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I got out three playing cards and did the experiment myself over and over. That was many years ago and now whenever I see something on the Monty Hall problem it

by c3534l 6y ago

I got out three playing cards and did the experiment myself over and over. That was many years ago and now whenever I see something on the Monty Hall problem it seems so obvious as if I can no longer even see why people think its unintuitive. The reason its unintuitive is that people don't really have experience with anything that works this way. I'm not sure that the linked explanation will be that helpful to people because of that, although it does make it plainly obvious that Monty Hall knows which door has the prize and he's telegraphing to you which door it might be by only opening doors which don't have it in there. I suppose if that's the missing part of the puzzle for you, then that will help.

OJFord 6y ago
I've always thought of it like the 'denominator' of probability for the second stage became 2/3 (vs. the usual/first stage 1). So when he reveals a goat door you know that door is certainly not a winner, 0 probability (of 2/3), and the switch door is certainly (if it were one of those) the winner, 1 probability (of 2/3). But I still used to manage to confuse myself thinking it's intuitively 'more likely' to be the original door 'now' that it isn't one of the others. Until I studied information theory at university and it really clicked - Monty's door choice has lower entropy than your initial pick! (But I acknowledge you can't say to the masses 'look look let's simplify this, if we just step back and take an information theoretic approach -')
dingaling 6y ago
> I suppose if that's the missing part of the puzzle for you, then that will help. Of course it will help, because it changes the entire scenario from two people playing a game of chance to one person playing a game of deduction against secret knowledge. The odds don't change because of some quirk of the Universe, they change because one player is changing the parameters due to his knowledge. > I got out three playing cards and did the experiment myself over and over. How did you do that when you didn't know which cards were 'goats'?

by c3534l 6y ago

OJFord 6y ago

I've always thought of it like the 'denominator' of probability for the second stage became 2/3 (vs. the usual/first stage 1). So when he reveals a goat door you know that door is certainly not a winner, 0 probability (of 2/3), and the switch door is certainly (if it were one of those) the winner, 1 probability (of 2/3). But I still used to manage to confuse myself thinking it's intuitively 'more likely' to be the original door 'now' that it isn't one of the others. Until I studied information theory at university and it really clicked - Monty's door choice has lower entropy than your initial pick! (But I acknowledge you can't say to the masses 'look look let's simplify this, if we just step back and take an information theoretic approach -')

dingaling 6y ago

> I suppose if that's the missing part of the puzzle for you, then that will help. Of course it will help, because it changes the entire scenario from two people playing a game of chance to one person playing a game of deduction against secret knowledge. The odds don't change because of some quirk of the Universe, they change because one player is changing the parameters due to his knowledge. > I got out three playing cards and did the experiment myself over and over. How did you do that when you didn't know which cards were 'goats'?