If you have to pick one of three doors then the probability of being right is 1/3.
If you have to pick one of two doors then the probability of being right is 1/2.
Later I might write a script to test the Monty Hall problem. I believe prior experiments have supported the hypothesis. — Michael
Or what if we wake her on Monday if it's heads and wake her on Tuesday and Wednesday if it's tails, ending the experiment on Thursday? Is P(Awake) = 3/6? — Michael
If whether each happens is determined by the toss of a fair coin, they're all equal and 3/6 is right. (If the subspaces had unequal probabilities, you might still be picking half of them numerically but they wouldn't add up to half the space.) — Srap Tasmaner
What do you think this changes? — Srap Tasmaner
Dude, if she's not sleeping we wouldn't call the condition "awake". What matters is when she's asked. — Srap Tasmaner
Here's the result of 100,000 games where you don't re-guess. 1/3 games are winners
Here's the result of 100,000 games where you re-guess. 1/2 games are winners
Go down towards the bottom and hit "Execute". — Michael
The probability of winning, when you choose 2 doors is 2/3. That is all there is to it. The fact that one of these doors is open is irrelevant. One door can always be opened, because one of them is certain to be empty. — tom
If the door you had selected has the car behind it (call this event C) then it is not relevant information.You, as the contestant, know for certain that one of the other two doors is empty, once the door is opened, you still know for certain that one of the doors is empty. OK, you now now which one is empty, and that IS information of sorts, but is it relevant information? — tom
She is never told it is Monday, each awaking is the same, there is no hint as to which day it is; temporally she is uncertain of her location. — Jeremiah
You, as the contestant, know for certain that one of the other two doors is empty, once the door is opened, you still know for certain that one of the doors is empty. OK, you now now which one is empty, and that IS information of sorts, but is it relevant information? — tom
All she knows is that she is awake, and that is twice as likely to be associated with tails. — tom
P(Heads|Awake) = P(Heads) * P(Awake|Heads) / P(Awake)
If she applies this before the experiment then she knows that P(Heads|Awake) = 0.5 * 1 / 1 = 0.5. — Michael
↪Andrew M She is never told it is Monday, there is no relevant self-locating information, and she knew there were only three possible awake periods before the experiment. Everything we know is everything she knows before the experiment therefore 1/3 is a prior. We don't have any privy information here. — Jeremiah
Oh good. I thought the OP didn't make this clear. The halvers would have it right if she went into this thing blind.She is never told it is Monday, there is no relevant self-locating information, and she knew there were only three possible awake periods before the experiment. Everything we know is everything she knows before the experiment therefore 1/3 is a prior. We don't have any privy information here. — Jeremiah
You're doing it wrong. You are mixing probabilities from different times, different points of view, or from positions with different information.I’ve showed the reasoning multiple times. It’s the Kolmogorov definition of conditional probability. — Michael
Do you agree? — Andrew M
Beauty conditionalizes on being awakened, so the values change to 1/4 / 3/4 = 1/3. — Srap Tasmaner
Oh good. I thought the OP didn't make this clear. The halvers would have it right if she went into this thing blind. — noAxioms
Read the article to find out why it has a plus. — Jeremiah
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