correct = numpy.random.choice(3, 10000)
(correct[correct != 1] == 0).mean()
This gives 0.49 or similar when I run it.
- 1/3 it is behind your door
- 1 host picks a goat
- switching always loses -> 1/3 chance of losing
- 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
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.
Let’s turn this around: explain why I should switch doors, but starting from scratch with the new problem, instead of by reference to the original. I think you’ll end up thinking the original solution is wrong, based on your no-telepathy rule, or you’ll see how they are different.
Or try this out. Let’s have another variant: before you pick the door, the host picks one that turns out to be a goat. Now you pick a door, and _then_ you have the option to switch. Do you still switch? Does that make sense? The situation is exactly the same (a goat behind one door and two closed doors). If both actions so far are random, it doesn’t matter what order they go in.