The Monte Hall Problem....again

[PantsOnFire]

Maybe you have heard of this problem, maybe you haven't.

The point of my post is to show a proof that is one of the most clear that I have heard (and there are several).

Again, here is the problem:

"Suppose you're on a game show (e.g. Let's Make a Deal), and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the other two doors, opens another door, say No. 3, which has a goat. He then says to you, 'Do you want to pick door No. 2 or keep door No. 1?' Is it to your advantage to take the switch?"

[jay_shark]

This is the best explanation to this problem .

Lets say you always select door A . Then 100% of the time , there will be a goat behind one of the two remaining doors (pigeonhole principle) . Since the host knows what's behind both doors , he will show you the goat which would be either in door b or door c . If you never switch , then the probability you win is clearly 1/3 . This means that 2/3 of the time , the prize belongs to the other door that has been unopened so you should switch !!

[Silent A]

Variation on the above ...

- Your initial pick will be correct 1/3 of the time.
- Monte never reveals a car.
- So, whenever your initial guess is wrong, the car will be behind the other door.
- Since your first guess is wrong 2/3 of the time, the car is behind the other door 2/3 of the time.

Therefore, you should always switch.

[pzhon]

If you think you have something new to add that wasn't stated in the over 100 times this has been discusssed on the 2+2 forums before, just say it. And please give a complete statement of the problem. If you don't understand that your problem statement was not complete, then you don't understand the Monty Hall problem, and you may want to read the rec.puzzles (usenet) FAQ.

[PantsOnFire]

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If you think you have something new to add that wasn't stated in the over 100 times this has been discusssed on the 2+2 forums before, just say it. And please give a complete statement of the problem. If you don't understand that your problem statement was not complete, then you don't understand the Monty Hall problem, and you may want to read the rec.puzzles (usenet) FAQ.

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You're right, I mentioned that Monte knows what's behind each door but I didn't mention that he will always reveal a goat using that knowledge.

Proof: The player picks one of three doors. There is a 1/3 chance of the car being behind his door and a 2/3 chance that the car is behind the other two doors. When Monte exposes a goat behind one of the other two doors, there is still a 2/3 chance that the car is behind the other two doors but we now know that of those two, one doesn't have the car so the door he didn't expose now has the 2/3 chance.

I see some earlier posters made very similar proofs so I guess I had nothing new to present so I slink away in shame now.

[pzhon]

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You're right, I mentioned that Monte knows what's behind each door but I didn't mention that he will always reveal a goat using that knowledge.

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Right, that assumption is important. For example, if Monty will only let you switch if you chose correctly the first time, then seeing that you get a choice means switching never wins. If Monty opens a random door, and shows you whatever is behind it, then switching wins 1/2. Only if he always shows you a goat does switching win 2/3--and his choice of door may provide more information, even in that case.

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Proof: The player picks one of three doors. There is a 1/3 chance of the car being behind his door and a 2/3 chance that the car is behind the other two doors. When Monte exposes a goat behind one of the other two doors, there is still a 2/3 chance that the car is behind the other two doors but we now know that of those two, one doesn't have the car so the door he didn't expose now has the 2/3 chance.

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Ok, that proof may be convincing to some, but how did it use the above assumption? It looks to me like there is a missing step, or at least a missing reference to the assumption.

[R Gibert]

Consider a Monty Hall Problem generalized to a zillion doors where Monty shows you that there is a goat in each of the zillion - 2 doors, etc. It gets pretty clear then to most people.

This and a few other explanations can be found here:
http://en.wikipedia.org/wiki/Monty_hall_problem

[PantsOnFire]

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Proof: The player picks one of three doors. There is a 1/3 chance of the car being behind his door and a 2/3 chance that the car is behind the other two doors. When Monte exposes a goat behind one of the other two doors, there is still a 2/3 chance that the car is behind the other two doors but we now know that of those two, one doesn't have the car so the door he didn't expose now has the 2/3 chance.

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Ok, that proof may be convincing to some, but how did it use the above assumption? It looks to me like there is a missing step, or at least a missing reference to the assumption.

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I'm not 100% certain but I took it as implicit. With Monte knowing where the car is, his revealing of the goat out of those two doors does not change the 2/3 chance for those two doors.

In other words, Monte's reveal is a freebee which does not change any odds. I couldn't do the math for that but I understood immediately when I heard that proof.

[bigpooch]

Monty (right spelling?) DOESN'T have to know where the car
is. If the contestant picks a door and then Monty picks
another door and if he finds a car, too bad! If not, and
Monty gives you a choice to switch, switching now gives you
a probability of 2/3. As you should see, a contestant would
usually prefer if Monty doesn't know where the car is after
a second door is opened and there is no car seen (unless he
happens to be a "blood relative" of Monty or other
circumstances!).

Now instead, suppose Monty knows where the car is.

Unfortunately, as pzhon mentioned, Monty could be mean:
if you pick incorrectly, you get a goat and if you pick
correctly, switching is never right. [ Perhaps the
contestant showed up in a costume he really dislikes! ]

Monty could also be more subtle and make it so that if he
shows another door, switching has a probability p of success
for any p in the interval [0,1] and it's possible that p
could depend on Monty's assessment of the originality of
the costume or whether the contestant is a cute girl, yada
yada yada!

Of course, this has been mentioned and rehashed ad infinitum
ad nauseum.

[pzhon]

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Monty (right spelling?) DOESN'T have to know where the car is. If the contestant picks a door and then Monty picks another door and if he finds a car, too bad! If not, and Monty gives you a choice to switch, switching now gives you a probability of 2/3.

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He needs to know. Your analysis is flawed. Let's suppose you always choose door 1. There are three equally likely cases, CGG, GCG, and GGC.

If Monty always opens door 2, then you get more information when you see what is behind it. This information has a 1/3 chance of updating your estimates of the probability a car is behind the first door to 0, (GCG, and you lose) and a 2/3 chance to update the probability to 1/2 (CGG and GGC). If you see a goat behind door 2, you should be indifferent to switching.

If Monty knows where the car is, and randomly (on the basis of a fair coin flip) opens a non car door other than the first, then you get no information about whether your choice was good or bad, and so your estimate stays at 1/3. Then switching is right. You are in case CGGH, GGCH, or GGCT, and these are equally likely.

If Monty knows where the car is, and opens door 2 if it contains a goat, but door 3 if the car is behind door 2, then you should be indifferent to switching if Monty opens door 2 (CGG and GGC are equally likely), but switching wins 100% of the time if Monty opens door 3 (GCG).

I hope that helps to illustrate that it is crucial to understand Monty's plan, since that determines the information you get from his actions. Monty's plan must be given in the problem statement, and used in the proof. The three prisoner version is less ambiguous, but still requires a bit of extra information:

You are one of 3 prisoners. Two prisoners will be executed, but you don't know who. You say to the guard, "Tell me the name of one of my fellow prisoners who will be executed." He says, "Smith will die." Did this response increase your probability of surviving to 1/2?

The answer depends on the guard's actions if given a choice. If the other prisoner is named Tarquin Fin-tim-lin-bin-whin-bim-lim-bus-stop-F'tang-F'tang-Olé-Biscuitbarrel, a name the guard would attempt to say only if necessary, the you'll survive 1/2 of the time. If it is a name the guard would say equally often as Smith if given a choice between them, then your probability of surviving is still 1/3. If it is a name the guard would much rather say than Smith, your probability of surviving has dropped.