Sound and Fury

Signifying nothing

Why people are irrational and stupid. Part 1

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So this is the first in what I hope will be a regular feature on this blog highlighting some paradoxes and thought experiments that show that people are not very good at thinking rationally. I hope to convince you that people really do suck at probabilistic reasoning and sometimes even simple logical reasoning fails. This is motivated partly by misanthropy- I think people are stupid. And partly by my fascination with these kinds of puzzles. A third motivation is that I think epistemology based on rational degrees of belief (Subjective Bayesianism for example) is flawed. Particularly in cases like quantum mechanics. But more of that in a later post. For now I’ll stick to the Monty Hall problem.

Monty offers you three doors- A,B and C. Behind one door is a car and behind the other two doors are goats. The idea is to pick the door with the car behind it because then you win the car. But if you don’t want a car, or you do want a goat, imagine that behind one door is something you really do want and behind the other two doors are things you don’t want. Like one door could be a treasure chest full of gold doubloons and behind the other two doors are scurvy. Or behind one door is cake and behind the other two doors is death. Or behind one door is Hans Christian Andersen’s The little mermaid and behind the other two doors is Katie Price’s Mermaids and Pirates. Behind one door is Battleship Potemkin and the other two doors the Pokemon movie. You get the idea.

So you pick a door. WLOG assume you picked door A. Monty now opens one of the other two doors and reveals a goat (scurvy, whatever). Now, you are offered the chance to swap doors. Should you swap doors? The answer is that yes you should. The probability of your winning a car by swapping doors is higher than if you stick with your original door. That does seem a little counter-intuitive, does it not? Surely once one of the goat doors has been revealed, there are two doors remaining and one has a car behind it. 50-50? If you are thinking that, then you are irrational and stupid. That’s not how to think about it.

Think about it this way- 2/3 of the time, the first door you pick, door A, will have a goat behind it. In those circumstances, one of door B or C will have the car (doubloons etc) behind it. So Monty won’t be able to open that door, he’ll have to open the other one. Which means switching will give you the car door. That happens 2/3 of the time… So you should always switch.

To make this even clearer, imagine there are 100 doors. 99% of the time you pick a goat door. Now Monty opens 98 goat doors. To leave you with your door and one other door remaining. So switching doors will net you a car 99% of the time.

There are two strategies. Either always switch or never switch. (OK, there are strategies where you randomly switch with probability p, but trust me setting p = 1 is optimal…) So two options. If you decide to never switch, you get the car 1/3 of the time – those times you pick the right door first time. Always switching, you’d think, would get you the car at least 50% of the time. In fact it’s even better than that. It gets you the car 2/3 of the time. Because if the only strategies are switch or don’t switch, and the only outcomes are win or don’t win; if one strategy wins 1/3 of the time, the other strategy has to win the other two thirds of the time.

Why isn’t it 50%? Because having Monty open a door tells you more about the door you haven’t picked than it does about the door you have picked. If you still aren’t convinced and think it is still 1/2, I suggest we meet up and simulate the game with playing cards, or play the card version of Bertrand’s box paradox. For money. I promise you, if we play for long enough, I will be able to buy myself a car with the money I swindle out of you. Except that I can’t drive. So I’d be looking to buy a chest full of gold. YARRR.

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Written by Seamus

August 9, 2008 at 5:33 pm

Posted in maths, paradox

Tagged with , ,

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