Tuesday, April 15, 2008
Scott Adams gets it wrong
Scott Adams, creator of Dilbert, makes up funny sounding BS for a living. Not a bad job, if you can get it. But sometimes he get carried away on his blog. He describes the Monty Hall Problem thusly:
The set up is this. Game show host Monte [sic] Hall offers you three doors. One has a car behind it, which will be your prize if you guess that door. The other two doors have goats. In other words, you have a 1/3 chance of getting the car.
You pick a door, but before it is opened to reveal what is behind it, Monte opens one of the doors you did NOT choose, which he knows has a goat behind it. And he asks if you want to stick with your first choice or move to the other closed door. One of those two doors has a car behind it. Monte knows which one but you don’t.
The trick here is that most people assume it makes no difference if they stick with their original choice or move to the other door. They believe the odds are 50% either way, since there are only two choices and you don’t know anything about either choice. But mathematicians say that is wrong. You substantially increase your odds by switching doors.
That is interesting enough on its own. (I’ll give a link later that explains the math of it.) But here is the freaky part. You only improve your odds by switching doors if Monte Hall knows what is behind each door. If he simply got lucky and opened a door with a goat behind it, your odds are unchanged. In other words, your odds are changed by Monte’s knowledge, and your knowledge that Monte has that knowledge.
If reality were objective, statistics wouldn’t be influenced by knowledge. That means your world is either partly created by your mind, or you are a hologram created by some other mind, and there are a few bugs in the software.
Nope, sorry, wrong. The odds do not change based on Monty's knowledge. Let's do the math, and for simplicity, let's assume that you picked door #1 initially and that, in each case, Monty "got lucky" and opened a door with a goat behind it without knowing beforehand:
|Door #1||Door #2||Door #3||No Switch Result||Switch Result|
If you happened to pick the car initially (1/3 chance), switching to the other (unopened) door causes you to lose. In the other two cases, when you picked a goat initially, switching causes you to win. There is no difference in the odds between Monty choosing a door with a goat behind because he knew there was a goat there or because he got lucky. As long as you know that Monty will always open a door, and that the door that he opens is a goat, (either through luck or foreknowledge) you should switch. The odds would be different, if, for example, Monty only offered a chance to switch if the player initially picked the car, thus rendering the switch-strategy a lose-only proposition.