tldr: Always flip the switch

Edited with some of TauZero's suggested changes.

• 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.

• P(C|R∈{1,2,3,4,5}) = 1/10
• P(C|R=6) = 1
• P(C) = P(C|R∈{1,2,3,4,5})P(R∈{1,2,3,4,5}) + P(C|R=6)P(R=6) = (1/10)(5/6) + (1)(1/6) = 1/4
• P(R=6|C) = P(C|R=6)P(R=6)/P(C) = (1)(1/6)/(1/4) = 2/3
• P(R∈{1,2,3,4,5}|C) = P(C|R∈{1,2,3,4,5})P(R∈{1,2,3,4,5})/P(C) = (1/10)(5/6)/(1/4) = 1/3
• If you flip the switch, you have a 1/3 chance of dying.
• If you don't flip it, you have a 2/3 chance of dying.

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:

• If you flip the switch, the expected number of deaths is (1/3)*1+(2/3)*0 = 0.33.
• If you don't flip it, the expected number of deaths is (1/3)*0+(2/3)*10=6.67.

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.