The Monty Hall Problem- Revision and extensive analysis

I will demonstrate here that the usual resolution of the Monty Hall problem is incorrect. The problem is stated as follows. In a game show, the player is shown three identical doors. Behind two of the doors, there are goats, and behind the third, there is a car. The game host asks the player to choose one door. After he has chosen one, the host opens one of the remaining doors and reveals a goat (since <I>he</I> knows which contains what). Now, the player is given the option of switching to the other door. The questions are:

*Does switching improve his chances of getting the car?
*Does it decrease his chances?
*Or will the chances be unchanged whether he switches or not?

The obvious answer is that it does not make any difference. Since there are two doors (the unopened ones) from which he can choose, and since one door contains a goat and the other the car, there's a 50% chance of either door containing the car.

But the answer which is claimed to be correct says that if he switches, the chance of winning is 2/3, and therefore, it is better to switch. This, allegedly, shows that statisitics can sometimes be counter-intutive. The justification is given thus:

There are three possibilities:

*The player has chosen Goat 1, the game host shows Goat 2. Switching wins.
*The player has chosen Goat 2, the game host shows Goat 1. Switching wins.
*The player has chosen the car, the game host shows either of the goats. Switching loses.

Since switching wins two out of three times, the probability is 2/3 in favor of it. This is the standard solution.

I propose to show that this is flawed. For, the events haven't been completely listed. The complete list is as follows:

*The player has chosen Goat 1, the game host shows Goat 2. Switching wins.
*The player has chosen Goat 2, the game host shows Goat 1. Switching wins.
*The player has chosen the car, the game host shows Goat 1. Switching loses.
*The player has chosen the car, the game host shows Goat 2. Switching loses.

So the third event from the previous case really consists of two possibilities because the game host can choose either of the goats. So both these cases should be listed, and since there is no preference for any goat, both of the last events have equal probability. So now there are two cases where switching loses. Which means that now there's only 50% chance of winning when you switch.

Of course, you don't have to go to all this trouble if you address the problem more directly. Forget that the player has chosen any door. The question whether to switch or not, is the same as the question which door should he choose. there are two doors. One contains a goat. The other contains a car. The probability that a given door contains the car is 1/2.

 

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In another thread, Funkstar challenged the above solution by saying that it could be wrong to assign equal probabilities to all the four cases. If I understand correctly, what he was trying to say was this. The last two cases apply when the player has chosen the car. At the start of the game, he has three choices to make, each with a probability of 1/3. So the probability that he has chosen the car is 1/3, and hence, the last two cases should not be given an equal consideration. So essentially, we coud say something like this:

*Chance 1/3: The player has chosen Goat 1, the game host shows Goat 2. Switching wins.
*Chance 1/3: The player has chosen Goat 2, the game host shows Goat 1. Switching wins.
*Chance 1/6: The player has chosen the car, the game host shows Goat 1. Switching loses.
*Chance 1/6: The player has chosen the car, the game host shows Goat 2. Switching loses.

But I want to remind you that probability deals with the given and the unknown. We have to base predictions on what is given, or what has already happened. We don't have to consider what the chances of happening were for an event which has already taken place. (You could even say that the chance for any event which has already occurred is 100% in a Deterministic universe. No, I'm not going to discuss Determinism here, for the appropriate place for that is in the Philosophy section where I've already done so a number of times). And probability is meaningful only when you have partial knowledge. When you toss a coin, you don't know all the different factors which influence its motion, and <I>therefore</I> you have to use probability. If you knew all the factors, (and if you have enough computational power) you would be able to predict exactly what would happen. For example, if you knew what was behind which door, you can say exactly what will happen if you switch doors. If you had chosen the car, and you know that now, you can say that probability of winning if you switch is 0. The game host knows that exact outcome of each choice. So there are no probabilities as far as he is concerned.

There is no point in considering the game from any other point of view than that of the player. And there is no point in thinking about the probabilities of what has already happened. That is, it does not matter whether the probability that the player has already chosen the car is 1/3 or 2/3 or whatever. He has already done so. It does not figure in further calculations. Because we are not trying to predict the outcome of the whole game. We are only trying to choose the door <I>after</I> one of them has been opened. So it makes sense only to talk about the chances available now.

 

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What's with all this babbling nonsense about determinism and computational power and whether or not we know something a priori? This is a pure logic puzzle, man, and and it's dirt simple.

Draw a decision tree for the case where the player switches doors. Starting from the top, the player has three choices, all of which are equally probable. The probability of each branch is exactly 1/3. Two of those branches lead to immediate success, as you noted. The other branch leads to failure, regardless of which door Monty opens. That's a total of 2/3 probability of success. Done. Even if you don't like the answer, that's the answer.

If you don't believe it, write a computer program which chooses doors at random, and watch it happen. Spooooooky.

- Warren

 

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Oh no! We're not trying to predict the outcome of the whole game as such. We're trying to choose a path when we've already come half-way through the game. Which is quite different.

 

And instead of explaining the thing from the start, try to point out exactly where my analysis is wrong!

 

At the point in question the player has two choices.
Assuming Monty shows door numer 3 has a goat...
A.) Door number 1 has a car.
or
B.) Door number 2 has a car.

The contestant has a simple choice to make.
The third door might as well not even exist, since it ALWAYS comes down to one of two doors.
Going into the game, the person has a choice between two doors (since one will be eliminated after his first decision), therefore a 50% chance of winning.

 

You went wrong in applying the principle of indifference. This principle is weak, has no formal justification, and is not a part of mathematical probability theory.

Consider tossing a coin. A coin can land heads up, tails up, or on-edge. Blindly applying the principle of indifference would indicate the odds of these three outcomes to be 1/3 each. I can only recall two times in my entire life when a coin landed on-edge. The principle fails because the underlying assumption that the outcomes are equally likely is not correct in this case.

Closer to the heart of the problem, consider counting the number of heads resulting from two tosses of a coin. Since there are three outcomes (0, 1, and 2), the principle of indifference suggests odds of 1/3 for each outcome. The principle of indifference is wrong once again.

Your application of the principle of indifference to the Monty Hall problem is wrong, and for exactly the same reasons that it is wrong in assigning odds of 1/3 to tossing 2 heads in a row. The fact that Monte can choose goat1 or goat2 when you chose the door hiding the car does not change the odds of your choosing that door in the first place. Funkstar was correct. It is incorrect to assign equal probabilities to your four events.

 

Can I propose a modification? Same scenario, three doors, two goats, one car. But in this scenario, Monty lets the player choose TWO doors. If either door hides the car, the player wins. The player has a 66 2/3% chance of winning at the beginning of the game. As usual, Monty opens a door with a goat. What is the player's chance of winning at this point? Does it lower the player's chance of winning if he accepts the option of switching doors at this point and why?

 

Your new scenario doesn't work. What if I choose the two doors hiding the two goats? The game is rigged if Monte shows me a third goat.

 

I don't understand how this is proving the usual answer that uses all the available information wrong. Seems to me, that limitting the contestant to only a portion of the usable information is rather silly.

But here:


A UFO drops off two abductees on the stage after the player (contestant a) has made his first choice and the goat is revealed. Monty fills in one abductee (contestant b) everything that has happened during the game. The other abductee (contestant c) has no additional information. He gives all three contestants a choice of the remaining doors.

Contestant a always sticks and wins 33% of the time.
Contestant b always switches from a's first choice and wins 66% of the time.
Contestant c picks randomly and wins 50% of the time.

The probabilities for both contestant a and b were covered in another thread.

The probabilities for contestant c is a simple 1/2 door has car, and 1/2 door has goat.

However, if you were to combine the probabilities for contestant c with the probabilities with contestant a and b:

Contestant c:
1/2-- Picks same door as contestant a (1/3 win)
1/2-- picks same door as contestant b (2/3 win)

total odds of winning 1/6 + 1/3 = 1/2

The same coin flip.

 

Suppose instead of three doors, there was initially a million doors. The player picks one randomly, and the 999,998 doors that contain goats are eliminated. Hopefully you see that the player should switch, because the car is almost certainly behind the door that wasn't eliminated. There are still only two doors, but the player's current door will only have the car if he initially made a correct 1-in-a-million guess.

It's exactly the same with the normal monety hall problem, but on a smaller scale. The player's current door will only have the car if he initially made a correct 1-in-3 guess.

Ah, but here is your mistake in an otherwise good analysis: you assuming that there is an equal probability of all four possibilities. That's not true; the first two possibilities that you list each have a 1/3 probability, while the last 2 each have a 1/6 probability. Your first and second possibilities must occur if the player initially picks the door with goat 1 or goat 2 (which is each a 1/3 possibility, because there are three doors). But your third possibility will only happen if the person initially selects the door with the car (1/3 probability) and the host reveals goat 1 (a 1/2 possibility, since there are two goats that could be revealed). Your 4th possibility has a 1/6 probability for the same reason. But then, this has already been explained to you.

But if you forget that the player initially picked a door, you lose valuable information. Again, I will refer to my above example of the same game with a million doors. After the 999,998 goat doors have been eliminated, there are two doors and one contains the goat and one contains a car. But it should be clear that the car is almost certainly not behind the door that the player initially picked. Now, if a second contestant were to come in after the 999,998 wrong doors had been eliminated and didn't know that one of the two remaining doors was only present because it had been randomly selected by the first player to not be subject to elimination if it was a losing door, then from this second person's perspective you would be correct - he only knows that there are two doors and one has a car, so it would be a 50/50 choice. But the first player has more information; he knows that his first choice of door is probably only still around because his initial selection of it prevented it from being subject to elimination, and that unless he initially made a correct 1-in-a-million guess he should switch doors.

 

I forgot to mention it, but Monty opens one of the two doors picked by the player. I thought that much would be obvious, since opening the unpicked door would determine the outcome of the game at that time. Not interested if Monty Hall or anyone else thinks this version is INTERESTING, I want to know the reasoning behind odds changing after a door is opened. The player starts with a 2/3 chance of winning by getting to pick two doors. If Monty opens one of the doors THE PLAYER PICKED to expose a goat, does the player's odds of winning change and why? Two doors are left, same as before, either one could hide the car.

 

chroot is correct. The decision diagram covers all possible scenarios, and unequivocally gives switching the edge.

Rosnet, you didn't answer my question in the other thread - if there's a chance that I might do a little dance if the coin lands heads up, and otherwise not, are the following scenarios equally likely to happen?

1. The coin lands tails up.
2. The coin lands heads up, and I do nothing.
3. The coin lands heads up, and I do a jig.

If you think the answer is yes, I would like to ask you why you think this is so. Please think about it, before you answer.

 

Well, I might have lost count myself. So, I would just have to ask myself if I felt this was my lucky day.

 

Okay everyone. My mistake. The fact is that additional information <I>is</I> given. Sorry! Thanks. My apologies for not posting this earlier. I figured it out (by thinking of Bayes theorem) the day I made the original post. Sorry for making everyone go to all this trouble.

 

An interesting thing about the "Monty Hall" problem that people seem to forget is that in the original show Montey would only offer you the chance to switch 50% of the time when you picked the wrong door, so you couldn't actually use statistics to increase your chance of winning - even if you switched, you still only had a 50% chance of winning.

 

You brought this up in the last thread and my explanation remains the same, what's become known as the "Monty Haul" problem is a simplification of the real game. No one who watched Let's Make a Deal thinks this is how Monty behaves, no one is forgetting anything.

"Montey would only offer you the chance to switch 50% of the time" is not true from my memory either. Monty did what Monty did. He was an unpredictable wacko. The simplifications made for an interesting problem with a seemingly strange result and it was based on the show, albeit loosely.

 

Yes, I just wanted to use it as one more piece of evidence that the under the classic statement of the Montey Hall problem you would usually win by switching doors; the show producers were aware of it.

I remember that sometimes he would let you switch and sometimes he wouldn't...I just assumed that he made the switch ofer 50% of the time.

 

I think he would sometimes offer you to exchange some amount of cash for your door. Sometimes he'd do this after revealing a door? I haven't seen it in some time. If monty always did as the "monty hall" problem assumes the show would have been pretty dull.


I'll mention this paper again:

See H. Bailey, "Monty Hall Uses a Mixed Strategy", Mathematics Magazine, vol. 73, No. 2, April 2000, pages 135-141

which gives some variations of this problem.