And since there are initially two wrong doors to choose from your initial choice is more likely to be wrong, so switching reverses those odds and makes you more likely to win.

Haha!

Okay, I want to believe, especially with all the nice programs apparently showing the TRUTH. However, I came up with the following simple diagram. Please show me where I went wrong.

Door 1 is always the winner (as I see it, having a random door be the winner means just increasing the number of cases by the number of doors with no appreciable effect on the odds)

Key: CASE = Reference Case #
1, 2, 3: Door #'s
I = Initial Pick
F = Final Pick
M = Monty's Choice
X = Unchosen Door
IR = Initial Result
FR = Final Result

CASE 1 2 3 IR FR
---- --- --- --- ---- ----
1 I/F M X WIN WIN
2 I/F X M WIN WIN
3 I M F WIN LOSE
4 I F M WIN LOSE
5 M I/F X LOSE LOSE
6 X I/F M LOSE LOSE
7 M I F LOSE LOSE
8 F I M LOSE WIN
9 M X I/F LOSE LOSE
10 X M I/F LOSE LOSE
11 M F I LOSE LOSE
12 F M I LOSE WIN

Overall, I win 33% (4 of 12) of the time with my initial pick, and 33% (4 of 12) of the time with my final pick. However, 4 of these cases are not possible: 5, 7, 9, 11 (Monty can't open the winning door). Initial pick wins 50% of the time (4 of 8) and final pick wins 50% of the time (4 of 8). If I keep my initial pick, I win twice and I lose twice (50/50). If I don't keep my initial pick, I win twice and I lose twice (50/50). So I don't see any advantage to changing.

Of course, I believe I am missing something...I just don't know what.

Let's say that Monty presented the game is this way:

There are 10 doors.
Pick one, any one, it doesn't matter.

You now have two choices:

Choice A) Open the original picked door.

Choice B) Open all 9 remaining doors.

What would you choose? A or B?

Choice B gives you a 9 times greater chance of hitting a prized door.

Now suppose Monty presents the game this way:

There are 10 doors.
Pick one, any one, it doesn't matter.

You now have two choices:

Choice A) Open the original picked door.

Choice B) Open all 9 remaining doors, BUT, Monty will open them in stages so that he opens 8 first and then the 9th. In effect you still get to open all 9 remaining doors, but Monty pauses for 10 seconds before opening the 9th door to make it look like a two step process instead of a one step process.

What would you choose? A or B?

Choice B gives you a 9 times greater chance of hitting a prized door.

Now suppose Monty presents the game this way:

There are 10 doors.
Pick one, any one, it doesn't matter.

You now have two choices, BUT YOU DON'T TELL MONTY YOUR CHOICE YET:

Choice A) Open the original picked door.

Choice B) Open all 9 remaining doors, BUT, Monty will open them in stages so that he opens 8 first and then the 9th. DURING THE PAUSE BEFORE THE 9TH DOOR, MONTY ASKS YOU FOR YOUR CHOICE. In effect you still get to open all 9 remaining doors, but Monty does it in such a way, by asking for your choice during the pause, as to give the illusion that the 9th door is seperated (isolated?) from the group of 9.

What would you choose? A or B?

Choice B gives you a 9 times greater chance of hitting a prized door.

Even though it appears that you were choosing between 2 doors at the end, you were in fact choosing between opening 1 door or opening 9 doors.

The principle is the same when you play with only 3 doors to begin with. You are choosing between opening 1 door or opening 2 doors.

That Monty fellow is one smooth slickster. I happen to have an opening in sales at my used car lot. I wonder if he'd be interested . . .

Dr Boom, sorry, I don't buy this explanation. My initial reaction was to reject it. Then I thought maybe I was wrong to reject it. Now I'm back to rejection.

The examples don't follow the rules of the original problem. None of the examples addresses the fact that Monty can't open the winning door and that you get to decide whether to switch AFTER all the other doors are open, choosing between two unopened doors. They are two separate events.

Reverse your example(s):
You have ten doors. You pick one. Monty then says you can...
(A) stay with the one you picked, plus any other 8 doors
(B) or you can pick another single door.

Monty then opens 8 of the 9 "Choice A" doors, which are guaranteed to not be winners. Which is the better choice. Of course (A) is in this setting because you have a 90% chance of being right.

However, this is not the situation under question.

The doors are opened BEFORE you get to decide whether to switch or not. So, you pick one door in 10. You have a 10% chance at this point. Monty opens 8 of the 9 remaining doors which are non-winning doors, then asks if you want to switch. Are you saying I have a 90% chance if I switch at this point?? If so, why do you assign the Monty-opened doors to the unchosen-door group rather than to the originally-chosen-door group?

It seems to me that the chances are 50-50, not 90-10, because the choice is 1 in 2, not 1 or 9 in 10.

Still want to believe, but finding it harder.

Well, I'm much closer to being a believer now even if it's all on faith. Wrote my own program in VB with nice output to 10000 rows of an Excel spreadsheet. Visually it's convincing, but I still don't understand it (despite the numerous explanations, including Dr Boom's). It may not have changed my world, but it's certainly giving me a headache. =^P

TO: Dean
RE: Proof Positive

"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." -- Dean

The most simple-minded way of proving a systems effectiveness is the ability to win money based on it.

Regards,

Chuck(le)

Last post on this...promise...

I tweaked my Excel VB program, allowing up to 100 doors and over 10000 repetitions, with Monty opening all but two of the doors in order (or randomly, doesn't make a difference), skipping the Winning Door and/or my initial pick. As stated before by others, switching doors at the end makes all the difference (especially with a large number of doors), essentially inverting the odds.

Very disturbing (as was Dean's point). Thanks to everyone for their patience as I beat this dead horse.

"Door 1 is always the winner (as I see it, having a random door be the winner means just increasing the number of cases by the number of doors with no appreciable effect on the odds)"

This is where you first went wrong. Not because it's inherently flawed, but because it is contributing to your confusion about the effects of Monty's door opening.

Try approaching the problem from a different angle:

At the outset, which door has the prize is completely random. To reduce the number of cases we need to examine, let's assume that you always choose Door A initially.

There is a 1/3 chance that Door A is the winner. Monty opens a random losing door. You will win if you keep your initial pick, and lose if you switch.

There is a 1/3 chance that Door B is the winner. Monty opens Door C. You will lose if you keep your initial pick, and win if you switch.

There is a 1/3 chance that Door C is the winner. Monty opens Door B. You will lose if you keep your initial pick, and win if you switch.

If you consider the cases where you choose Doors B or C first, you'll find the same results.

There is a 1/3 chance you will win by sticking with your initial door. There is a 2/3 chance that you will win by switching to whichever door that Monty has left for you.

If you make a new and completely random pick after Monty opens a door, you will indeed have 50/50 chance of winning. But that doesn't mean that first door has a 50% chance of having the prize at that time - we already know it doesn't. Even if we knew for a fact that the first door you chose was NEVER the winner, a random choice would still have a 50/50 chance. There was a 1/3 chance your first choise was the winner when the game began, and nothing has happened to change that.

To summarize:

Keeping your original door: 1/3 chance of winning
Switching to other door: 2/3 chance of winning
Picking a random door after Monty eliminates one: 1/2 chance of winning

I'm really surpr the that this puzzles so many bright people...

That's supposed to be "I'm really surprised that this puzzles so many bright people". Oh, and "choice", not "choise". Arrgh.

All the discussion so far misses the fact that the puzzle presents two independent events but asks you to treat them as related. They are not.

I just tried a number of iterations of this puzzle with my two sons, using 3X5 cards. They could not see which card was marked with a dollar sign and which two were marked with a "G" (for goat, one of Monty's favorite booby prizes). I could see, though.

So when one or the other selected a card, I showed a goat card and gave son the chance to switch, or not.

Result: whether they switched or kept, they won half the time.

The fallacy of the solution presented is treating the revealed door as bearing on the problem. It is actually irrelevant.

Consider: Suppose there are two boxes to begin with, one with the money. Obviously, your odds are 50-50 of choosing the money. You choose and then Monty tells you he has another, third box, which the pretty girl wheels onto the stage and opens to reveal a bag of hay.

Now Monty says that you can choose the other box that you did not choose to begin with. But there's no reason to, mathematically.

The revealed door is always a losing door, no matter what you choose. That means that the choosing problem always begins anew at that point. Hence:

Event A: Choose door # 1, 2, or 3. A losing door (say No. 3) is revealed and thus removed from consideration.

Event B: Choose door # 1 or 2. This is an independent event from Event A.

Look at it another way, which makes it obvious that the opened door is a red herring, just an irrelevancy. Suppose there are the three doors, but without you choosing, Monty opens one, behind which is a goat. Then Monty says that behind one of the other two doors is a goat and behind the other the money. Now you can choose. Odds? 50-50, of course.

It holds true for any number of doors, say 500 doors. You choose one and then Monty takes away 498 of the other doors - but all are losing doors. At the end, you are always left with two doors, one a winner and one a loser, and told you can make a new choice. There's no difference between that long process and just starting with two doors to begin with.

The two times you choose a door are independent events. The odds of the first choice have no bearing on the second choice.

Thundar,

I believe you misthought about your cases: specifically are they all equally probable?
If you drop your 4 cases, then you should always pick door 1 -- you have a 50% chance with door 1, and a 25% chance with door 2 or 3.

Perhaps you can't eliminate the cases you think you can.

Perhaps this will help:

Suppose Monty Chooses Door # 1.

Case # -- C
Prize Door -- $
Your choice -- U
Monty's Choice -- M
Win if you switch? -- ?

C - $ - U - M - ?
1 - 1 - 1 - 2 - N
2 - 1 - 1 - 3 - N
3 - 1 - 2 - 3 - Y
4 - 1 - 3 - 2 - Y

So the odds are 50/50?

No -- Case 1 and 2 together are as likely as either case 3 or case 4.

Properly, the above table should look like:

Case # -- C
Prize Door -- $
Your choice -- U
Monty's Choice -- M
Win if you switch? -- ?
Probability of occurring -- P

C - $ - U - M - ? - P
1 - 1 - 1 - 2 - N - 1/6
2 - 1 - 1 - 3 - N - 1/6
3 - 1 - 2 - 3 - Y - 1/3
4 - 1 - 3 - 2 - Y - 1/3

hope it helps.

Aha! I asked earlier how bright people could fail to grasp this problem, and Donald Sensing has provided an excellent example.

His analysis of the problem is absolutely incorrect, of course, but it does give insight into where he (and others) have gone wrong. At least in his case, the root of the problem is a poor understanding of what is an "independent event".

"All the discussion so far misses the fact that the puzzle presents two independent events but asks you to treat them as related. They are not."

"The revealed door is always a losing door, no matter what you choose. That means that the choosing problem always begins anew at that point."

This is completely wrong.

You seem to think that your making a choice is an "event". It is not. An event, in this context, is the determination of a random occurrence.

Picking an unrevealed door is not an event. Changing your mind about which door you want is not an event. Most importantly, Monty giving you a hint by revealing a losing door is not an event - it is information about an event that has already occurred.

In this problem, the only independent event is the original placement of the winning prize. That only happens once. There are no other random occurrences. Even if you want to consider the determination of whether each door is a winner as an event (which is a dubious expression of the problem, since there is only 1 winning prize which is placed at random behind 1 of the 3 doors), the 3 "events" would clearly be completely conditional on each other.

"Event B: Choose door # 1 or 2. This is an independent event from Event A."

If the prize location were re-randomized at this point, you would be correct. But it is not.

When you initially choose a door, there is a 1/3 chance that you have chosen the correct door. Since the prize is never moved, nothing that happen afterwards changes that initial probability.

If you initially choose Door A, there is a 1/3 chance you are right, and a 2/3 chance that the winning door is B or C. Thus, once Monty reveals 1 of the other 2 doors as a loser, there is a 2/3 chance that you will win if you trade door A for the remaining door.

If you initially choose Door B, there is a 1/3 chance you are right, and a 2/3 chance that the winning door is A or C. Thus, once Monty reveals 1 of the other 2 doors as a loser, there is a 2/3 chance that you will win if you trade door B for the remaining door.

If you initially choose Door C, there is a 1/3 chance you are right, and a 2/3 chance that the winning door is A or B. Thus, once Monty reveals 1 of the other 2 doors as a loser, there is a 2/3 chance that you will win if you trade door C for the remaining door.

Q.E.D.

JB's explanation is the best so far (the 22-April, 2:06pm one, that is), IMHO. And I feel better knowing Mr. Sensing's thoughts pretty much mirrored my own initial thoughts.

Well, the quarter is in the payphone, but I don't get the dial tone yet.

Why is it that the 1/3 probability from the revealed door is added to the unchosen door rather than distributed evenly to the two remaining unpoened doors?

The real question in this scenario is one of two: “Will I get the money?” or “What are my chances of getting the money?” The answer to “what are my chances of getting the money” will have to do with statistical odds, and the stats tell me I should probably switch doors. However, at any given moment in time (once my choice is already made) I will either have the money or I won’t. That is a truism, not statistical odds. So, if my goal is to insure I get the money, I cannot do that -- because I either will or I won’t have the money after my fateful choice -- no matter what my choice is. However, if my goal is to stack the odds against losing, I can do that. I can choose the other door, given the opportunity to change my original choice. However, once the final choice is made I will either be very glad I switched doors, or very sorry I didn’t stick with my original choice. All statistical odds aside, life still happens. From that perspective, every choice is 50-50. Stats never tell the story for the individual -- only for the group. Think of it this way: given the fact that you are a contestant, regardless of how many various choices you may be given by Monty (and regardless of how many times he allows you to change your choice) you can be sure you either will or will not be leaving with the desired prize in your hand. 50-50

I bred cats for six years, starting with four cats who each had a Siamese colored sire. When bred together, according to the stats, I should have gotten one Siamese colored kitten out of every four kittens born from these breedings. Forty-eight kittens later, I didn’t have any Siamese colored kittens. I brought in a new cat, whose lineage had no Siamese colored kittens for multiple generations. The stats led me to believe I would have a snowballs chance on a hot summer day. In fact, I had given up hope. But, in the first litter out of the new cat, I got a Siamese colored kitten. All I can say is that when waiting for each litter to be born, I could be sure I either would or would not have a Siamese colored kitten in the litter. Fifty-one kittens weren't Siamese colored, and one was.

Monty would have just loved me ….

This way of looking at it seems the clearest to me.

You know that if you pick a wrong door to start, and you ALWAYS switch, then you will ALWAYS end up with the correct door at the end.

So, you have a 66% chance of picking the wrong door on the first guess and since you know that this will get you a win if you ALWAYS switch, then you should hope to pick the wrong door from the start, then ALWAYS switch.

Contrast that to the case where you NEVER switch. It is pretty clear that you have a 33% chance of being correct.

Given the choice of always switching or never switching, you double your chances by always switching.

Does anyone find fault in that logic?

I'm not sure that this will help Jim get the quarter into the phone, but let me take a crack at it.

When you first make your pick, you have a 1/3 chance of getting it right, a 2/3 chance of getting it wrong. When Monty opens a losing door, it doesn't do anything to increase the chance there's a prize behind your door - if you were wrong before you're still wrong now.

So, if there is a 1/3 chance you have the right door then there has to be a 2/3 chance that the only other remaining door is correct (given that the odds have to add up to 100%). 1/3 chance to win if you stay, a 2/3 chance to win if you switch.

Dean, very interesting post by the way. Bet you never thought you'd be getting comments on it more than a year later.

Gee, I come over here, vaguely recognize I've seen this post before, and find I've commented. Weird sensation, sorta like answering the phone and the caller is you.

Jeffrey Lewis:
OUCH! I should have written "The laws governing me and my fellow USAmericans are not made by the British Parliment . . .", to make it clear what I meant.

Donald Sensing:
You did this how many times? And the written records showed what results?

I am, btw, astounded that so many people kept stubbornly insisting Dean was wrong, even after the multiple clear explanations of why the person who switches will win two times out of three. And you keep insisting, without proposing any procedure to settle the matter. If nothing else, don't you realize how silly it looks to say, "I know I'm right, but I won't put it to the test?"

As I said in the main article, the most astounding factor in this entire riddle--and the reason it fascinates me so--is that (a) most people, including very intelligent people like Reverend Sensing, get it wrong, and (b) there is a certain core segment of people who will not only refuse to believe the right answer, but will refuse to look at evidence that challenges what they "know" to be right.

The riddle therefore ends up being a lesson in humility--if you're one of the majority who gets it wrong--and a lesson in the limitations of humanity itself. (Lord, what fools these mortals be. Uh-oh, that includes me? Whoops, guess it does!)

I have an open offer of money to anyone who wants to prove me wrong. Here's the offer again for anyone who wants it:

We'll set up the game using a mutually-trusted third party to keep us both honest and legit. You just have to agree to never switch doors from your original guess. I'll pay you $100 every time you win. You pay me $75 every time you lose.

I will boost those quantities to $1000 and $750 if you like.

Or, if the stakes are too big for you and you'll feel guilty, we can make it $1.00 vs. 75 cents. I don't really care.

My only requirement: we play a minimum of 20 times. Then we play as long as you want after that, or until I'm out of money.

How long do you want to play? Come on guys, I'm a sucker here! You'll make a killing! When do you want to play? We just agree on a third party that we both trust to arbitrate.

I'll just warn you in advance: I will not cheat you, I will not deceive you. I will simply keep taking your money from you until you concede defeat. ;-)

Ah, Dean, you give yourself away! Now you want to cheat! Read the problem again - you don't get 20 times! You get one time. That's all.

Believe me, I really have done the math and worked the problem. I readily concede that out of 20 attempts, mathematically the switcher will win about 13 or so.

But probabilities are not certainties. To answer Mr. St. Onge's question, I played the game with my sons 18 times - easily divisible by three - and they won 10 times and lost eight. That's beating the odds. The next day I played the game by myself (it is possible to do so fairly) 54 times, and switching won 32 times, not 36 as math predicts.

In that series I actually had eight straight losses by switching, followed obviously by an over-sample string of wins. Such is real life. I presume that another set of 54 tries could result in, say, 42 wins rather than the mathematically correct 36.

However, with a universe of only one event, which is what the problem actually requires, the probability chain simply breaks down. The fact that you have a 1/3 chance of correct selection on the first choice is a chimera. When the extraneous losing card is removed, the problem starts anew. You have one winner, one loser, and can freely choose either one. 50-50.

Dean, my friend, could it be you betray the weakness of your position by stacking the bet's ratio so heavily in your favor? Why did you stack the bet 3/4 to 1 in your favor on a putative 2/3 to 1 advantage? Hedging, perhaps?

How 'bout this bet: you pay me a dollar every time I win, I pay you 33 cents every time I lose. That's just as fair as the wagers you propose. (Work it out.)

But we don't play 20 times. We play once. That's what the problem actually demands.

As for third-party, I would be quite comfortable with Joe Katzman, Geitner Simmons, Steven Den Beste, Austin Bay, Bill Hobbs.

Finally, I do not actually make money wagers (never have), so I propose we agree that the loser contribute volunteer hours to a worthy local cause - you donate 10 hours if you lose, I donate 3.3 (call it 3.5) if I lose, to be served before the end of May.

Remember - one shot. Deal?

Dean, rereading my last comment, I wish I had rephrased my opening sentences. I do not mean by any means to imply you are a cheater, and I apologize for using that terminology.

Donald, do you see the knots you're tying yourself into? "I readily concede that out of 20 attempts, mathematically the switcher will win about 13 or so. ... When the extraneous losing card is removed, the problem starts anew. You have one winner, one loser, and can freely choose either one. 50-50." You don't see any contradiction between those two statements? The odds are 50-50 ... yet switching still wins two times out of three?

Try this: Enumerate all the possible cases, and number them: 1.1, 1.2, 1.3, 2.1, etc. The first number is the door the prize is behind, the second is the door number you chose. Nine cases, each equally probable (assuming you're not predisposed to pick a particular number).

In three of the nine cases -- 1.1, 2.2, and 3.3 -- you got the right choice the first time, Monty opens one of the other doors, and the winning move is to stick with your original choice.

In the other six cases -- 1.2, 1.3, 2.1, 2.3, 3.1, and 3.2 -- you got the wrong choice the first time, Monty opens the remaining door (not the one you picked, and not the one with the prize) and the winning move is to switch.

Nine cases, each equally probable. In three you win by standing pat, in six you win by switching. Odds favor switching by 2:1.

Or try this: You are shown three dorrs, and you choose one. Monty then asks you, "What are the odds that you chose the right door the first time?" Answer: one in three, of course.

Monty then opens one of the other doors, with nothing behind it. He then asks you exactly the same question again: "What are the odds that you chose the right door the first time?"

Same answer. The probability that you guessed right the first time was, is, and forever will be the same: one in three. Odds do not change retroactively.

But the odds that you guessed right the first time are the same as the odds that you have the right answer now. In effect, that is exactly what Monty is asking you if he asks whether you want to switch: what are the odds that you got it right the first time?

The odds do not get "reset" when Monty opens one of the doors; that's the real chimera. He has showed you that one of the doors you didn't pick has nothing behind it; but you already knew that. He hasn't given you any new information, and his actions have not changed the probabilities. You either got it right the first time (1 in 3) or you didn't (2 in 3).

Any of this making sense?

Heh. Methinks the good Reverent has figured it out. The math says you should switch.

I read this a time ago and wanted a SIMPLE explanation of the answer. Here is attempt #1:

There are 36 permutations for this situation:
  3 doors the money can be behind
x 3 choices for you
x 2 doors Monty can eliminate
x 2 doors left to choose from
= 36 possible scenarios

Note: the REAL chimera here is that Monty always has two doors to choose from, but MUCH of the time (2 out of 3 actually, whenever you pick a loser) he is NOT ABLE to choose one because it has money behind it. THIS is the "information" that he provides you with his choice, thus upping your odds.

There are 3 identical sets of 12 permutations, 1 set for each door the money is behind. Therefore:

Door $ is Behind 1 1 1 1 1 1 1 1 1 1 1 1
Your Choice 1 1 1 1 2 2 2 2 3 3 3 3
Monty's Choice 2 3 2 3 1 3 1 3 1 2 1 2
Monthy's Action 2 3 2 3 3 3 3 3 2 2 2 2
Switch? Y Y N N Y Y N N Y Y N N
Win/Lose N N Y Y Y Y N N Y Y N N

You win 4 times for every 2 times you lose when you switch.

You win 2 times for every 4 times you lose when you stand pat.

Solution #2: My wife and I just had dinner with 2 engineer friends. The first one very simply said:

"There's a 1 in 3 chance of the door you picked being the money and a 2 in 3 chance that the money is behind one of the other doors.

Eliminate one of the other doors and you still have a 1 in 3 chance that the money is behind your door but there is now a 2 in 3 chance that the money is behind the remaining door."

Master of simplicity.

The math says to switch, every time, whether you play once or a million times, your odds are best that way.

Thank you for the problem!

George

Hi there,

I just wanted to thank you for the post and the discussion. I'd heard this puzzle years ago, but never really understood the counter-intuitive solution until today. Thanks! Rob Bray's wrong-but-quickly-corrected explanation was what made the penny drop for me! Ta! :-)

Here's my version of the solution based on Rob Bray's. I personally find this explanation easier to understand as it makes a distinction between the two "goats" (booby prizes) and it accurately describes the odds of the situation at the moment that Monty asks the question "Wanna switch?", which I think is a more useful and natural way of thinking about it (I've never been comfortable with advanced probability problems). Perhaps it will help Donald Sensing see the light. Perhaps not! :-) Anyway, here goes:

As everyone seems to agree, when you pick one of the three doors, you're faced with the odds that you only have a 1/3 chance that you're right, but a 2/3 chance that you're wrong. Then Monty says he's going to show you one of the goats and let you either switch your choice to the other remaining closed door or keep your original choice. At the moment that Monty reveals one of the goats to you and you must finalise your decision, there are only three possibilities (because you only had three choices in the first place):

A. You originally chose Goat1, Monty reveals Goat2 - if you switch you'll WIN!
B. You originally chose Goat2, Monty reveals Goat1 - if you switch you'll WIN!
C. You originally chose correctly, Monty reveals a goat - if you switch you'll LOSE!

Monty won't reveal the prize, so the above are ALL of the possibilities. What initially threw me was not realising that the third possibility (C - Monty reveals one of the goats) counts as only ONE possibility not two, even though Monty has a choice of two goats to reveal. Monty can only reveal ONE goat - the fact that Monty can choose which goat to show you doesn't change anything about YOUR choices. So switching your choice gives you twice the chance of being right.

The situation is even clearer if there were five doors. The possibilities are:

A. You originally chose Goat1, Monty reveals Goat2&3&4 - if you switch you'll WIN!
B. You originally chose Goat2, Monty reveals Goat1&3&4 - if you switch you'll WIN!
C. You originally chose Goat3, Monty reveals Goat1&2&4 - if you switch you'll WIN!
D. You originally chose Goat4, Monty reveals Goat1&2&3 - if you switch you'll WIN!
E. You originally chose correctly, Monty reveals three goats - if you switch you'll LOSE!

Switching becomes pretty good odds. Or have I misunderstood?

IMHO, Donald Sensing's suggestion that any practical demonstration of this puzzle be restricted to only one round is not exactly a fair or scientific approach - how can a one-round experiment prove the ODDS of the situation? No matter what the outcome, either party could claim victory while the other cries "fluke!". But after a hundred rounds, the truth of the odds would become more apparent. :-)

Thanks again!

Skev

I'm surprised this confused so many bright people too, especially after it got explained to them. I drew the experiment on paper and it is bright as day.

Scenario: Door1=Prize, Door2=Booby, Door3=Booby

Case 1: don't switch doors
Trial 1 - pick door1 => Win
Trial 2 - pick door2 => Loose
Trial 3 - pick door3 => Loose
Result=Win 1/3 times

Case 2: switch doors
Trial 1 - pick door1 => Loose
Trial 2 - pick door2 => Win
Trial 3 - pick door3 => Win
Result=Win 2/3 times

Why? Same reasons everybody else has given. Chances of picking the right door off the get-go is 1/3. Chances of picking a looser is 2/3. Essentially by switching after a looser door is opened, you're betting that you origianally picked a looser, so your odds are 2/3.

You can apply the same logic to the 1000 doors. Chances of picking correctly the first time is 1/1000. Chances of picking a looser is 999/1000. So if you switch your choice after 998 loosers are revealed, you're odds of winning is 99.9%. CONCLUSION: If we play Let's Make a Deal we should deal for more doors.

Ok, so what everyone is saying here is that it is actually better to start out with three doors, and after having made your choice, have one eliminated by Monty; than to just simply start out with two doors from which you can choose only one. This means that you will have better odds of 2/3 (66%)when there are three doors than the 1/2(50%) odds. So in theory the more doors the higher the percentage of winning.

NO NO NO!
Everyone must realize that the first choice is absolutely meaningless!
If you pick one of three doors, and then one Booby is opened, That door is now out of the question.
What remains are two doors, one with a prize, one without. This is an entirely new game, completely unaffected by the first. Your odds in the first situation can be thrown out because you no longer have the third door to deal with.
New Game: Two equal doors. The only difference is that one had been chosen before, but this does not affect it's contents.

What I think is going on here is that the question has been poorly presented. If the choice is WHETHER OR NOT TO HAVE THE FIRST DOOR OPENED, with a double-or-nothing element, THEN the choice is better to open. After that, the odds of the two doors are reset to 50/50.

FANTASTIC. It is fascinating how both sides of the argument appear correct. My first impulse was to dismiss this teaser as one of semantics or numerical confusion. It seemed so obvious. I will sleep on this one. I like it a lot. - wayne

FANTASTIC. It is fascinating how both sides of the argument appear correct. My first impulse was to dismiss this teaser as one of semantics or numerical confusion. It seemed so obvious. I will sleep on this one. I like it a lot. - wayne

What a wonderful puzzle! It appears obvious that the odds must be 50/50. I did a 180. The odds favor switching. Math and logic support this conclusion.
The illusion is that Monty appears to reduce one of three whereas he is actually reduces one of two on one side of the ledger.

Humbled in Canada

Ps: Is there 2nd puzzle that changed North America? Hungry for more.

I can't believe some people are stupid enough to think that chances will change when Monty opens a door. Monty can always open a door, and always will, this doesnt' change the probability of you first choice. The brainteaser is a bit dull, and not too deep.

i agree that this teaser is dull. if there were 2 players with doors A, B, and C, and player 1 chose A, and Player 2 chose B, and Monty opened C, should they switch? No, they both are stuck at a 1 in 3 chance. To say that Player 1 should switch to B or Player 2 should switch to A is pointless. To say that the teaser's solution is true is to say that Player B for example had the right choice the whole time.

Actually, I have a 2nd brainteaser that will certainly change MY world if I can ever figure it out. I can't remember where I first heard this puzzle and I'm still not convinced of the answer, so please read on and tell me what you think.

Let's say you're attending a convention of people who have exactly two children. You're talking to someone else at the convention and she mentions that one of her children is a girl.

The question is: What are the odds that her other child is a boy?

I'll follow Dean's example and give you some lines to think about it.

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When I first heard this puzzle, I was sure that the odds were 50/50 that the other child was a boy. After all, if I asked you what are the odds that some randomly-selected kid was a boy, you'd say 50/50 and you'd be right (ignoring slight differences in biological likelihoods and life expectencies). However, I've been told (and I now mostly believe) that there are actually 2 chances in 3 that the other child is a boy.

The explanation is that there are 4 equally likely combinations for people with two children. Listing the two children in birth order, the possibilities are girl-girl, girl-boy, boy-girl and boy-boy. In learning that one of the children is a girl, the boy-boy possibility is ruled out and you're left with girl-boy, boy-girl and girl-girl. So, given at least one girl, the chances are 2 in 3 that her other child is a boy.

According to the argument I heard, the odds would have been different if the parent had said that her OLDEST child was a girl. This information would have ruled out the boy-boy and boy-girl possibilities and the odds would then have been 50/50 that the other child was a boy.

Obviously, similar logic would apply if she had said that her YOUNGEST child was a girl. Again the odds of a boy would be 50/50.

So, here's where I get stuck.

If you accept everything that's been said so far... this women says she has a daughter and you think to yourself, "Aha! That means there are 2 chances in 3 that her other child is a boy!" Then, too foolish to leave well enough alone, you ask, "Oh really? Is she your oldest or your youngest child?"

The woman answers and now the odds seemingly change to 50/50, regardless of her answer!

I feel certain I've gone wrong somewhere in my thinking, but where? Please help!

Tom Z said:

* Actually, I have a 2nd brainteaser that will
* certainly change MY world if I can ever figure it
* out. I can't remember where I first heard this
* puzzle and I'm still not convinced of the answer,
* so please read on and tell me what you think.

* Let's say you're attending a convention of people
* who have exactly two children. You're talking to
* someone else at the convention and she mentions
* that one of her children is a girl.

* The question is: What are the odds that her other
* child is a boy?

50-50. I THINK I see the mistake you're making (or else I'm making a mistake!).

* When I first heard this puzzle, I was sure that the
* odds were 50/50 that the other child was a boy.
* After all, if I asked you what are the odds that
* some randomly-selected kid was a boy, you'd say
* 50/50 and you'd be right (ignoring slight
* differences in biological likelihoods and life
* expectencies).

I agree with this.

* However, I've been told (and I now
* mostly believe) that there are actually 2 chances
* in 3 that the other child is a boy.

* The explanation is that there are 4 equally likely
* combinations for people with two children. Listing
* the two children in birth order, the possibilities
* are girl-girl, girl-boy, boy-girl and boy-boy.

Here is where I think you're making your mistake. Why did you list the possibilities in birth order? Their birth order is not relevant to the question at stake as their order of birth was not mentioned. She didn't say her eldest was a girl, only that one of them was a girl. So "boy-girl" and "girl-boy" are the same possibility. So the possibilities when you first meet her are that her children are:

1) girl-girl
2) boy-girl
3) boy-boy

If you had tried to guess what the gender of both her children were at this point, you'd have a 1/3 chance of getting it right. Your chances of guessing correctly that at least one or more child is male is 50/50. Your chances of guessing correctly that ONLY ONE child is male is 1/3. Yes?

However, since she then gives you the information that one of them is a girl (ie, that option 3 is not possible), then at the moment you make your guess about the remaining child the options are either:

1) girl (plus the given girl)
2) boy ( plus the given girl)

So you have a 50-50 chance of getting your guess right. The question was never "what are the genders of my children?" (a question that was never actually asked) it's "what is the gender of my other child?". Which amounts to the same as "what is the gender of my single child?", since the gender of one child has no impact on the gender of the other child (for the purposes of this problem).

* In
* learning that one of the children is a girl, the
* boy-boy possibility is ruled out and you're left
* with girl-boy, boy-girl and girl-girl.

No, you're stating ONE possibility twice, giving you a skewed result. Order of birth is not relevant to the problem.

Have you simulated this problem with software? I started to make an attempt, but it became so obvious that the random gender outcome of one child had no impact on the random gender outcome of the other that I didn't know how to write the program without effectively hard-coding my own assumptions into the program. This probably says more about my programming skills than anything else.

Think about it this way: You've got two coins (male on one side, female on the other). You toss both coins and record the results. You tell a guesser the result of one of the coin tosses and ask them to guess the other one. See it? No matter what you tell the guesser about the results of one of the coin tosses, it has no impact on their ability to guess the other coin.

* So, given at
* least one girl, the chances are 2 in 3 that her
* other child is a boy.

I don't see it that way. :-)

* According to the argument I heard, the odds would
* have been different if the parent had said that her
* OLDEST child was a girl. This information would
* have ruled out the boy-boy and boy-girl
* possibilities and the odds would then have been
* 50/50 that the other child was a boy.

Although I agree with the odds of 50/50 in this new problem, the order of birth is still not relevant, as the fact that the oldest child is female has no impact on the gender of the younger child. Or to put it your way, the possibilities are:

a) girl then boy
b) boy then girl
c) girl then girl
d) boy then boy

If she says her eldest is a girl, then options b and d are gone, giving you a 50/50 chance of guessing boy correctly. But ultimately no connection is made between the gender of one child affecting the gender of the other - so it's always 50/50 in both these problems.

It's quite different to the two-goats-one-prize problem (which has three possibilities).

* Obviously, similar logic would apply if she had
* said that her YOUNGEST child was a girl. Again the
* odds of a boy would be 50/50.

* So, here's where I get stuck.

* If you accept everything that's been said so far...

I don't, of course... :-)

* this women says she has a daughter and you think to
* yourself, "Aha! That means there are 2 chances in 3
* that her other child is a boy!" Then, too foolish
* to leave well enough alone, you ask, "Oh really? Is
* she your oldest or your youngest child?"

* The woman answers and now the odds seemingly change
* to 50/50, regardless of her answer!

But as I think I've explained, the odds were always 50/50 in the first place.

* I feel certain I've gone wrong somewhere in my
* thinking, but where? Please help!

I hope I've explained it clearly, though I too may have made a mistake! Thanks for the problem though - it wasted my time (ahem) exercised my mind for a few minutes! :-) I'd LOVE to see an explanation different to mine!

All the best,

Skevos Mavros
mavart@mavart.com
http://www.mavart.com

Skevos, thank you for your response! You may be right that it's the first part of the puzzle where I've gone wrong, but I don't see it yet.

You said:
* So the possibilities when you first meet her
* are that her children are:
*
* 1) girl-girl
* 2) boy-girl
* 3) boy-boy
*
* If you had tried to guess what the gender of
* both her children were at this point, you'd
* have a 1/3 chance of getting it right.

I disagree.

At a convention of people who have exactly two children, half the people there have a boy and a girl, 25% have 2 boys and the remaining 25% have 2 girls.

* Your chances of guessing correctly that at
* least one or more child is male is 50/50.

Nope. Using your coin-toss analogy, flip two coins and 75% of the time you'll get at least one heads.

* Your chances of guessing correctly that
* ONLY ONE child is male is 1/3. Yes?

I don't think so. The chances of guessing that ONLY ONE child is a boy is actually 50/50. Back to your coin-toss analogy... if I flip two coins, what are the odds that I'll get exactly one heads?

Regards,
Tom

if you know one is a girl, and ask if it is the oldest or the youngest, chances will off course not change!!! you here again forget that there can be two girls, and then the parent can say either (oldest or youngest) and no additional info is given. it is different when they tell you 'oh, i have two girls, youngest and oldest have the same sex.' then you get more information and chance will change.

important thing when calculating a probability is getting your initial prerequisites right, then the probability will be constant, if no more (or irrevelevant, which is the case her, off course) information is given.

better take up a course of statistics, dear Tom Z

if you know one is a girl, and ask if it is the oldest or the youngest, chances will off course not change!!! you here again forget that there can be two girls, and then the parent can say either (oldest or youngest) and no additional info is given. it is different when they tell you 'oh, i have two girls, youngest and oldest have the same sex.' then you get more information and chance will change.

important thing when calculating a probability is getting your initial prerequisites right, then the probability will be constant, if no more (or irrevelevant, which is the case her, off course) information is given.

better take up a course of statistics, dear Tom Z

Amazing - it certainly appears that some fairly bright people here have lost sight of a fundamental law of probability! If you flip a coin 1,000 times, 1,000 times (no I am not being redundant) the chances of it landing exactly 500 times on heads, and 500 times on tails are astronomical. The law of probability says that out of the 1,000 times you flip it 1,000 times you will probably yield closer to a 40/60 ratio. That's pretty darn close to the two out of three times that changed Dean's world. Just because there are only two choices, doesn't mean it is really a 50/50 chance. Therefore, the correct answer is, sometimes you'll win, sometimes you'll lose.

Just wanted to drop by and wish you a Happy Holidays!

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greetings

Well thought out blog entry, though I'm not sure I entirely agree with the point being made! However saying that everyone has their own point of view and is entitled to it.

nice site and very informativ topics. worth tostay here. wisch you a nice day and health for the coming year.

best wisches
berta