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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”
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:
Yeah, if I woke up tied to train tracks and had someone explain that to me, I’d zone out and then panic because I had no idea what the fuck was going on
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.
Good question to ask, since specifics of selection process may affect the decision outcome! Other variants include growing humans in a vat from scratch on demand, using Star Trek transporter clones, or abducting the necessary number of people from a pre-selected list where your name happens to be the first one. For now, imagine the potential population as the 5 billion living cognizant adults.
as X approaches infinity, the difference becomes negligible
It may be negligible to the 4.999… billion adults sleeping comfortably and securely in their beds tonight, but the problem presupposed that you have already been abducted. It remains underdefined whether you refers to you the specific person reading this meme, or a more general you-the-unfortunate who has been chosen and is now listening to this on the headphones.
Noise cancelling earphones sucks at blocking voices. Just yell and ask if there are others.
Or spit, or blow air at your potential neighbor, or fart in their general direction!
If I’m tied to a train track any potential fart risks coming with a little extra mustard on it.
That’s assuming the villain who is trying to deny you information by the blindfold and earplugs was dumb enough to put them close together that a spit would reach a neighbor.
Exactly! Trying to think outside the box in a trolley problem is like wishing you could wish for more wishes in a genie problem.
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.
I am always surprised how my first guess gets wrecked by Bayes rule. I would have thought that there is 5/6 chance I am on side track and 1/6 that I am on the main track.
Excellent excellent!
If 6 is rolled, then P(X|R=6) = (N-1 choose 9)/(N choose 10)
Might as well reduce that to 10/N to make the rest of the lines easier to read.
If you don’t flip it, you have a 2/3 chance of dying.
There is also a chance that your switch is not connected and someone else has control of the real one. So there is an implicit assumption that everyone else is equally logical as you and equally selfish/altruistic as you, such that whatever logic you use to arrive at a decision, they must have arrived at the same decision.
No matter what your goal is, given the information you have, flipping the switch is always the better choice.
That is my conclusion too! I was surprised to learn though in the comment thread with @pancake that the decision may be different depending on the percentage of altruism in the population. E.g. if you are the only selfish one in an altruistic society, you’d benefit from deliberately not flipping the switch. Being a selfish one in a selfish society reduces to the prisoner’s dilemma.
There is also a chance that your switch is not connected and someone else has control of the real one. So there is an implicit assumption that everyone else is equally logical as you and equally selfish/altruistic as you, such that whatever logic you use to arrive at a decision, they must have arrived at the same decision.
Ah, yes. I forgot to account for that in my calculations. I’ll maybe rework it when I find time tomorrow.
I know it’s not the point of the question, but remember whatever happens is 100% the villain’s fault, not yours.
Spit to the left and right. If it spits back then there is someone else and I’m on the main track.
I don’t get it. What is the downside of pulling the lever? It’s if you don’t pull you die. If you do, you may not die.
It’s not like anyone else is laid on the other side of the tracks in either scenario so what’s even the dilemma here?
Also how are the dice relevant?
You are on the side track in scenario A. You die if you pull. Ironically, you’d be killing yourself. The dice are to make the two scenarios not equally likely.
The original phrasing could use some work
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there’s no way to know which track the trolley is on
It’s a standard trolley meme problem, the trolley will keep going on the main track unless the lever is switched 😁. I thought !science_memes would be familiar with trolley problems, but I guess I get to introduce some of you! You might want to start off on some easier trolley memes first, this is advanced level stuff.
where the real lever sends it
There is not usually ambiguity with the lever. If you wish, you can have an announcement in the headphones “main track… side track…” every time you flip the lever. Your only uncertainty is which track you yourself are bound to, given how you’re blindfolded.
there’s a 0.017% chance
1/6 * 10% = 1/60 = 0.01666… = 1.666…% ~= 1.7%! Careful there!
It’s not really a trolley problem, because in both scenarios a track is empty,
Everything is a trolley problem.
How are they going to explain it to me when I am blindfolded with noise canceling headphones?
Through the headphones :P
And dumb.
The time I understand the rules, the trolley is not there anymore.
The trolley is like really slow.
The real question is whether the song being played on the noise cancelling headphones is a foot tapper or not.
I’ll flip the switch back and forth as quick as I can in hope of catching trolly as it changes and causing it to derail.
Kills all 11 of em
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.
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.
Can I vote to move the one person to join the ten. Misery loves company.
I don’t get it. Do we know that the trolley is heading for the people or not? Do we know if flipping the switch moves it away from whatever track that the people are on? Or is it going in the main track in all instances unless you hit the switch?
I assume a villain would aim the trolley at the people, regardless of what track they’re on. That’s why they’re villains. So I would always flip it.
It’s a standard runaway trolley problem. The trolley is traveling down the main track unless the switch is flipped to send it down the side track. The lever is labeled such that there is no ambiguity which way it is set, the blindfolds notwithstanding. The villain is pernicious and will be equally (though not exceedingly so) delighted to see you die by your own action where inaction would have had saved you. You can somehow trust that the announcement in the headphones is true and not a lie. Such as, for example, you have seen this exact situation happen many times before on TV and survivors/witnesses have described the villain to be truthful every time.
Best change to get hit by the trolly is to not attempt to flip the switch. So I choose to do nothing
There’s a 5/6 chance someone is put onto the side rail (the one the trolley won’t go down without interfering with it)
There’s a 1/6 chance someone is put onto the main rail (the one the trolley WILL go down)
You’re more likely to be on the side track if you’re involved in this scenario, so if you wanna get hit you SHOULD try to flip it (if you’re the one on the side track, it guarantees a hit. If you’re one of the 10 people on the main, you have a 90% chance of having a dead switch and should try anyway)
Unless being tired at work is making me miss something
No, your math is wrong. The chance you’re on the main track is actually twice as high. Imagine it like this: When all numbers 1-6 would come up once, there have been 10 people overall on the main track and only 5 people on the side track
But 9 of those people can’t affect the outcome either way. So the chance that you’re on the main track and can affect the outcome in a positive way by flipping the switch is only 1/15.
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