Coding Horror: Monty Hall, Monty Fall, Monty Crawl

Re:Coding Horror: Monty Hall, Monty Fall, Monty Crawl by lacker on Monday June 22, 2009 @02:47PM (#669229)

I doubt Erdős was really confused. The Monty Hall problem is complicated because usually the person explaining it tries to make it complicated by leaving out necessary information. Specifically, they often leave out that Monty always opens an empty door and his choice of which empty door to open is done at random. Without those facts (which some but not all people take for granted) there is not enough information to solve the puzzle.

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Re:Coding Horror: Monty Hall, Monty Fall, Monty Crawl by callahad on Monday June 22, 2009 @03:31PM (#669304)

Why is Monty's choice of which empty door being random important? Isn't the fact that he always opens an empty door alone sufficient?

If you select the correct door, switching will yield an empty door.

If you select an empty door, switching will yield a prize.

You have a 2/3 chance of selecting an empty door, thus, 2/3 of the time, switching will grant you the prize.

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Re:Coding Horror: Monty Hall, Monty Fall, Monty Crawl by anatoly on Monday June 22, 2009 @05:19PM (#669460)

The reason it's important is that the problem, as usually phrased, is a conditional probability problem. Say you picked door number 1. Given that Monty opened door number 3 and it's empty, what's the probability you'll win by switching?

Now suppose that Monty will always open the empty door as advertised, but he will also always prefer to open a door with the smaller number, if he has a choice of two empty doors to open. If that is how Monty always acts, then given that he's opened door number 3, your probability to win by switching is 100%, and not the usual 2/3. If he's opened door number 2, then your probability to win by switching is 1/2, not the usual 2/3.

If Monty will always choose which of the two empty door to open by random, then the probability to win by switching is indeed 2/3.

There is a way to phrase the problem differently, so that it becomes an unconditional probability problem, and then it doesn't matter how Monty chooses the empty door to open. One way to enforce this interpretation is to phrase it as follows: Suppose some player always chooses to switch. Over many attempts, what will be the approximate proportion of his wins? The answer will be 2/3. But that's not how the problem is usually phrased.

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Re:Coding Horror: Monty Hall, Monty Fall, Monty Crawl by callahad on Monday June 22, 2009 @11:38PM (#669826)

Oh, wow, you're completely right. That connection eluded me; thank you!

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Re:Coding Horror: Monty Hall, Monty Fall, Monty Crawl by nollidge on Monday June 22, 2009 @04:26PM (#669388)

Why has no one phrased the answer like that before? That makes way more sense then talking about "adding information". I never grokked the solution until you described it that way.

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Re:Coding Horror: Monty Hall, Monty Fall, Monty Crawl by lacker on Monday June 22, 2009 @06:13PM (#669510)

Your logic is correct but you are a making a different assumption, that Monty treats each door symmetrically. For example let's say Monty's strategy is to always open door #3 if he has a choice. Then if your choice is to switch to door #3 or to stay, you should choose "stay".

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