Monty Hall Problem

[Gordon B]

If you have read the answer somewhere, please don't vote.

You are a contestant on the old game show "Let's Make a Deal" with Monty Hall.

Monty tells you that behind one of three doors is a Toyota Prius. A cd of Kenny G's greatest hits is behind each of the other two doors.

You choose door #1.
Monty, who knows the location of the car, opens the curtain on door #3. There's a Kenny G cd there.

Monty offers you the choice of switching from door #1 to door #2.

Do you switch?

[Monte Smith]

I don't get it. Aren't we supposed to want the Kenny G CDs?

[Brian Olewnick]

It is an excellent, counter-intuitive problem!

[Monte Smith]

I read the initial post poorly. I won't change my answer, but I should have paid closer attention. Maybe Brian knows why.

[bluenoter]

It doesn't matter- the odds are 50/50 either way.

I may be a birdbrain at this, but how could it be otherwise?

[John L]

Sorry, I answered before I saw the qualifying comments. If you really think about it, there is nothing counterintuitive going on.

[crawjo]

I am not great at these sorts of logic problems, but I am sure that the answer is 2/3rds for a very simple reason: the obvious answer is 50/50. But why would Gordon go through the trouble of starting a thread about a question for which the answer is obvious? So it has to be something other than what you would expect. And it can't be "There is not enough evidence to answer this question," because in his initial post Gordon instructs those who have seen the answer not to participate in the poll. So, there is an answer. By process of elimination, it has to be 2/3rds.

[Brian Olewnick]

Damned meta-problem solvers!

[walto]

Originally Posted by bluenoter

I may be a birdbrain at this, but how could it be otherwise?

It might be easier if you consider an example where you have to pick from 100 doors instead of 3. Let's say you pick door 11. You know it has only a 1 in 100 chance of winning. This time, Monty opens all the doors except yours and door 57 and, again, they're all losers. If he lets you take #57 rather than keep 11, I'll bet you'd do it. The principle is the same. Your door's odds haven't changed, he's just given you some info about a likely winner.

[Lenny D.Guitarist]

Originally Posted by Gordon B

If you have read the answer somewhere, please don't vote.

You are a contestant on the old game show "Let's Make a Deal" with Monty Hall.

Monty tells you that behind one of three doors is a Toyota Prius. A cd of Kenny G's greatest hits is behind each of the other two doors.

You choose door #1.
Monty, who knows the location of the car, opens the curtain on door #3. There's a Kenny G cd there.

Monty offers you the choice of switching from door #1 to door #2.

Do you switch?

Your odds are better if you switch. This is a well-known probability question that has been debated for many years, the most recent time being in Parade Magazine's "Ask Marilyn" column. Marilyn vos Savant, the Parade's weekly columnist, is reputed to have an IQ at or above 200. I was one of the readers who offered a somewhat cogent explanation of why this works.

Check out Marilyn's explanation(click on archives)
http://www.parade.com/current/columns/askmarilyn.html

[John L]

It is not so difficult to understand. Compare two strategies: one where you choose curtain 1 and stay with it, and another where you choose curtain 1 and then switch.

With the first strategy, you win only if it is behind curtain 1 (1/3 probability).

With the second strategy:

If it is behind curtain 1 (1/3 probability), you lose

If it is behind curtain 2 (1/3 probability), Monte Hall will show you curtain 3 and you will choose curtain 2 (You win)

If it is behind curtain 3 (1/3 probability), Monte Hall will show you curtain 2 and you will choose curtain 3 (You win).

Thus, with the second strategy, you only lose if it is behind curtain 1 (1/3 probability)

[Gordon B]

Originally Posted by John L

It is not so difficult to understand. Compare two strategies: one where you choose curtain 1 and stay with it, and another where you choose curtain 1 and then switch.

With the first strategy, you win only if it is behind curtain 1 (1/3 probability).

With the second strategy:

If it is behind curtain 1 (1/3 probability), you lose

If it is behind curtain 2 (1/3 probability), Monte Hall will show you curtain 3 and you will choose curtain 2 (You win)

If it is behind curtain 3 (1/3 probability), Monte Hall will show you curtain 2 and you will choose curtain 3 (You win).

Thus, with the second strategy, you only lose if it is behind curtain 1 (1/3 probability)

Good explanation, John. Marilyn Vos Savant posed this problem in her Parade column on 9.9.90.. After she published the answer a week later she got many angry and insulting letters, including some from professional mathemeticians and statisticians. They were wrong and she was right. Even the great Paul Erdos got it wrong when fellow mathemetician Andrew Vazsonyi posed the problem to him.

[Pete C]

What are my odds of getting lovely Carol Merrill?

[Monte Smith]

This problem is discussed in the excellent book The Curious Incident of the Dog in the Night Time by Mark Haddon. I read that and it should have disqualified me from voting; Brian O. read it as well and he did not vote (damn your ethics, Brian).

[Brian Olewnick]

Originally Posted by Monte Smith

This problem is discussed in the excellent book The Curious Incident of the Dog in the Night Time by Mark Haddon. I read that and it should have disqualified me from voting; Brian O. read it as well and he did not vote (damn your ethics, Brian).

Yeah, but I encountered it a long time back, maybe in a Hofstadter column for Scientific American?

It sorta reminds me of the omniscient alien problem. Know it? Not directly related, but....

There's this alien who has devoted countless millennia to the discreet study of humans and their psychology. He's gotten extremely good at it, enough so that if he observes a human presented with a given choice, he will always be able to predict that human's decision. 100%--he's never wrong.

So he calls you in one day and presents you with the following scenario. There are two boxes in front of you, one transparent, one opaque. In the transparent box, you see a $1000 bill. The alien says, "You have two choices. You can either take both boxes and their contents or you can take only the opaque one. However, I know what you're going to do. If I know that you're going to take both boxes, then the opaque box is empty. If I know that you're only going to take the opaque box, then that box contains $1,000,000." [The problem uses money for convenience--feel free to substitute any two things of vastly different value] The alien then announces that his work on Earth is done and instantly transports himself back to his home world, leaving the two boxes in front of you.

What do you do? (ie, what choice maximizes your chances for greatest gain?).

[jesus marion joseph]

My head hurts.

[Monte Smith]

Originally Posted by Brian Olewnick

What do you do?

Hit the bottle. Because at this point it becomes the Monte Alban problem.

[Brian Olewnick]

Just found my alien problem stated a bit more cogently (and, maybe, interestingly with the tag at the end)

Further discussion on it here:

http://forums.topcoder.com/?module=T...c=72&view=flat

*******************8

Someone posed this to me recently. I thought I'd post it here and see what people think because people have been discussing, e.g., the Monty Hall problem. This scenario isn't really a probability puzzle, however.

You're sitting in your living room, minding your own business, when there's a flash of light and suddenly you're in a room with an alien. The alien explains that he's been doing research (on the behavior of complex systems, say) and you've been one of his test subjects. He tells you that he's been monitoring you from afar since birth and making predictions about decisions that you would make. He has made tens of thousands of predictions about you, and every single one has been correct. He says that you've been a useful test subject, and he's brought you here both to reward you and to make a final prediction. After this, he will travel to a distant planet, never to return.

He shows you a transparent box with $1,000 in it. He then shows you an opaque box and pulls $1,000,000 from his pocket. He explains that you will have the option of either taking both boxes or taking only the opaque box. He will predict which option you will choose, taking into account what he is about to tell you. He says that if he predicts that you will take both boxes, he will put nothing in the opaque box, whereas if predicts that you will take only the opaque box, he will put $1,000,000 in the opaque box. Whether he is right or wrong, you get the contents of the box or boxes that you pick -- he assures you that the contents of the boxes will not be altered in response to your actual decision. He makes his prediction, turns his back to you, and either puts the money in the opaque box or back in his pocket. You have absolutely no way to tell which he did.

He tells you that you can push a button next to the opaque box, which will return you and only the opaque box to your living room in the manner by which you came, or you can push a button next to the transparent box, which will return you and both boxes. In any event, it will be utterly impossible for you to examine the contents of the opaque box until you decide, at which time it will open.

Having explained all this, the alien disappears in a flash of light, leaving you to make to your decision.

Which do you pick, and why?

Perhaps more interestingly, approximately what is your cut-off point for the value of the transparent box in this scenario? That is, almost everyone will take only the opaque box if the transparent box only has $1 in it, and almost everyone will take both boxes if the transparent box has $999,999 in it. Therefore, almost everyone has a point (or at least a fuzzy boundary) at which their decision changes.

[John L]

Both boxes obviously maximize the gain. Specifically, you get $1000 more in that case. The alien cannot change the contents of the opaque box in response to your decision. For that reason, it is probably empty. If it is not, all the better.

[Brian Olewnick]

Originally Posted by John L

Both boxes obviously maximize the gain. Specifically, you get $1000 more in that case. The alien cannot change the contents of the opaque box in response to your decision. For that reason, it is probably empty. If it is not, all the better.

But he knew you'd say that! You get only $1000. If only you were the type to pick the opaque box, you'd have gotten $1,000,000. Or, can you flip-flop your decision in your head before choosing, the $1,000,000 flitting in and out of existence as you do? Obviously not. I think....

[John L]

You mean that there is no right or wrong answer?

Actually, the way that you presented the question the first time made me unsure. You wrote that the alien is always, without exception, 100% correct in his predictions.

In the second case, you wrote that the alien could be wrong and cannot change the contents of the box in response to the decision. In that case, it is clear that taking both boxes is the best strategy.

In the first case, given the context, one might suspect that something supernatural is going on. If the alien can't be wrong, perhaps he can manipulate fate? Perhaps the contents of the box do change in response to the decision?

[Brian Olewnick]

Originally Posted by John L

You mean that there is no right or wrong answer?

Actually, the way that you presented the question the first time made me unsure. You wrote that the alien is always, without exception, 100% correct in his predictions.

In the second case, you wrote that the alien could be wrong and cannot change the contents of the box in response to the decision. In that case, it is clear that taking both boxes is the best strategy.

In the first case, given the context, one might suspect that something supernatural is going on. If the alien can't be wrong, perhaps he can manipulate fate? Perhaps the contents of the box do change in response to the decision?

Oh, I don't think there's a right or wrong answer and you're correct, if you dwell on the (extremely) unlikely particulars in the set-up, you can find all manner of issues. But I do think it's an interesting thought experiment that, like others derived from Prisoner's Dilemma, Cooperate/Defect problems, appears in real-world decision making all the time.

Hofstadter conducted a wonderful one in 1983. This appeared after a great deal of discussion on the ethics of Cooperate/Defect scenarios and the difficulty of doing the "right" thing (Cooperate) because of the interdependence of your decision with that of others and the trust issues vs. practicality it raises. In this case, there is a (roughly) correct answer:

***************8
Douglas Hofstadter, in a series of essays on the Prisoner's Dilemma published in Scientific American, devoted one essay (Scientific American, June, 1983 and reprinted in his book, Metamagical Themas:) to the games based on Prisoner's Dilemma. At the end of the essay he makes the following offer to the readers:

"This talk of holding back in the face of strong temptation brings me to the climax of this column: the announcement of a Luring Lottery open to all readers and nonreaders of Scientific American. The prize of this lottery is $ 1,000,000/N, where N is the number of entries submitted. Just think: if you are the only entrant (and if you submit only one entry), a cool million is yours! Perhaps, though, you doubt this will come about. It does seem a trifle iffy. If you'd like to increase your chances of winning, you are encouraged to send in multiple entries without limit. Just send in one postcard per entry. If you send in 100 entries, you'll have 100 times the chance of some poor slob who sends in just one. Come to think of it, why should you have to send in multiple entries separately? Just send one postcard with your name and address and a positive integer (telling how many entries you're making) to:

Luring Lottery
c/o Scientific American
...
... You will be given the same chance of winning as if you had sent in that number of postcards with ‘1' written on them. Illegible, incoherent, ill-specified, or incomprehensible entries will be disqualified. Only entries received by 5:00 PM on June 30, 1983 will be considered. Good luck to you (but certainly not to any other reader of this column)!"

[Rob C]

Originally Posted by Brian Olewnick

The alien then announces that his work on Earth is done and instantly transports himself back to his home world, leaving the two boxes in front of you.

Just a little something to say "Thanks for the anal probes"?

[John L]

Originally Posted by Brian Olewnick

In this case, there is a (roughly) correct answer:

Now you've got me confused, Brian. The optimal strategy in this case obviously depends on what conjectures you are making about the behavior of others. If you expect that 1000000 other people are going to send in 100 each, then the optimal answer is 0. The expected value of the lottery is not worth the postage or service fee of Luring Lottery.

The best solution would be to collude with all other possible participants, buy only one ticket, and then split the $1 000 000. However, that is not a strategically stable solution, i.e. people will have an incentive to cheat and buy other tickets secretly.

[walto]

FWIW, two of your other Gods have discussed that paradox in their books, Brian.

Nozick....and me.

[crawjo]

Originally Posted by John L

It is not so difficult to understand. Compare two strategies: one where you choose curtain 1 and stay with it, and another where you choose curtain 1 and then switch.

With the first strategy, you win only if it is behind curtain 1 (1/3 probability).

With the second strategy:

If it is behind curtain 1 (1/3 probability), you lose

If it is behind curtain 2 (1/3 probability), Monte Hall will show you curtain 3 and you will choose curtain 2 (You win)

If it is behind curtain 3 (1/3 probability), Monte Hall will show you curtain 2 and you will choose curtain 3 (You win).

Thus, with the second strategy, you only lose if it is behind curtain 1 (1/3 probability)

Reading this again, I'm not sure if I agree with the logic. The second strategy answers don't really define the question. Because the choice of whether or not to switch doors doesn't occur until AFTER Monte Hall has already eliminated one of them as a possibility. Thus, when you get to a choice between door 1 or door 2, all you know is that it is behind one and not the other. I think your chances are 50 percent. Why am I wrong?

[Gordon B]

Originally Posted by Brian Olewnick

But he knew you'd say that! You get only $1000. If only you were the type to pick the opaque box, you'd have gotten $1,000,000. Or, can you flip-flop your decision in your head before choosing, the $1,000,000 flitting in and out of existence as you do? Obviously not. I think....

For John, the alien, knowing that John was a good utility maximizer would put nothing in the opaque box.

For a regular lotto player, the alien would put $1,000,000 in the opaque box.

The way you first phrased the question, where the alien was omnipotent, it would always be correct to choose the opaque box, given that his omnipotence was beyond question. Perhaps the alien was really the all-k-p-l God we talked about in the other thread

[John L]

Originally Posted by crawjo

Reading this again, I'm not sure if I agree with the logic. The second strategy answers don't really define the question. Because the choice of whether or not to switch doors doesn't occur until AFTER Monte Hall has already eliminated one of them as a possibility. Thus, when you get to a choice between door 1 or door 2, all you know is that it is behind one and not the other. I think your chances are 50 percent. Why am I wrong?

When you have chosen curtain number 1, you know that there is a one third probability that you are right and a two thirds probability that it is behind either 2 or 3. A strategy to switch gives you both 2 and 3. Receiving an (independent) additional piece of information that it is not behind curtain number 2, for example, makes the conditional probability that it is behind curtain number 3 equal to two thirds. Another way to think about it is that Monte Hall is going to reveal a curtain in any case. So that event cannot increase the conditional probability that it is behind curtain 1. It remains 1/3. The only other possibility (2/3 probability) is that it is behind the other remaining curtain.

You seem to be thinking of the case when Monte Hall reveals information before you make an initial choice. In that case, the odds are 50-50 between the two remaining curtains. Once you have made an initial choice, the elimination of a curtain reveals more information, i.e. the elimination of one of two curtains is more informative than one of three.

[crawjo]

Originally Posted by John L

When you have chosen curtain number 1, you know that there is a one third probability that you are right and a two thirds probability that it is behind either 2 or 3. A strategy to switch gives you both 2 and 3. Receiving an (independent) additional piece of information that it is not behind curtain number 2, for example, makes the conditional probability that it is behind curtain number 3 equal to two thirds. Another way to think about it is that Monte Hall is going to reveal a curtain in any case. So that event cannot increase the conditional probability that it is behind curtain 1. It remains 1/3. The only other possibility (2/3 probability) is that it is behind the other remaining curtain.

You seem to be thinking of the case when Monte Hall reveals information before you make an initial choice. In that case, the odds are 50-50 between the two remaining curtains. Once you have made an initial choice, the elimination of a curtain reveals more information, i.e. the elimination of one of two curtains is more informative than one of three.

I'm still not sure that I agree. Because the problem poses the question of what do you do AFTER Monte has already eliminated one of the options. At that point, there are two options left, door 1 and door 2. You have no information to suggest that one choice is better than the other. The fact that Monte revealed door 3 tells you absolutely nothing about door 1 or door 2, other than that one of them is the right door. There is no 2/3rds option, because there are no longer three doors to choose from. There are only two.

It just occurred to me that the initial choice of a door is irrelevant. If it is the case that door 1 and door 2 have a Kenny G cd, and door 3 has the car, no matter which one I choose, Monte is going to open one of the Kenny G doors. So, to take the above example, that I choose door 2 or door 1 doesn't matter at all. Because Monte will ALWAYS open a door with a Kenny G cd, there is always a 50 percent chance that I have chosen a door that has the other cd, and a 50 percent chance that I have chosen a door that has the car. Therefore, there shouldn't be any probability of 1/3 or 2/3rds, because every single time Monte is going to open one of the doors with the cd. As such, my choice is between two doors, not three. Three is irrelevant.

[Gordon B]

Originally Posted by crawjo

I'm still not sure that I agree. Because the problem poses the question of what do you do AFTER Monte has already eliminated one of the options. At that point, there are two options left, door 1 and door 2. You have no information to suggest that one choice is better

First of all, it's Monty Hall, not Monte Smith
Secondly, you do have information suggesting that door 2 is more likely.

Let me try a different way.

Mr. Hall says "Crawjo, the car is equally likely behind doors #1, #2, #3. You can choose #1 or you can choose both #2 and #3. If you choose 2&3, I'll eliminate one of them before we tell the producers which door you picked. If either #2 or #3 has the car, I'll eliminate the one with the cd. If #1 has the car, I'll eliminate either #2 or 3 but you will lose."

The only difference between my wording and the original is that in one case, you choose to switch after Monty eliminates a door and in the other case you agree to choose either #2 or #3, Monty's choice, provided that he agrees to eliminate a door with no car.