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Hmmm...we're getting no where with Joe using words, Mister Faldage. What say we dazzle Friday with numbers?  Hey Joe, like you, a guy on the net couldn't see beyond the gumshoe logic that is endemic to those who are blessed with being average. But he was a hardheaded fellow and was dogged to try. These are his results.... _______________________________________ He did the sequence of two steps a billion times. And here are his results: " Number of times I stayed with the choice in step (2): 499,990,387
Number of times I chose to switch: 500,009,613
Number of times I was right when I stayed: 166,585,940
Number of times I was right when I switched: 333,391,697
Fraction right when I stayed: 0.333
Fraction right when I switched: 0.666." Yet he still can't intuit the process. ________________________________________ 
Last edited by themilum; 11/10/06 07:35 PM.
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Carpal Tunnel
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like I said, them, I follow the logic (even *with your overkill of that dead horse); it's just so dern counterintuitive (sympathy for those who still doubt).
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Let's try looking at the problem backwards to see if we learn anything.
Say we have 2 cups. One of the cups is on your right and one on your left.
Prior to your walking into the room, I have placed a gold coin beneath one of the cups. I tell you that l rolled a "fair" die and if it came up 1 or 2, I put the coin under the left cup; otherwise, I put it under the right one.
So if
L = coin is under left cup, and
R = coin is under right cup, then we conclude:
P(L) = 1/3 and P(R) = 2/3.
Question 1: Does this conclusion make sense? If not, then don't proceed to the next part.
Note: If you have gotten this far, it means you realize very clearly that the probability of the coin's being under each of the two cups is not the same.
Unfortunately, while I was distracting you with that question, someone decided to be clever. He moved the cups around and around and around. You turn your head back suddenly and you see the cups are jostled from their original positions.
You can no longer tell which cup is which.
What now is the probability that the coin is under the left cup?
I assert that P(L) = 1/2 and P(R) = 1/2.
Question 2: What is different about this situation from the initial state?
Answer: We have lost information.
Conclusion: Our calculations of probability are based on how much information we have about the situation.
Gaining or losing information changes the probability.
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In that example we have lost information and the probablities (may) have changed. In the three door monte problem we (may) have gained information. But we must remember, Monty can always open a door with a non-prize behind it. We don't know if he had a choice in which door to open.
And if I keep beating this dead horse it's only because it insists on getting up and whinnying all the time.
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"But we must remember, Monty can always open a door with a non-prize behind it. We don't know if he had a choice in which door to open."
Very astute observation - and this was one of the reasons that MVS had such trouble. People did not assume this game was exactly like the game show. In the game show, Monty Hall has perfect knowledge and we KNOW that he never turns over the grand prize - only the goat!
If MH's guess is random, the problem doesn't work out like this. It depends on the fact that his guess is not random. If his guess *IS* random, then the probability is what everyone else thinks it is.
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But if Monty opens a door with the prize behind it it is eliminated from the universe of discourse since the problem states that he reveals a non-prize when he opens the door.
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"But if Monty opens a door with the prize behind it it is eliminated from the universe of discourse since the problem states that he reveals a non-prize when he opens the door. "
Precisely. That is why it is prohibited in the correct statement of the problem that he should do so.
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Pooh-Bah
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Quote:
If MH's guess is random, the problem doesn't work out like this. It depends on the fact that his guess is not random. If his guess *IS* random, then the probability is what everyone else thinks it is.
If MH shows a door with the prize behind it then the contestant would pick that door! What makes the problem intersting is the counter-intuitive nature of the result favored by most mathematicians. The facile case in which MH reveals the prize thereby giving it away to the contestant (who after all may pick any door they wish) has no interest. Why do people bring it up?
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If Monty doesn't know and *does open the door with the prize and then those instances are eliminated, you *do have a fifty-fifty chance. If you assume he *does know then half the time he will have a choice of what door to open and half the time he won't and you're better off switching.
So, in a sense, either side of the argument is correct, depending on which assumption you make about Monty's knowledge.
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