In the car today, I figured out that the Monty Hall Problem is correct. Obviously it is correct, but what I mean is that I figured it out for myself. I had never really believed it myself, even though I knew people smarter than I had figured it out.
Suppose you ask someone to choose between a, b, and c. They have a one third chance to choose correctly. Now, suppose a is the correct answer, you reveal that one of the incorrect answers they haven't chosen is, in fact, incorrect. (So if they choose a, you say that b or c is incorrect. If they choose b, you say that c is incorrect. If they choose c, you tell them that b is incorrect and let them guess again.)
Should they, at this point, change their answer? They have a one third chance of having correctly guessed a in the first place. So people who do not switch at this point have a one third chance of getting the right answer. People who have already chosen a wrong answer, though, will automatically win by choosing the other example. So 2/3 of people who switch will get the right answer.
In other words, you should switch on the assumption that you are more likely to be in the 2/3 of people in the same situation who guessed wrong than in the 1/3 who guessed right. You are exactly twice as likely to be in that position.
This is counter-intuitive to a lot of people, including me, before today. It seems that eliminating one possibility makes it a 50/50 guess. Since you've already made a guess, switching back doesn't seem to improve your chances. After all, you had no reason to prefer a to b or b to c before one of these possibilities was eliminated.
I found a way of making it much clearer, though.
Suppose you ask someone to guess a number between one and ten, a number you have already written on a piece of paper. They guess the number, and you tell them that it is either the number they have guessed, or "6." Should they change their guess at this point? (What you have done is thrown out 8 of the remaining wrong answers. If they guessed right, you throw out 8 numbers. If they guessed wrong, then you throw out 8 numbers. 6 is either the right answer or the one wrong answer you didn't toss out.) If they guessed right the first time, then switching to 6 will make them lose. But if they were wrong, then 6 is necessarily the right answer. So not switching their answer gives them a 10% chance of winning, and switching gives them 90%.
The numbers are different, but the logic is the same as in the Monty Hall problem. It is just easier to see when the numbers create an even less advantageous position for the person who doesn't want to switch.
The Wikipedia article has a version with a million doors. You pick one and the host now opens all the remaining doors except one. Would you switch? Of course you would. The three door version of the problem is the same, though even after seeing the connection I have to admit that my "intuition" that switching in the three-door case is pointless remains. It's a very interesting example of the tension between statistical and commonsense reasoning.
Long way from Lorca, though! What got you thinking about it?
I am interested in things that are counter-intuitive. How our intuitions mislead us. It doesn't seem like giving a wrong answer that one did not choose gives us any insight into whether we might have chosen right or wrong the first time, but it does.
I am interested in all kinds of cognitive biases. I'm not sure why I was thinking of it yesterday, per se. I guess it bothered me that I did not understand it so I had to work it out myself. I am also interested in Zeno's paradox and the prisoner's dilemma. Go figure.
This is pretty cool simulator:
Yes, the prisoner's dilemma is an amazing (sometimes amazingly depressing) puzzle. It is especially interesting (sometimes distressing) that social scientists/engineers have understood it for decades. ("It was originally framed by Merrill Flood and Melvin Dresher working at RAND in 1950.") Sometimes I'm convinced we live in that prison. No amount of poetry will get us off its horns.
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