The stupid 3-doors problem

DWarrior · #1 07-12-2005, 07:33 PM

Someone threw this at me a year ago, and I just rememberd it now again.

You're on a show and you have to pick one door out of three. Behind 1 of the prize, behind the other 2 is nothing. The agreement is that you pick your door, then the show host reveals one empty door and you have the option to switch to another.

Should you switch?

Obviously the odds of picking the first time are 1/3, but my friend and the CS professor claimed that the second pick is actually 2/3, because the remaining door somehow "combines" probability with the door that the host opened.

I argued that there is no way to know what's behind the second door, so as soon as the host opens the empty door, you have the option of choosing one door out of 2, so your odds are 1/2, which means that there is really no difference between switching and staying the on the same door.

So, 3 doors, A B and C, you pick A. Game host reveals door B as the empty one. Are the odds of door C having the prize 1/2 or 2/3 (B and C combined)?

Back then, I was awed by my CS professor, and considered him a wise man, so I did not argue (though I kept my theory), but a year later, I still think the odds are 1/2 on both doors.

BruceZ · #2 07-12-2005, 07:48 PM

I'm tempted to write a script to remove any posts with the words "3 doors".

Archives, search function, Monty Hall.

GTSamIAm · #3 07-12-2005, 07:52 PM

I think there is some confusion with the sequence of events. If you pick the right door the first time, if you automatically win, the chance before you pick the first time is 2/3. The 1/3 chance the first time plus the 1/2 chance the second time. If you don't automatically win, it looks like the chances are 1/2. You always have two doors to choose from. If he never opens the door you have picked when it is empty then obviously you have a 100% chance to win, as you can switch.

BruceZ · #4 07-12-2005, 08:04 PM

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I think there is some confusion with the sequence of events. If you pick the right door the first time, if you automatically win, the chance before you pick the first time is 2/3. The 1/3 chance the first time plus the 1/2 chance the second time. If you don't automatically win, it looks like the chances are 1/2.

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That's wrong. You don't automatically win, and it is 1/3 if you don't switch, so it is 2/3 if you switch.

Here is another way to solve problems like this without having to invoke the equations of Bayes' theorem directly. Say you pick door A, and the host, who must reveal an empty door from one that you did not pick, reveals that door B is empty. There are 2 things that could have happened. Either the prize is behind door C, and the host was forced to open door B (since he can't open your door or door C), or the prize is behind door A which you selected, and the host made a random choice of door B from doors B and C, which are both empty. The first case is twice as likely as the second case, since in the second case the host will choose door B only 1/2 of the time, while in the first case he will choose door B all of the time. So it is twice as likely that the prize is in C than it is in A, so the probability is 2/3 for C and 1/3 for A. So we switch to C. Note that we also need to know that the host chooses randomly when he has a choice of 2 doors. This assumption is almost never stated in the problem, and the lack of this makes the problem ill-posed and subject to debate.

GTSamIAm · #5 07-12-2005, 08:10 PM

It never said the host opens a door that you don't pick.

BruceZ · #6 07-12-2005, 08:15 PM

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It never said the host opens a door that you don't pick.

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Right, the problem was not worded correctly, which is the cause of most debate about these problems. You have to know how Monty Hall was normally played. If the host can open your door, then the answer indeed becomes 1/2.

BruceZ · #7 07-12-2005, 08:19 PM

BTW, the prisoner problem is worded correctly, but you got it wrong anyway.

PowerHouse · #8 07-12-2005, 08:19 PM

If you have a problem understanding why it is 2/3 probability then think if you had to choose between 100 doors you choose Door#1 and then the host opens all but your choice and Door#73. There would most likely be a reason he didn't open Door#73, so with switching you have 99% of being right because all the odds for Door#2-100 are 1% each they all combine into Door#73. Would you still now think that your odds are 50/50 between 1 and 73?

GTSamIAm · #9 07-12-2005, 08:24 PM

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BTW, the prisoner problem is worded correctly, but you got it wrong anyway.

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Yeah, rub it in. But I don't understand exactly how Monty Hall applies to that problem.

BruceZ · #10 07-12-2005, 08:26 PM

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BTW, the prisoner problem is worded correctly, but you got it wrong anyway.

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Yeah, rub it in.

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OK. [img]/images/graemlins/smile.gif[/img]

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But I don't understand exactly how Monty Hall applies to that problem.

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Replace being being killed with winning a prize, and they are exactly the same. Sheesh.