Let N be the size of the population that the villain abducts from
Let X be the event that you are abducted
Let R be the outcome of the villain’s roll
Let C be the event that you have control of the real switch
If 1-5 is rolled, then the probability that you are abducted is P(X|R∈{1,2,3,4,5}) = 1/N
If 6 is rolled, then P(X|R=6) = (N-1 choose 9)/(N choose 10) = ((N-1)!/(9! * (N-10)!)) / (N!/(10! * (N-10)!)) = 10/N
The probability of getting abducted at all is P(X) = P(X|R∈{1,2,3,4,5})P(R∈{1,2,3,4,5}) + P(X|R=6)P(R=6) = (1/N)*(5/6) + (10/N)*(1/6)
The probability that a six was rolled given that you were abducted: P(R=6|X) = P(X|R=6)P(R=6)/P(X) = (10/N)*(1/6)/((1/N)*(5/6) + (10/N)*(1/6)) = 2/3
So as it turns out, the total population is irrelevant. If you get abducted, the probability that the villain rolled a 6 is 2/3, and the probability of rolling anything else is its complement, so 1/3.
Let’s say you want to maximize your chances of survival. We’ll only consider the scenario where you have control of the tracks.
tldr: Always flip the switch
Edited with some of TauZero’s suggested changes.
So as it turns out, the total population is irrelevant. If you get abducted, the probability that the villain rolled a 6 is 2/3, and the probability of rolling anything else is its complement, so 1/3.
Let’s say you want to maximize your chances of survival. We’ll only consider the scenario where you have control of the tracks.
If you want to maximize your own probability of survival, you flip the switch.
As for expected number of deaths, assuming you have control of the tracks:
So to minimize the expected number of casualties, you still want to flip the switch.
No matter what your goal is, given the information you have, flipping the switch is always the better choice.