So let’s see if I can reason through this: A is steak, B and C are not. We pick A, wife picks B, we switch: lose if switch. A, C: lose if switch. B, A: wife informs us of steak, we find steak. B, C: win if switch. C,A: excluded. C, B: win if switch.
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So if we exclude the world states in which this tweet doesn’t happen because your co-experimenter just hands you the steak, “stay” and “switch” are both right 50% of time. *Not* the Monty Hall problem.
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What they could have done doesn't matter? Initially, 1 in 3 chance of getting steak. After the bean reveal, 1 in 2. Isn't that enough to justify switching as a good strategy?
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No, each of his possible picks if 50% possible. In original, switching is win 2/3, vs staying is 1/3. So, in this case, same advantage not checking.
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There’s no co participant here? Seems like exactly the Monty Hall problem
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I think it is after she tells him.... His odds go from 1/3 to 2/3
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But obviously if she had picked steak it wouldn’t be.
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The way to think of the Monty Hall problem is the fist pick is not of “C” but of “A or B”, knowing that the host will eliminate one of the two for you. I think this only works when the elimination is based on knowledge.
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If Monty doesn't know which door is the car, it is not the Monty Hall problem. If beans are correlated with not-steak then it is an even more Bayesian version of it.
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