Sunday, January 29, 2012

Monty Hall's flying risccu

Visiting one of the world's few remaining independent general bookstores prior to a concert, I found an entire book about The Monty Hall Problem (that's the book's title) by Jason Rosenhouse. I only had time to browse through it, and I decided not to buy it at this time, but it looks fascinating. Rosenhouse is interested in the mental blocks that cause even professional mathematicians not to get this one. And he's done diligent research, tracking down earlier formulations of the problem (not necessarily set in Let's Make a Deal) and reprinting most of the original Marilyn Vos Savant columns that raised the controversy. I was surprised to find that Vos Savant's original reply included what I've always considered the killer proof of the correct reasoning: Imagine that there are 100 doors instead of 3, and that Monty opens 98 of them to reveal goats before inviting you to switch. Now will you believe that the other closed door is more likely to hide the desired car than the one you picked at random out of a hundred?

Rosenhouse considers it necessary frequently to specify that, in the third of the cases where the contestant did pick the right door originally, Monty chooses which of the two wrong doors to open at random. I don't see why it makes any difference whether he picks it at random or not; it's going to be one of the wrong doors either way. But Rosenhouse waits for quite a while before he makes the one absolutely necessary non-obvious specification (the obvious ones being that Monty knows what's behind each door, that the contestant is not Porgy and does not prefer a goat to a car, and that nobody behind the scenes is switching the prizes during one iteration of the game) to make this work, which is that Monty makes the offer to switch and opens a door for every contestant. Because if he does it only with selected contestants, he could select for the ones who chose the right door originally, and the probability assessment goes out the window.

It appears that Rosenhouse is also going to discuss other similar problems. At the point I had to put the book down, because it was time to get where I was going, he was just about to explain why, if you have two puppies of sex unknown to you, the question, "If at least one of the puppies is male, what's the probability that the other one is also male?" has a different answer from the question, "If [specified] Puppy A is male, what's the probability that [specified] Puppy B is also male?", because in the first question, the second puppy is contingent on the first puppy in a way not the case in the second question, where they're totally independent variables. That's similar to the situation in the Monty Hall Problem, where the reason that switching will win 2/3rds of the time is that the probability is contingent on whether you picked the right or a wrong door to begin with.

Anyway, interesting-looking book.

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