Okay, apparently everyone in the world is aware of the Monty Hall Paradox except for me. Warning: If you're a Perplex City player, the link above and the discussion below will spoil card #174 Jaunty Paul for you, which is what this is coming out of. But.. if you're not a probability nut, and I mean a nut, and you haven't seen this before, you might do yourself a favour to read on anyway.
It goes like this: You're on a gameshow, and you're asked to choose between three doors. Behind two of the three doors are goats. Behind the other door is a yacht. You get to choose one door, and you win whatever is behind it. You pick door A. The gameshow host steps foward before you open the door, and says, "Are you sure about that choice? Do you want to change your mind?" He opens door C, and shows that there is a goat behind door C. The gameshow host always opens a door and reveals a goat at this stage in the program, regardless of whether the player has chosen the correct door.
The question: What are your chances of winning the yacht if you stick with door A? What are your chances if you switch to door B?
So my first answer was, "That's easy, it's a 1 in 2 chance no matter whether you stick with A or go with B." Wrong.
So my second answer was, "Well, you could say it's a 1 in 3 chance, no matter which you choose. But it's still even odds." Wrong.
The correct answer is: you have a 1 in 3 chance if you stick with door A, but your odds improve to 2 in 3 if you switch doors. To put it another way, you are more likely to win the yacht if you switch. A player who always switches at this point in the routine will win the yacht 2 out of 3 times.
That is, on your first guess, you have a 2 out of 3 chance of being wrong, so when the choice is offered, and you have (more likely than not) chosen the wrong door, it makes sense to jump ship to the other one.
Huh? Arrgh! My brain hurts.
The Wikipedia entry gives a good illustration:
It may be easier to appreciate the result by considering a hundred doors instead of just three. In this case there are 99 doors with goats behind them and 1 door with a prize. The player picks a door; 99% of the time, the player will pick a door with a goat. Thus, the chances of picking the winning door at first are very small: only 1%. The game host then opens 98 of the other doors revealing 98 goats and offers the player the chance to switch to the only other unopened door. On 99 out of 100 occasions the other door will contain the prize, as 99 out of 100 times the player first picked a door with a goat. At this point a rational player should always switch.Well, yes, that sounds good when you say it like that, but I still don't see how it squares up with the fact that there are two doors, and the yacht is just as likely to be behind either one. The probability that I've chosen the wrong one doesn't affect the probability that a given door hides the yacht.
And this is why I parted ways with maths in the eighth grade and have never seen eye to eye with it since. Oh, and I solved the damn Perplex City card. Grr.