Monty Hall Variants

Speaking of the Monty Hall problem, I recently came across this terrific essay (PDF format), by Jeffrey Rosenthal, a professor in the Department of Statistics at the University of Toronto. Rosenthal discusses several variants of the Monty Hall problem, and shows how a clever application of Bayes’ Theorem helps to distinguish between them. Here are the variants he considers:

• Monty Hall Problem: A car is equally likely to be behind any one of three doors. You select one of the three doors (say, Door #1). The host then reveals one nonselected door (say, Door #3) which does not contain the car. At this point, you choose whether to stick with your original choice (i.e. Door #1), or switch to the remaining
door (i.e. Door #2). What are the probabilities that you will win the car if you stick, versus if you switch?
• Monty Fall Problem: In this variant, once you have selected one of the three doors, the host slips on a banana peel and accidentally pushes open another door, which just happens not to contain the car. Now what are the probabilities that you will win the car if you stick with your original selection, versus if you switch to the remaining door?
• Monty Crawl Problem: As in the original problem, once you have selected one of the three doors, the host then reveals one non-selected door which does not contain the car. However, the host is very tired, and crawls from his position (near Door #1) to the door he is to open. In particular, if he has a choice of doors to open (i.e., if your original selection happened to be correct), then he opens the smallest number available door. (For example, if you selected Door #1 and the car was indeed behind Door #1, then the host would always open Door #2, never Door #3.) What are the probabilities that you will win the car if you stick versus if you switch?
• Monty Small Problem: In this variant, the host is only somewhat tired. If he has a choice of doors to open, then he has a small probability p of opening the largest number available door, otherwise (with probability 1-p) he opens the smallest number available door. What is the probability that you will win the car if you then switch to the third door? (The case p = 1/2 is the original problem, while p = 0 is Monty Crawl.)
• Incidentally, variants on the Monty Hall problem are something of a hobby of mine. Feel free to suggest other variants in the comments.

1. #1 Michael Ralston
December 26, 2006

In Hall, of course, it’s 2/3 that switching wins, and 1/3 that sticking wins.

In the Fall, I’m pretty sure the odds are 1/2 for both.

In the Crawl … I’d have to say it depends. Hmm. Let’s examine that.
First of all, it doesn’t actually matter which door you pick for the subsequent analysis; for simplicity, then, assume you pick door 3.
In that case, there’s (obviously) three possibilities for the car’s location … and what door will be opened is fixed by where the car is. Numbering cases by the door the car’s behind, Monty opens:
Case 1) Door number 2.
Case 2) Door number 1. (these apply as per the monty hall problem).
Case 3) Door number 1.

Therefore, if Monty opens door number 2, you /must/ switch (100% chance of winning by doing so), but if he opens number 1, it doesn’t matter what you do; 1/2 for sticking or switching.

If you picked door one, then if he opens door three you must switch and door two means it doesn’t matter. Likewise, picking door two means that if he opens door three, you must switch, otherwise it’s 50-50.

So your total odds of winning if you decide to stick or switch before he opens his door work out to be 2/3 switching, and 1/3 sticking – but when he opens his door you get a bit more information and will either be certain to win by switching (1/3 chance), or will have a 50% chance to win no matter what. (2/3 chance of that scenario).

For Monty Small: Analysis…
well, use the cases from before.
1) Door 2. (1/3)*1
2) Door 1. (1/3)*1
3) Door 1 (1/3)*(1-p), Door 2 (1/3)*p.

So, if he opens door two, then probability 1/(1+p) that it’s behind door one, and probability p/(1+p) that it’s behind door three – or 1/(1-p) that you win by switching, and p/(1-p) that you win by sticking.
If he opens door one, then probaiblity 1/(2-p) that you win by switching, and (1-p)/(2-p) that you win by sticking.

And that makes sense – if p=.5, then it works out that 2/3 of the time you win by switching no matter what door he opens. likewise, if p=0, then you win by switching on an opening of door two, and it doesn’t matter what you do on an opening of door one.

Of course, I haven’t actually read the PDF yet. I should go do that.

2. #2 David D.G.
December 26, 2006

I seem to be stuck between the Monty Wall problem, just seeing this as a completely insoluble barrier, and the Monty LOL problem, simply laughing at all of this.

~David D.G.

3. #3 windy
December 26, 2006

We were discussing the Monty Hall problem at coffee break one day. A few workmates who hadn’t heard it before were skeptical at first, but after a while all agreed that it was better to switch doors (in the classic version.) However, one person still held that switching doors would be more profitable in the long run (over several tries), but that the better odds wouldn’t “matter” if you only played the game once. Mhuh? 🙂

I can’t think of another version of the problem, but what about the psychological dimension? Isn’t it more embarrassing to lose the car by switching doors (since you already “had” the right door once), than to lose by steadfastly sticking to your first choice? 🙂

4. #4 Greta Christina
December 27, 2006

Important question: Does Monty know which door the car is behind — and is he trying to psych you out into not picking it? In other words, is this a psychology/game theory problem, or just probability?

I spent waaaay too much time trying to figure out the write-up of the original version of this problem in the Straight Dope, and finally realized that I was thinking of it as pure probability when it was actually game theory. (Neither of which I’m great at, but at least with probability you don’t have to second-guess yourself ad infinitum…)

5. #5 Mustafa Mond, FCD
December 27, 2006

Who knew you could cheat at chess?

NEW DELHI, Dec 27 (Reuters Life!) – An Indian chess player has been banned for 10 years for cheating after he was caught using his mobile phone’s wireless device to win games, chess officials said on Wednesday.

6. #6 Dave S.
December 27, 2006

Mustafa:

There are lots of ways to cheat at chess. A buddy of mine at the World Open in Philly was playing a Grunfeld and watching a nearby game game where white was winning handily in a Kings Gambit. He noted the white player would get up after every move or so and walk over to his girlfriend to chat while black was thinking. Between moves, his girlfriend was engrossed in a chess book. My buddy decided to see what she was reading while her boyfriend was at the board, and sure enough, she was studying exactly the opening he was playing. He didn’t rat them out because he couldn’t tell if she was passing along info or merely just following along the game, since he didn’t hear what they were saying and couldn’t be sure she could not see his board. They immediately stopped whatever they were doing however when he smilingly asked her if she had a Grunfeld book too.

There’s also many other instances of covert computer use and assorted rigged draws/wins. Mantulovic was reputed to have sold his final round game against Tiaminov for a rumoured \$400 in 1970 Interzonal in Palma de Mallorca. What goes around comes around though…as Taimanov was blasted in the Candidates with 6 losses in a row against Fischer, a result which did not please the Russian federation.

7. #7 Zehra
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July 22, 2012