Noise cancelling earphones sucks at blocking voices. Just yell and ask if there are others.

Most people would fail to understand the question. So many will flip the switch a bunch of times randomly. In other words: This would be super frustrating for the villain.

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.

There surely must’ve been a more comprehending way to phrase the dilemma.

Info: How many people live in the kingdom?

There’s a five in six chance you are picked 1 out of X, and a one in six chance you are 10 out of X.

If you’ve been picked, there are three possible outcomes.

Flipping the lever kills you. 5/6 x 1/X

Flilling the lever saves you and 9 other people. 1/6 x 1/X

Flipping the lever does nothing at all. 1/6 x 9/X

From a purely statistical standpoint, you’re five times more likely to die flipping the lever, but the expected value, measured in lives saved, for flipping the lever is twice as high as not.

From a purely altruistic measure, you should always flip the lever, because at worst you kill yourself, at best you save 10 people, and you can do it with significant confidence that it doesn’t actually matter.

But back to my original question, 5/6X vs 1/6X vs 9/6X where as X approaches infinity, the difference becomes negligible.

From a purely utilitarianism perspective, assuming all utility is linear and unscaled:

5/6 chance I’m on the side track * 1 person saved = 5/6

1/6 chance on the main track * 1/10 chance my switch is real * 10 people saved = 1/6

Seems pretty clear that you should not flip the switch. However, if I am on the main track, this thinking will lead to no-one flipping the switch and no lives saved whereas everyone thinking it will lead to a guaranteed save -> utility of 10/6.

If I can assume more than half the people can be rational and will think like me then I should flip the switch.

If I cannot, I should not flip the switch.