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Comment by icambron | Hacker News Reader

The choice happens after the door is opened. If in this variant you are allowed switch to the revealed prize door, you win 100% of the time the prize turns out to be behind the door they open. If you aren't allowed, you lose 100% of the time that happens. So that new "branch" doesn't affect any decision you could make.

But given they do open a door with a goat, the probability is the same as the regular Monty Hall problem. So the parent's point stands: it's not the host's knowledge that makes the game "work". He's leaking information by not opening your door, thus "concentrating" the 2/3 chance it wasn't behind your door into the one remaining door.

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It’s not the same. Try stimulating it out if my argument above didn’t convince you. WLOG if you pick door 0 out of 0,1,2 and host picks door 1 always, then the the relevant value is in numpy:

correct = numpy.random.choice(3, 10000)

(correct[correct != 1] == 0).mean()

This gives 0.49 or similar when I run it.

Then you aren't simulating the problem statement.

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.

That person, and, sadly, now you too, are completely wrong. What the host knows or does not know is completely irrelevant. Imagine:

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.

You make the decision based on your priors: the conditional probability of the prize being behind a certain door is updated by the new information. The information content of that update can certainly be affected by your knowledge of his constraints. In the original Monty Hall problem, that knowledge you have is “he can’t reveal the prize”. There is nothing magic about “the rules were x, and given that the rules were x, my update of the probabilities is y”. It has nothing per se to do with his mental state; it has to do with the rules he had to follow, and what you can infer from them.

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.

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