Schrödinger was a physicist who devised a paradox to illustrate what he saw as a problem in a certain interpretation of quantum mechanics. It postulates that a cat is inside a sealed box along with poison and a radioactive source. If an internal Geiger counter detects radiation, the flask is shattered, releasing poison that kills the cat. The quantum mechanics implication is that the cat is simultaneously alive and dead until we look into the box.
The first time I learned of the paradox I thought, "Ho ho. What a doofus! Obviously, the cat is either alive or dead, not both." Since then I have learned not to mock people who are way smarter than I am, and Monty Hall had a lot to do with that.
The Monty Hall problem is simpler, I think. You're on "Let's Make a Deal". There are three doors. Behind two of the doors are goats, and behind one of them is a Subaru Outback. You hopefully choose Door #1. Then Monty opens Door #2, and you see that a goat was behind that door. Monty says that you can stay with Door #1, or switch to Door #3. What should you do? Does it make any difference? (Assume that you want the Subaru, not the goat.)
I thought that my extensive experience in games would help me solve the Monty Hall problem easily. (I actually developed my own "game theory" in college. It was, basically, "I can develop my mind better by playing this game than by studying.") I'd learned about a priori expectations while studying backgammon, and the principle of restricted choice while playing bridge. It turned out that both of those could be applied, but I failed to do that accurately. Even worse, after I learned the correct answer to the problem, I thought, "Ho, ho. What a doofus! Obviously, that answer is wrong!"
One serving of humility pie, coming right up.
The correct answer to the Monty Hall problem is that you double your chance of winning the car if you switch to Door #3. Specifically, if you stay with Door #1, your chance of winning the car is 1/3, the a priori expectation. But if you switch to Door #3, your chance of winning is 2/3. You are, in effect, choosing Door #2 and Door #3.
There are plenty of places on the web where you can learn why that answer is correct, including Wikipedia. Just for fun, though, I'll throw in a bit of a kicker.
The principle of restricted choice states that the play of a particular card is an indication that the player does not have an equivalent card. For example, suppose you need to know how the king and queen of a suit are distributed between your two opponents. Since both the king and queen will lose to the ace of the suit and beat all other cards, they are equal. If the opponent on your right (RHO) plays the queen, the odds become 2 to 1 that the player on your left has the king. That's because your RHO might have played either the king or the queen if he'd held both, but if he held just one of them, his play was restricted to that particular card. That's kind of a mind bender in itself, I suppose, but very handy when playing bridge.
Anyway, here's the kicker. Suppose Monty himself doesn't know which door the car is behind, so he opens one of the remaining doors randomly. Does that affect the solution to the problem?
(Yes, I realize that this is kind of a boring post, but it makes an important point in my ignorance series. I'll try to get a picture of Twila picking her nose or something, to liven up my next post.)
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