The Monty Hall Problem

Originally posted by lloydk
"Argument is argument. You cannot help paying regard to their arguments, if they are good. If it were testimony you might disregard it…Testimony is like an arrow shot from a long bow; the force of it depends on the strength of the hand that draws it. Argument is like an arrow from a crossbow, which has equal force though shot by a child."

Intelligence is recognizing the accuracy of this statement.
Knowledge is knowing who said it.

😏

Very nice. Was it not Socrates who always suggested, when someone called him a fool, that his indictors "explain to this poor fool the error of his ways"? And they never could.

Originally posted by Acolyte
I'm assuming here that the numbers 1 and 3 are arbitrary (otherwise why would you ever choose 3?) and that probabilites are equal, in which case I disagree. You have a 1/3 chance of picking the correct door first time. Clearly if you switch, you lose. However there's a 2/3 chance that you've picked the wrong one. The host then eliminates the other wro ...[text shortened]... you switch, you have a 2/3 chance of winning. I think this problem has been posted here before.

Aco:

royalchicken chose Door Nr. 1 out of three. Monty eliminated Door Nr. 3 as cack and asked royalchicken to choose again if he so wished. (His odds are best, he feels, if he now chooses Door Nr. 2.) But, at this same moment, Monty pulled Waldorf's friend, IQ156, from the audience and asked him, or her, to choose either Door Nr. 1 or Nr. 2. What is the population now? Which door should he, or she, choose? Does royalchicken's historical perspective affect IQ156's odds? Are royalchicken's odds better?

Arbeiter, I fear your profile is accurate. I'll have to agree that switching is better:

Door:
_ 1 2 3

A w w r

B w r w

C r w w

"w" indicates "wrong", "r" indicates "right". If case A is true, and I have chosen door 1, Monty will eliminate the other wrong door (in this case it must be what I have called door 2), which he will apparently call door 3. By switching to what I call door 3, I will get the prize. In case B, Monty will eliminate what I call door 3, and I will switch to door 2, getting a prize. In case C, Monty will eliminate one of the two wrong doors, and I will switch to the other, giving up the prize. So in 2 of three cases, I get the prize by switching. Only once in three cases will staying be beneficial. So I will have to agree with Acolyte and bbarr (although some behind the scenes argument bears out the utility of this question in the applications bbarr has suggested-see previous post).

... Arbeiter, I fear your profile is accurate... So I will have to agree with Acolyte and bbarr...

roiledpullet:

But, what about IQ156's chances - two doors, two choices?

What if IQ156 and myself are flipping a coin-first player to flip heads wins. I flip first. So my chances right there of flipping a head are obviously 1/2. However, if I flip a tail, I'll be in the same boat that IQ156 was before I went. Thus my chances of WINNING are exactly twice as good as his, and since the probabilities sum to 1, the probability that I win this game is 2/3.

I know I'm not intelligent 🙄

Originally posted by Acolyte
You have a 1/3 chance of picking the correct door first time. Clearly if you switch, you lose. However there's a 2/3 chance that you've picked the wrong one. The host then eliminates the other wrong one, so if you switch, you're guaranteed to win.

In conclusion: if you stick, you have a 1/3 chance of winning. If you switch, you have a 2/3 chance of winning. I think this problem has been posted here before.

There is no advantage to be gained by switching.

When we speak of probabilities of this sort, it is necessarily subjective. Whie I may have only a 1 in 3 chance of correctly picking the door with the prize behind it, the host, who knows where the prize is, has a 1 in 1 chance of picking the correct door.

After I've picked door #1, and the host tells me that door #3 is wrong, I have some extra information at my disposal. I now know that the prize is behind either door #1 or door #2. I have no information which allows me to differentiate between #1 and #2. Accordingly I have a 1 in 2 chance of picking the correct door.

"But wait," you exclaim, "you've just switched the probability of door #1 being correct from .33 to .50, you can't do that!"

Can't I? If that is true, then how is it OK to switch the probability of door #2 being correct from .33 to .67?

The probability that the prize will be behind one of the doors is 1, and the individual probabilities for all of the doors must add up to 1. By assigning a probability of 0 to one of the doors, the probability of the prize being behind each of the other doors is 0.5, in the absence of something to differentiate between the two.

Your error arises from failing to recognize that, by eliminating the possibility that the prize is behind #3, the host has also eliminated one of the two chances that your original pick was wrong.

Originally posted by richjohnson
There is no advantage to be gained by switching.

When we speak of probabilities of this sort, it is necessarily subjective. Whie I may have only a 1 in 3 chance of correctly picking the door with the prize behind it, the host, who knows where the prize is, has a 1 in 1 chance of picking the correct door.

After I've picked door #1, and the host ...[text shortened]... ehind #3, the host has also eliminated one of the two chances that your original pick was wrong.

You double your chances of winning by switching.

Suppose I have three cards, only one of which is the Joker. Suppose I deal one card to you and two cards to me. At this point the probability that your card is the Joker is one-third. The probability that the Joker is one of my cards is two-thirds. This is, if I were to deal the cards n times, the probability of my having the Joker will converge to two-thirds. This means that the probability that you have the Joker will converge to one-third. If it really made no difference whether you switch, then it shouldn't make any difference if you always keep your card. But if you always keep your card, then in n trials, your probability of winning will converge to one-third.

If I picked one of the three cards at random and flipped it over, and it happened to not be the Joker, then the probability of either of the remaining cards being the Joker would, indeed, be one-half. But I do not show a card at random, my choice of which card to flip is dependent upon the identity of my cards. I only flip cards that are not the Joker. If prior to my flipping a card the probability of my having the Joker is two-thirds, and the probability of my having at least one card that is not the Joker is one, then that means that regardless of whether I have the Joker, I'll always have a card to flip that is not the Joker. But the probability that one of my cards is the Joker is still two-thirds.

Originally posted by richjohnson

Your error arises from failing to recognize that, by eliminating the possibility that the prize is behind #3, the host has also eliminated one of the two chances that your original pick was wrong.

Acolyte is correct. To see this, I worte a QuickBasic program to simulate it I did 1.3 million simulations, and came up with a probability of winning of 66.58%. To really convince the empiricists, I simulated staying, 724773 times-33.28% wins.

Arbeiter, you do ask a good question. I believe we have discussed it, we'll see what others think. I don't think your question is relevant, for one key reason. Monty knows the whereabouts of the prize. Thus this is not an question of random guessing, but of picking a stratagey, because he has not changed the size of the problem space.

Richjohnson, Monty Hall, in eliminating one wrong choice, has eliminated 2 of the three ways in which the contestant can be wrong, and thus by switching one can double one's odds. For a good explanation, you can see

www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

Finally, in 1990, someone sent this question to the newspaper column of Marilyn vos Savant, reputed possessor of the world's highest IQ (of course beside the point here). She correctly answered that switching doubles one's odds. She received about 10 000 letters, many from well-reputed mathematicians, saying essentially what richjohnson and wladorf said. After much discussion, 10 000 letter-writers were found with egg on their faces.

To make it really clear if your intuition is still rejecting it, think of what would happen if there are 1000 000 doors and Monty Hall takes away 999 998 wrong doors. Clearly, the one you picked is, at the beginning, incredibly unlikely to be correct. The other one left is very likely correct. The situation has not changed, and, in fact, switching increases you're odds astronomically.

Now this just doesn't make any sense guys. You can explain it through any faulty mathmatical equation you desire to design, but the odds are still 1/2. When there were three doors, the prize was behind door x. After one door was eliminated, the prize was still behind the same door x. No matter if door x is door 1 or door 2, the prize does not move and the first guess is just as good as changing your decision.

Originally posted by Omnislash
Now this just doesn't make any sense guys. You can explain it through any faulty mathmatical equation you desire to design, but the odds are still 1/2. When there were three doors, the prize was behind door x. After one door was eliminated, the prize was still behind the same door x. No matter if door x is door 1 or door 2, the prize does not move and the first guess is just as good as changing your decision.

It's not faulty, though. It is perfectly logical. If your machine can run DOS applications I will send you my simulations; apparently you are an empiricist.

Originally posted by Omnislash
Now this just doesn't make any sense guys. You can explain it through any faulty mathmatical equation you desire to design, but the odds are still 1/2. When there were three doors, the prize was behind door x. After one door was eliminated, the prize was still behind the same door x. No matter if door x is door 1 or door 2, the prize does not move and the first guess is just as good as changing your decision.

Actually, I thought it over again last night, and it does make sense to me now. What was giving me problems before was the following:

If Alice, Bob and Charles pick doors 1, 2 and 3, respectively, and then Monty tells them that Charles was wrong and the prize is not behind door #3, clearly, neither Alice nor Bob could gain by switching. (If one could, so could the other, and the totatl probability would exceed one).

However, the original question is different. I thought there was nothing to differentiate #1 and #2, but there is. Monty selects one of the two UNpicked doors to open. Door #1, being picked, is not subjected to this selection and its chances of having the prize are unaffected by it. The other doors' chances become bound together by being subjected to Monty's selection, and revealing the door without the prize shifts all of the 67% chance that one of these doors will have the prize to the other door. Thus, the difference between #1 and #2 is that #2 has been subjected to Monty's selection, and #1 has not.

Bonus question: Does the answer change if Monty selects one of the two unpicked doors to open at random (i.e. Monty has no knowledge of the prize position), and it happens to be one of the wrong doors?

Refering to the inital post, we don't have all the info we need to apply probablilty theory sensibly. Our main missing fact is that we don't know whether the host actually knows, whether they do but are lying, whether they do and are being truthful, whether they think they do.... blah blah etc. To summerise: INSUFFICIENT DATA: DOES NOT COMPUTE!

We do know that the host has perfect information.

We do know that the host has perfect information.[/b]

at the first choice, he apriori probablity of a given door being correct is 1/3. The host's information changes the apriori probabilities for the second choice: Had one picked door 3 initially, switching door would clearly have improved the odds of winning (from 0 to 1/2). If either door 1 or 2 had been chosen (as in the example), the host's new information improves the apriori probability of door #2 being correct, but it equally improves the apriori probablity of door #1 being correct. Both improve from 1/3 to 1/2. Try it with a random number generator if you don't believe it. Its just 50/50 on doors 1 and 2.