|
|
|
|
|
by icambron
1475 days ago
|
|
Hmm, you are right. If the host chooses randomly, we get this tree: - 1/3 it is behind your door - 1 host picks a goat
- switching always loses -> 1/3 chance of losing
- 2/3 it is behind one of the other two doors - 1/2 host picks the goat
- switching wins always -> 2/3 * 1/2 = 1/3 chance of winning
- 1/2 host picks the prize
- result undefined -> 1/3 chance of ending up here
So you get 1/3 wins, 1/3 losses, 1/3 undefined. So given the goat is revealed, you have a 1/2 chance to win by switching.On reflection, this makes sense. The host is indeed leaking information by not be able to pick the around the prize. If the prize is not behind your door, his choice is constrained. So his revealing it supplies information about that constraint. If he is not constrained, then you are both picking doors at random. I'm sold. |
|
You are playing the classic Monty Hall problem. You pick door 1. Monty reveals a goat behind door 2. Monty asks if you want to switch. You know the correct strategy is to switch, so you are about to say "Yes - switch" when suddenly, the lights dim and announcer's voice booms over the speakers "You thought Monty was knowingly revealing something to you, but actually Monty just revealed a door at random. He had no foreknowledge." Are you now ambivalent about switching or staying?
If you still want to switch, and you should, that's because obviously it does not matter what Monty knew going into the problem. If you don't want to switch, please explain how the contents of Monty's brain affect the probability of which door conceals a car and which a goat.