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Sleeping Beauty Trolley Problem
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this post was submitted on 06 Nov 2023
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Except that if people are chosen randomly there is 2/3 chance that you are on the main track according to Bayes. Let's assume there are 10 people.
The probability to be chosen is 1/6 (all are chosen if 6 is rolled) + (5/6) × (1/10) (only one is chosen to go to the side track if 1-5 is rolled) = 15/60 = 1/4.
The probability that you are on the side track knowing that you have been chosen is the probability that you have been chosen knowing that the side track is selected (1/10) × the probability that the side track is selected (5/6) divided by the probability for you to be selected at all (1/4), so (1/10)×(5/6)/(1/4) = 20/60 = 1/3. So there is a 2/3 chance that you are on the main track.
If you do not flip the switch, (2/3)×10 = 20/3 people die.
If you flip the switch, 1/3 (you if on side track) + 10 × 2/3 × 9 / 10 (switch misfires 9 out of 10 times if on the main track) = 190/30 = 19/3 die. This is slightly better than not flipping the switch, you save 1/3 people more. That's an arm and a leg.
I wouldn't assume anything about the number of people and how anyone was chosen. All I know is I'm on the track.
But if you say you are on each track with p=1/2 then you also assume (different) details about how you were chosen
The task is unclear here, you have to make an assumption. I don't know which is intended.