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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:
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.
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.
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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.
I know it’s not the point of the question, but remember whatever happens is 100% the villain’s fault, not yours.
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
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.
Can I vote to move the one person to join the ten. Misery loves company.
Thank you! <3 My guess is that no to the first, since I have a 1/3 chance of being in the forked path, vs 1/15 of being in the straight path and my lever being connected. However, in the second situation I would flip it, since I’d only kill 1/3 of all people (myself every time), versus 2/3 (myself included) if I don’t flip it.
My guess is that no to the first, since I have a 1/3 chance of being in the forked path, vs 1/15 of being in the straight path and my lever being connected.
Suppose you live in a kingdom where everyone is as selfish as you, and you’ve seen on TV many situations exactly like this one where people were tied to the tracks - usually one at a time and occasionally 10 at a time. (The villain has been prolific.) You’ve seen them all follow this logic and choose not to flip their switch, yet out of ~1500 people you have seen in peril this way, ~1000 of them have died. If only their logic had convinced them (and you) otherwise, 1000 of them could have selfishly survived! Doesn’t seem very logical to follow a course of action that kills you more often than its opposite.
(If you don’t want to imagine a kingdom where everyone is selfish, you can imagine one where x% are selfish and (100-x)% are altruistic, or some other mixture maybe with y% of people who flip the lever randomly back and forth and z% who cannot even understand the question. The point is that the paradox still exists.)
Edit: I can see now how in a 100% altruistic kingdom, where you are the only selfish one and you know for sure that everyone else will logically altruistically pull the lever, it makes sense for you to not pull the lever. Presumably there is some population x% split (44% selfish/56% altruistic?) where your selfish decision will have to reverse. Weird to think that your estimate of the selfishness of the rest of the population has a relevance on your decision!
It’s a really cool puzzle, nice job! The solution being a huge prisoner’s dilemma makes it all the more interesting and deep. I guess an iterated version resulting in collaboration would be difficult in this particular case, though ;)
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.
There’s a 1/6 chance that scenario B happens, but that scenario involves 10 people, and the only thing you know is that you are one person strapped to the rail, so the chance that you are strapped to the main rail is P(main) = P(B|abducted) = P(abducted|B) * P(B) / P(abducted) = 10 * P(abducted|A) * 1/6 / P(abducted).
We can do the same for P(side) = P(A|abducted) = P(abducted|A) * P(A) / P(abducted) = P(abducted|A) * 5/6 / P(abducted). Then, P(main) / P(side) = 10 * 1/6 / 5/6 = 2. Since P(main) + P(side) = 1, then P(main) = 2/3 and P(side) = 1/3.
Edit: typo
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I am confused as to why anyone would not flip the switch? Flipping the switch seems to have somewhere between a 10% and 100% chance of saving your life, and not flipping the switch seems to guarantee death?
Is there some kind of penalty to flipping the switch that I am missing?
Or is the drawing misleading, and in Scenario B, there is also supposed to be a person drawn on the other track?
I think in Scenario A you die if you flip the switch.
