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165

Jigsaw Trolley Problem (lemmynsfw.com)

submitted 1 week ago by [email protected] to c/[email protected]

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[–] [email protected] 1 points 1 week ago (1 children)

Knowing that the ball was gold gives you Bayesian knowledge about the boxes behind the door, since the prior probability of the host pulling a gold ball from a 6-gold door is different than from the 3/3 door. So you have to multiply Monty Hall probabilities and Bayesian probabilities together.

That assumes the host pulled a ball at random, of course, and not a deliberately gold ball.

[–] [email protected] 2 points 1 week ago (1 children)

Hmm, you're quite right. My intuition is that the Bayesian portion would exactly offset the Monty hall portion. I think, at a glance, Bayes would give door 1 a 2/3 probability of having 6 gold, but Monty Hall would give door 2 the same probability, so we can effectively cancel these out and just consider a raw probability

You either have 5 gold or 2 gold 3 silver behind door 1, and 6 gold or 3 and 3 behind door 2, which gives door 2 a very slight edge. Does that check out?

[–] [email protected] 2 points 1 week ago (1 children)

Yes! Cancels out, leaving only a very slight edge on door 2. All that work for only... 2.77% edge over picking at random. What a troll problem, huh?

[–] [email protected] 3 points 1 week ago (1 children)

Heh what a trolly problem indeed

[–] [email protected] 1 points 4 days ago

Speaking of trolls I assume the madman has perfect knowledge of the layout, hence did not pick a ball out at random.