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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.
I love the idea of the villain explaining the whole thing to the captive and at the end being like, “okay, I’m about to put the blindfold and noise canceling headphones on, so this is the last chance for any other questions about how this works”
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.
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.
There surely must’ve been a more comprehending way to phrase the dilemma.
Half the fun of trolley problems is adapting them to puzzles for which they are utterly unsuitable:
