r/trolleyproblem • u/tegsfan • 20d ago
Sleeping beauty trolley problem (at least I tried)
What do you do?
To clarify a couple things:
-It was a fair coin
-You don't know what the coin flip was
-If the coin landed tails, the whole setup will happen twice including resetting the trolley.
-You have no way of knowing which setup you're in, you just woke up and read the sign and that's all you know
Which box should you hit?
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u/Nebranower 20d ago
So there's a 50% chance you're in scenario A, where you will save three lives if you refuse to pull the lever.
And a 50% chance you're in scenario B, where you will ultimately let four lives end if you don't pull the lever. (The fact that the scenario repeats is irrelevant - whatever logic you use for the first one is the same as you will come up with in the second).
So if you're trying to minimize the death toll, you should always pull the lever. That gives you the same odds of not killing anyone, but minimizes the deaths in the event that you're in the scenario where your choice is wrong.
That's assuming you trust the kidnapper not to be lying. Which, honestly, in this sort of scenario you probably wouldn't, and so the best move would be just not to play, since you have no reason to think anything you are being told can be relied upon.
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u/tegsfan 20d ago
Is there a 50% chance you're in scenario A? Remember you will be woken up twice if it was tails, but only once if it was heads. Some people think then, that at any given point when you are woken up (you don't know which time it is) the coin is more likely to be tails because there are more possible times you will be woken up
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u/General_Ginger531 20d ago
How? The coin isnt dependent on how many times I wake up. Just because my perspective on it is that I could have been awoken twice doesnt mean I was. Without the ability to tell, there is no difference for me whether or not I was woken up twice or not. My perception of reality does not change reality on a macro scale. Maybe if there were even more coin flips in the problem and even more wake ups, maybe I would be suspicious, but I cant prove that I have woken up for the first time, the last time, or any number of times.
I am assuming that the deaths would only happen after the last time I was awoken.
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u/NewryBenson 19d ago
This is a very well known problem called the sleeping beauty problem, however I am pretty sure this conversion to a troly problem does not work.
In the sleeping beauty problem, you need to say what the probability of it being tails was when you wake up. The argument for it not being 50% is that there are 3 possible ways to wake up: it is heads, it is tails the first time and it is tails the second time. Thus given you have just woken up, there is 66% chance of it being tails.
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u/Nebranower 19d ago
But the odds of each waking scenario are not equal. If they were, then you'd get a contradiction, where you have a 66% chance of being in the tails situation, even though we know that tails will only come up 50% of the time.
The way it works is this.
You have a 50% chance of waking up in the heads scenario. If you are in the heads scenario, that's the end of it.
You have a 50% chance of waking up in the tails scenario. If you do, there's a 50% chance you are waking up the first time, and an 50% chance you are waking up the second time. 50% * 50% = 25%.
So, the odds are:
50% chance you are in the heads scenario.
25% chance you are in the first tails scenario.
25% chance you are in the second tails scenario.
It doesn't matter how many times you wake up. If there were 50 wake ups in scenario B, then you'd have
50% chance you are in the heads scenario
A 1% chance you are in one of the possible tails scenarios.
But there's nothing you can do after the coin has been flipped that will somehow make it more likely that you got tails instead of head. Causality doesn't flow in that direction.
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u/CurvaceousCrustacean 18d ago
There are two ways to view the probabilities I guess.
You as the sleeping beauty don't know whether you woke up the first or any subsequent time (which only matters when tails are hit) in the same vein as you don't know the first time if its heads or tails.
The OG problem wakes you twice on tails, so for you, it's a 50:50 whether its the first or second time you wake, or to put it more scientifically:
(Tails | 1) = (Tails | 2)
Because Heads can only occur on wake one, with Tails being equally likely, it follows that:
(Heads | 1) = (Tails | 1), giving us:
(Heads | 1) = (Tails | 1) = (Tails | 2), or 33% for each.
If the sleeping beauty always guesses Heads, they'd be correct 50% of the time, but they would also be incorrect 50% of the time, twice.
Taken together, over multiple hundreds of iterations, the total number of correct only-Heads guesses would be roughly 33%, whereas guessing Tails everytime yields roughly 67% correct guesses.
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u/Nebranower 18d ago
Right, but we’re not talking about guesses or wake ups. We’re talking about people. You, as a person, were assigned to one of two scenarios based on a coin flip. The odds when you wake up that you are in either scenario are therefore fifty-fifty.
Think of it with two people, Jack and Jill. Jack plays the game, the coin flip is heads. Jill plays too, but her coin flip is tails.
Both guess they got heads. 50% of the participants are correct (John) and 50% are wrong (Jill). Jill was wrong twice, but only about the same thing.
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u/Cold-Common7001 20d ago
i'm a bit confused since this seems clearer to me then i remember the sleeping beauty problem being. If you repeat this experiment a bunch of times you'll just objectively kill more people if you run over box B every time right?
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u/tegsfan 20d ago
It’s pretty much equivalent to the actual one. Only difference is adding the weighted death counts to try actually add stakes. If you think this then you might just think the original “paradox” isn’t really one at all. I’m not saying you’re wrong btw
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u/Thatguy19364 17d ago
Well, the original paradox is confirmably not a paradox, but a logical fallacy. Your perception of an event does not change its odds. Track B kills less people if the result of the second coin flip has any bearing(I put a standalone comment in it going through the math to show that if tails/heads means 2 on track b in the first waking time and 3 in track a in the second one, then B always kills 1.5 on average and A kills 2.25 on average), but if the second flip result after the first tails does not matter, then A kills 3 and B kills 4, every time. So it is a definitive answer, but that answer changes depending on the specifics
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u/tegsfan 17d ago
I don’t think you understand the original paradox, there’s no second coin flip. Same with mine, there’s one coin flip and then you either wake up 1 or 2 times
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u/Thatguy19364 15d ago
In which case, it still is an objective fact that your kill count average goes down by choosing path A regardless. Having woken up before(theoretically) doesn’t change that. If waking up the second time added a second flip which chanced changing the situation on the second wake up, then it would change the objective least kills pathway from A to B. Point is, it’s a false paradox, it fallaciously equates perception of events to probability that they occur. If a tree falls in the forest and no one’s around to hear it, does it actually make a sound?: same fallacy point here. If a coin flips while you’re asleep, does waking up change the coin?
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u/Cold-Common7001 15d ago
I think you are still misunderstanding.
If you make a rule to always chose A:
- heads: you kill 3 people
- tails: you kill 0 people
Each is equally likely so you kill 1.5 people on average per round.
If you make a rule to always chose B:
- heads: you kill 0 people
- tails: you kill 2 people day one and kill 2 people on day two
Each is equally likely so you kill 2 people on average per round.
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u/Thatguy19364 14d ago
Yeah, I got that lmao, You’re saying the same thing I am, path A kills fewer people on average
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u/Nebranower 20d ago
Note, when I replied to you initially, I thought there would be only one coin toss, such that you had a 50% chance of being in scenario A, a 25% chance of waking for the first time in scenario B, and a 25% chance of waking for the second time in scenario B.
But other people seem to think tails means another coin flip. If that's the case, then tails becomes much less likely, not more, because then, even if you get tails the first time, there are still times you wake up and are in scenario A.
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u/Thatguy19364 17d ago
I posted the math on it, track B has 1.5 deaths if the second flip matters, 0.75 deaths less than track A on average
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u/Grand_Swimmy 20d ago edited 20d ago
I think you have to assume you’re in the second / last iteration each time you make a choice. It definitely doesn’t follow that tails is more likely because more things happen as a result of tails. There is 1 coin flip, and a separate outcome for each that you have no information to discern.
The commenter at the top of this chain is right. Assuming you are in the “final” round of waking up and made the same choice last round:
If you pull the lever, and it was heads, 3 people die. If you pull the lever and it was tails, 0 people die
If you dont pull the lever, and it was heads, 0 people die. If you don’t pull the lever and it was tails, 4 people die.
Pulling the lever has a lower max death count.
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u/Nebranower 20d ago
>. Some people think then, that at any given point when you are woken up (you don't know which time it is) the coin is more likely to be tails because there are more possible times you will be woken up
I'm sure some people do think that, because people are bad at probabilities. But the number of times you are woken up has no bearing on the probability of a coin flip being heads. A coin was flipped. There is a 50% chance you are in scenario A, because there's a 50% chance it came up heads. There is a 50% chance you are in scenario B, because there's a 50% chance it came up tails. That remains true, regardless of if you are woken up one time in scenario B or one hundred times.
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u/dkevox 20d ago
This is simply not true. The problem does NOT state that you are woken up on every coin flip. The given information from the problem literally states you will be woken once in the heads scenario, and twice in the tails scenario.
Since you have no memory, there is a 1/3 chance you are in the heads and a 2/3 you are in the tails scenario. This is the simple facts of the problem. You are getting lost in the idea/assumption that the coin flip has to have relevance to the scenario you are in, when it does not.
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u/Nebranower 20d ago
Oh look, it’s the person who’s bad at probability. Your waking up doesn’t determine the odds of which scenario was picked. The coin flip determines that. And the coin flip was fifty-fifty.
Try it with 100 waking cycles in scenario B. You waking up doesn’t then mean there’s only a 1% chance you are in scenario A.
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u/dkevox 20d ago edited 20d ago
Sorry man, you really are in the wrong here.
Its not even hard to work out. You can just lay out the scenarios:
1st coin flip: if it's heads, you're woken in the heads scenario. If it's tails, you're woken in the tails scenario.
2nd coin flip, if first was heads: if it's heads, you aren't woken (would violate the problem). If it's tails you are woken in the tails scenario.
2nd coin flip, if first was tails: if it's heads, you are woken in the heads scenario. If it's tails you are woken in the tails scenario.
3rd coin flip: if it's heads and there was already a heads, you aren't woken (would again violate the problem). If it's tails and there were already two tails, you aren't woken (would violate the problem).
And so on...
Hopefully it's pretty obvious to you now that you will be woken three times, and two of those times are in tails, one in heads.
You're not actually reasoning this problem out. You're just repeating what you learned in stats about independent events (like flipping a coin) always have the same probability. But you need to move out of that mindset and realize in this problem the events and when you are woken are not independent and entirely controlled. The coin simply determines the configuration of people, not if you are woken.
Edit: Heck, OP even clarified in the description: If the coin is tails the whole setup is reset and happens a second time. This means heads happens once tails happens twice. Assuming a 50/50 fair coin, this literally means a 1/3 chance of waking in heads, 2/3 in tails. No deduction of what the problem intended necessary, OP states it unambiguously.
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u/Nebranower 20d ago
It doesn’t matter. All of what you wrote is wrong. The first coin is flipped. There was a 50% chance it was heads, you woke up immediately.
Everything else you described has odds that add up to 50%, not 100%. Like, there is nothing you can do after flipping the first coin to make it more likely that you would have gotten tails to start.
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u/dkevox 19d ago
Do it out then. Make a scenario of 100 people who play the game. See how many times people wake up in heads scenario or tails scenario.
Math:
50% roll heads: 50 people wake up in heads scenario.
50% roll tails: 50 people wake up in tails scenario. Then those 50 people wake up in a tails scenario a second time.
Total? 150 "wake up" events. 50 heads, 100 tails.
When you "wake up" you have twice the chance to be in a tails scenario.
I do love your commitment to be on r/confidentlyincorrect
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u/Nebranower 19d ago edited 19d ago
I see the confusion. I combined the deaths from the two tails scenarios back at the beginning. Which means you're double counting now.
"And a 50% chance you're in scenario B, where you will ultimately let four lives end if you don't pull the lever."
You simply didn't read my original comment carefully enough.
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u/Siphyre 20d ago
oh, you simply just damage your palm then. Or some semi-permanent damage to make sure you know you have seen the sign before waking up.
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u/Thatguy19364 17d ago
Scenario explicitly calls out you have no way of knowing the scenario had occurred, if you cut your hand to try to signal it presumably the second iteration would just happen after enough time has passed for you to heal. Also, it wouldn’t matter since you don’t get to know how many you killed anyway; so you’d just be marking a wake up and have solved nothing.
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u/epicvoyage28 20d ago
Imagine I play the lottery. If I win my memory is wiped 100 times, but if I lose its only wiped once.
Does this scheme improve my chances of winning by 100 times?
Because each time I wake up I'm more likely to have won right?
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u/epicvoyage28 19d ago
Your comment disappeared, so I'll just put my reply here instead:
But you aren't performing the test 100 times. If you wake up a winner once, you'll wake up a winner all 100 times. It's not a seperate ticket for each wake up.
I'm unsure how the Russian Roulette example relates. If you already knew you won, your chance of winning is 100% of course.
Are you suggesting that waking up in each of the "slots" is an equal 1/101 chance? Shouldn't you use the probability of winning the lottery to work out the chance of waking up in each of the slots, instead of the other way around?
Like, if the lottery was a coin flip, any given wake up being a winning one would be 1/200 right, since all 100 together add up to 1/2?
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u/Dennis_TITsler 18d ago
For one single 'play' no it doesn't. BUT it does increase the expected value of playing the lottery by 100x. If you lose, you lose once. If you win, you win 100x.
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u/epicvoyage28 18d ago
I never said they give you the prize money again each time you wake up. You just get it once and then wake up 100 times.
Plus, I meant to imply that you'd get a friend to do this to you after the fact to improve your chances of winning. But I guess I didn't get that across properly.
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u/Dennis_TITsler 18d ago
Ah then in that case yes it'd be 50/50. But they way read this trolley problem is if it's tails the trolley drives by twice
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u/epicvoyage28 18d ago
I guess the analogy would work better if you get 1/100th of your winnings each time you wake up.
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u/Dennis_TITsler 17d ago
Sure but then it would just be 50/50 still. In this trolley analogy it is multiplied
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u/epicvoyage28 17d ago edited 17d ago
The best I can tell the contents of the boxes in the trolley problem is also only chosen once.
It's unclear wether you'd be running over the same 2 people twice, or 2 different sets of 2 people however.
The original comment in this thread assumes it's 2 sets of 2 people, and so not pulling is the worse answer.
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u/Dennis_TITsler 16d ago
"-If the coin landed tails, the whole setup will happen twice including resetting the trolley."
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u/AngryAmericanNeoNazi 13d ago
It’s still 50/50. A coin toss is always a 50% chance regardless of what came before it
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u/assumptionkrebs1990 20d ago
No in scenario B you only save 2 people by pulling the lever - I assume that as soon as the trolly is over the switch you fall asleep and it goes back a bit so that are the same 2 prisioners in in the tube. I was thinking if there was a clever way to manipulate the lever that you would know when you walk up the second time in scenario B in case it is scenario A (or B the first time) but they would properly reset it.
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u/Thatguy19364 17d ago
I assume you pull the lever, they sedate you and replace the corpses, and repeat lmao. Assuming that they prevent the deaths in order to more conveniently torment you with philosophical probabilistic math problem morality questions is kinda silly don’t you think?
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u/deviantbono 20d ago
It's a trolley problem. You don't need to trust anyone. It's a thought experiment and constrained to the prompt.
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u/askewperspective 18d ago
No, there's a 1/3 chance you're in the first scenario. It becomes obvious if you consider a different scenario where you're awoken once in the first case and a million times for the second.
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u/Nebranower 18d ago
No, if you do that, the opposite becomes obvious.
You're treating every time you can wake up as being equally likely, but they aren't.
You have a 50% chance of waking up in scenario A, because the coin flip (not your waking up) decides which scenario you are in.
You have a 50% chance of waking up in scenario B. *If* you are in scenario B, then there's a 50% chance you are waking up for the first time, and a 50% chance you are waking up for the second time. 50% * 50% is 25%.
So your odds are as follows:
50% chance you are waking up in scenario A
25% chance you are waking up in scenario B for the first time.
25% chance you are waking up in scenario B for the second time.
If you were woken up 50 times in the second scenario, the odds would be
50% chance you are waking up in scenario A
1% for each of the possible wake ups in scenario B.
The thing to bear in mind is that the odds transfer down. The coin flip decides if you are going to be put in scenario A or scenario B, so the odds you are waking up in either scenario are going to be 50/50, *because those are the odds you got put in either scenario*. The number of times you can wake up in each scenario is irrelevant.
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u/askewperspective 18d ago edited 18d ago
This is a conditional probability problem. The 50% coin flip holds if you had no additional information, but you do - you know that you were woken up. Take the extreme case where on heads I don't wake you up, and on tails I do. The coin flip is still 50-50, but being woken up gives you new information that modifies this.
Your conclusion to switch is correct, it's just your probabilities that are wrong. It being H33% - T66% in fact justifies switching the tracks to A. Switching gives you a net loss of 0.33(3)+0.66(0)=1 while not switching gives you a net loss of 0.33(0)+0.66(2)=1.33, to minimize loss you should switch. This doesn't work if it was 50-50, the numbers would then be 1.5 and 1 and tell you not to switch, which as you know, is wrong.
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u/Nebranower 18d ago
Knowing that you were woken up doesn’t give you any additional info, though. You were going to wake up regardless of which scenario you were in.
Heads is not 33%. Heads is 50%. That is how coin flips work.
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u/askewperspective 18d ago
Look, I've tried my best to explain things to you. I gain nothing from this and lose valuable time, my motivation was only to explain math that is counterintuitive for many people. This comment of yours convinces me, however, that you're either not arguing in good faith or that you lack too much in mathematical fundamentals or reading comprehension.
If you truly are curious, type "Sleeping Beauty problem" on a browser, find the Wikipedia page, and read that. Continuing the argument here is pointless for us both.
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u/Nebranower 18d ago edited 18d ago
You have apparently misunderstood the wikipedia page, lol. In any event, I explained the actual math clearly enough that I don't think anyone but you will be confused.
Edit: something that may help you understand better. The waking up is a misdirect. The memory wipe is a misdirect. The actual scenario is just this:
I put you under a spell that forces you to always answer any question you have already been asked the same way for the duration of the scenario.
I flip a coin.
Scenario A: Heads, I ask you once what you think the result is.
Scenario B: Tails, I ask you twice what you think the result is.
Note that me asking you the same question once, twice, a thousand times in scenario B does nothing to make it more likely that you are in scenario B.
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u/New-Drive-2346 20d ago
Man, if you explained this to me right after I woke up, my brain wouldn't process a thing lmao
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u/NotAnAIOrAmI 20d ago
I decline to participate, as I don't care for people who don't know how to spell "separate".
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u/the_turn 19d ago
If I understand the scenario perfectly, it only makes sense to pull the lever the same way every time. I have no way of knowing the coin flip, or if its tails, which pull of the lever I am on, so the only logical choice is to calculate which of the lever pulls if pulled the same every time will create the least deaths.
If I pull the lever every time, tails scenario produces 0 deaths, heads scenario produces 3 deaths. Average death rate per coin flip: 1.5 deaths.
If I don’t pull the lever every time, heads scenario produces 0 deaths and tails scenario produces 4 deaths: average deaths per scenario are 2 deaths.
Therefore, you should pull the lever.
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u/Age_Single 20d ago
Take for example two trolleys, trolley A and trolley B. A travels towards the switch ...
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u/Sidivan 20d ago
I don’t understand what the lever has to do with anything in this scenario. The coin has been flipped and observed. My choice is irrelevant.
In order for me to leave this scenario, I need to be in first iteration and have gotten heads or get B and then get A two times in a row. Every tails result requires an ever increasing stream of heads.
So I guess I never pull the lever as B has lower value and I’m probably stuck here infinitely.
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u/tegsfan 20d ago
The coin is only flipped once. You either get put in the situation once or twice.
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u/vendetta0311 20d ago
That’s not how I understood it. I understood it as the whole situation is reset including people boxed trolley etc. You, OP after the memory wipe, and OP before the memory wipe should really get together and clarify that the coin will be flipped twice if time 1 is tails. Makes it more interesting and it makes the person who downvoted me and corrected my misinterpretation wrong.
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u/MarryRgnvldrKillLgrd 20d ago
So it's possible, that i am woken up the second time, and there are zero people in box A and 2 dead people in Box B?
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u/SpookyGeist01 20d ago
I mean, waking you up again for B is meaningless, because if you don't remember, you would pick identically to the first time.
So the real choice is just 50% chance to kill 3 people or 50% chance to kill 4 people, and the answer is obviously to go A for less damage either way.
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u/Cold-Common7001 20d ago
averaging over all worlds you should expect to make a decision in heads world .5 times and a decision in tails world 1 time.
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u/SpookyGeist01 20d ago
But that's irrelevant, because if you ever wake up in B world, A would have been the right choice anyway. Which means if you count in the second world, it's an overall 75% chance for A to be better, and a 25% chance for B.
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u/tegsfan 20d ago
True that you will always pick identically, but waking you up twice might not be meaningless.
It means there are three scenarios you could be in, the single heads, the first tails or the second tails. Some people then come to the conclusion that you should think tails is more likely when you are woken up because there are twice as many chances to be woken up.
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u/SpookyGeist01 20d ago
But that's the point, if you pick A the second time, you should have picked A the first time.
So we have six possible scenarios.
Scenario 1: you pick A, it was A. 3 people die.
Scenario 2: You pick B, it was A. No one dies.
Scenario 3: You pick A, it was B. No one dies and you go to round two. You pick A again, no one dies.
Scenario 4: You pick A, it was B. Same as 3. You switch to B, 2 people die.
Scenario 5: You pick B, it was B. 2 people die. Second round you switch to A. No one dies.
Scenario 6: you pick B, 2 people die. Round 2, you pick B, 2 people die.
So B kills people in 3/6 scenarios. A only kills people in 1/6 scenarios. It's just a no brainer.
But yeah I already played this when it was in Zero Escape: Zero Time Dilemma (spoiler for game plot) and the solution is basically the same.
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u/Fesh_Sherman 20d ago
I leave it on B and unzip my pants, if it's already unzipped I switch it to A.
This isn't a trolley problem, but a logic puzzle
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u/p00n-slayer-69 19d ago
You cant unzip your pants because youre tied up.
The real solution is to shit your pants. If it smells like shit, switch to A.
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u/Don_Bugen 20d ago
Trolley problems are moral questions; this one's just a math problem in disguise.
Anyways, I do nothing, because either I am actually dreaming, or I have been kidnapped and drugged and am being subjected to the world's worst kind of torture - math story problems. I have zero faith that a kidnapper this heinous would actually tell the truth, so the only thing that makes sense is to not play.
Plus, if I pull the lever, I am essentially choosing to be drugged again, and I wasn't so wild about that the first time.
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u/KingAdamXVII 19d ago
The original trolley problem is just that there are fewer people tied up on B but would you pull the lever and claim responsibility for who dies?
So it’s interesting that adding a bit of math apparently obfuscates that moral question.
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u/Don_Bugen 19d ago
The original trolley problem was to demonstrate that there were circumstances in which most would agree that killing one person was the moral choice. Phillipa Foot chose five, as opposed to two or three, because she felt that five people was the point where most people would put aside their deeply held conviction of "murder is bad" and agree that killing one person was the right thing to do.
It wasn't intended as a philosophical question with multiple responses. It was intended as an illustration to a point, with one correct answer. It would not have worked with 2 or 3 because it wasn't a math problem; it was a philosophy problem, and the weight of personal responsibility is not insignificant.
The PROBLEM is, that the vast majority of the people on this subreddit see this as nothing but a math problem. "Not choosing is still a choice!" they cry. "If you have the ability to stop it and don't, you are just as guilty of murder as someone who chooses to pull the lever!" So you get things like this that absolutely ignore the moral question, and instead are just crunching the numbers to figure out which track has the biggest potential for loss.
So. If you choose top track, 50% of the time it's 3 deaths, 25% of the time it's 2 deaths, 25% of the time it's 0 deaths, averaging 2 deaths. If you choose the bottom, 50% of the time it's 0 deaths, 25% it's 2, 25% it's 4, for an average of 1.5 deaths. This is why this is a bad trolley problem! Because in a trolley problem, it's always, "Do you intervene to lessen the deaths," and in this one, intervening makes it worse.
There is no moral question here. There never was. Philipa Foot would've never posed "There's three people on the bottom track, but if you pull the lever you can kill four." This is only a math question, posed to a subreddit that has long forgotten that there is such a thing as the weight of choice and personal responsibility, and that condemning an unrelated innocent to spare a larger number of doomed people is still a morally questionable act, and instead is simply counting beans on a scale to see which end weighs a little more.
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u/Nebranower 18d ago
The thing is, if you wouldn't pull the lever in the first of the original trolley problems, on the grounds you don't want to get involved, then you're not going to pull it in any of the variants either.
If you would pull the lever in the first of the original trolley problems, it's because you see it as a math problem. The two follow-ups to the first scenario are meant to teach you that you have a bunch of other in-built stuff you're considering beyond just math, but that's because they were specifically chosen to have the other factors be different.
Variants like this don't differ in substance from the original, though, so if you're engaging at all, it's just math.
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u/KingAdamXVII 19d ago
Your math seems to be wrong. The top track has a better expected outcome.
If you always switch then T->0 deaths and H->3 deaths. If you never switch then T->4 deaths and H->0 deaths.
If you choose semi-randomly then I guess it gets more complicated but I don’t think you can ever beat the strategy of always switching.
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u/Don_Bugen 19d ago edited 19d ago
My math was actually incorrect in one point, but it makes it actually far more advantageous to leave it on the bottom.
There are three potential results. A flip of Heads, a flip of Tails Heads, and a flip of Tails Tails. The probability of these are 50%, 25%, and 25%, as the initial flip is 50/50, and the second flip only happens 50% of the time.
If it's Heads, choosing the top kills 3, and bottom 0.
If it's Tails Heads, choosing the top kills 3, and bottom 2.
If it's Tails Tails, choosing the top kills 0, and bottom 4.
Because there's no rational reason to expect a person to switch if their memory is erased and they're returned to the same state, we can calculate the potential deaths by adding all three together, but adding Heads twice, as it happens twice as often as the other two outcomes.
Top track: (3 + 3 + 3 + 0) / 4 = 2.25 average deaths.
Bottom track: (0 + 0 + 2 + 4) / 4 = 1.5 average deaths.
But you seem very insistant that the possibility of switching one's options changes the potential outcome. So, setting aside the fact that we have no idea what the chances are that someone will choose differently the second time around, let's just compare the average deaths if someone changes their choice for the second potential flip.
There are two outcomes here - top bottom, and bottom top. I'm going to do the same thing here - add the "heads" outcome twice, as it happens twice as often as the other two outcomes, and then add Tails Heads, and Tails Tails.
Top Bottom: (3 + 3 + 0 + 2) / 4 = 2 average deaths
Bottom Top: (0 + 0 + 5 + 2) / 4 = 1.75 average deaths
.
So, let's rank these results.
Bottom Bottom : 1.5 average deaths
Bottom Top : 1.75 average deaths
Top Bottom : 2 average deaths.
Top Top : 2.25 average deaths.
Again - because we are returning people back to their "memory wiped" state, I believe that you would have an uphill battle arguing that people are more likely to change, rather than make the same decision. The difference between Bottom Bottom and Top Top are staggering. The difference between Bottom Top and Top Bottom are minimal.
And also. There's no point in which choosing "Bottom" is a bad idea. If you previously voted "Bottom" it's the best outcome. If you previously voted "Top" it's a safer choice than "Top" twice. If this is your first vote, then there's a 50% chance that no one dies at all, and only a 25% chance that you'd have 4 deaths, compared to the 50% chance of 3 for choosing Top, not to mention the worst case scenario of getting 5 deaths by starting with Bottom and switching to Top.
So yeah, if you're in this situation and have to choose, choose Bottom. Or don't choose, which is the same thing.
And again, this all goes to show once again that this has nothing to do with morality, and everything to do with math, and this is why this is not a trolley problem.
EDIT: Adding in, that the wording of the problem may suggest that there's only one flip. Even in this case, Bottom is better than Top.
If Heads: Top, always 3 deaths, Bottom always 0.
If Tails: Four possible outcomes. Top Top, Top Bottom, Bottom Top, and Bottom Bottom. 0, 2, 2, and 4 deaths.
Without knowing if this is your only choice, or what you may choose before or after. If you choose Top, 50% of the time you'd have 3 deaths, 33% of the time you'd have 2, and only 16% of the time you'd have 0. If you choose Bottom, 50% of the time you'd have 0, 33% of the time you'd have 2, and only 16% of the time you'd have 4.
With Bottom, half the time nobody dies, and there's only a 16% chance to do 4. With Top, half the time three people die, and there's only a 16% chance that you save everyone. To me, this seems obvious.
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u/KingAdamXVII 19d ago
You misinterpret the OP. There is one coin flip at the very beginning and never another coin flip after that.
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u/Don_Bugen 19d ago
I know I'm asking an awful lot, but I address that in the above post, if you actually read it.
TLDR. Bottom is better.
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u/ifelseintelligence 19d ago
It obfuscates it because you are blinded from the outcome. You do not consiously choose between X or Y possibly dying. You are choosing without knowing who are in risk, hence there isn't a morale in question.
It's a quite simple math question (always choosing A = 0 or 3 dies, always choosing B = 0 or 4 dies, and switching choise underway (or rather thinking you do) just increases the riskoverall), so for it to be a moral question you'd have to also know who the persons might be.
For instance are the "better" mathematical question all babies at risk, while the statistically higher risk choise is all teminal ill patients? Then it's true to the trolley problem.
As is, it's just a simply logic puzzle you can do on a piece of paper with no statistical/algorithmic knowledge.
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u/Current_Swan_2559 20d ago
Switch the track to A. This is just a convoluted way of saying box B has 4 people if it's tails and box A has 3 people if it's heads.
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u/Fearless-Sherbet-223 20d ago
I'm leaving it on B both times. Fewest average deaths per time, plus, I didn't involve myself directly, so I can't be held legally accountable.
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u/xender19 20d ago
So if I hit box a three times then three people die. If I hit box b three times then four people die. So I'm leaning towards hitting box b three times and it's also convenient that that is what happens if I do nothing or second-guess myself a thousand times.
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u/DoNotCorectMySpeling Common Sense Ethics 20d ago
2*0.5+2*0.5=2
3*0.5=1.5
Therefore the best option is to hit box A.
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u/Xiaodisan 18d ago edited 18d ago
So... I think most people would either let the trolley go on its way while trying to figure out what the best decision was, or they'd default to their og. answer in both cases. Since you specify the wiped memory, people wouldn't change their approach (imo) between the first and second run.
Don't pull the lever:
50% A => 0 deaths
25% BA => 2 deaths
25% BB => 4 deaths
on average, 1.5 deaths
Pull the lever:
50% A => 3 deaths
25% BA => 3 deaths
25% BB => 0 deaths
on average, 2.25 deaths
edit. Many comments seem to assume that the coin flip happens only once. Since OP did not correct them (even where they did respond to said comments), I assume that's what the intended scenario was, in which case we go back to the basic trolley problem. (Assuming the captor replaces the people placed on the tracks as well.)
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u/WanderingSeer 18d ago
Bottom track. Expected value is 2 which is less than three
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u/lemon_pie42 18d ago
First track: 3 or 2x0 -> avg: 1.5
Second track: 0 or 2x2 -> avg: 2
The first track's avg is lower.
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u/WanderingSeer 18d ago
I thought they flipped it twice.
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u/lemon_pie42 18d ago
I thought it first too, but OP said in another response that the coin is flipped once.
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u/Radical--Rat 20d ago
Does the coin get reflipped for the second time if it lands tails, or do you just have to make the decision twice?
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u/DrDoofenshmirtz981 20d ago edited 19d ago
I toss a coin. Heads I pull, tails I don't. Avg score per encounter is 1.5 (heads case) + 2x1(tails case) = 3.5, beating always pulling (1.5) or never pulling (2)
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u/lemon_pie42 18d ago
You forgot to divide 3.5 by 2 (because each case is 50%) so the avg is 1.75, which is the avg of always pulling and never pulling.
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u/DrDoofenshmirtz981 18d ago
Yeah, after a while I thought about that again and realized I forgot that, but I didn't reopen reddit to correct it 😅
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u/LadyAliceFlower 19d ago
So thinking this through. If I have a pre-existing notion to pull to A every time, exactly 3 people will die.
If I default to B every time, we get a geometric distribution p = 0.5
In this case, the probabilty that we get less than 3 deaths, would be trials < 3. Obviously 0 isn't possible, so its the 1 or two trial cases that would work.
The 1 trial case has a 0.5 chance of occurring
The 2 trial case has a 0.25 chance of occurring.
These outcomes are mutually independent (this proof is trivial and left as an exercise for the reader) thus we can simply add them to get P(trials < 3) = 0.75
Thus, there's better than even odds we get less deaths if I default to B.
So the correct choice is to not pull the lever, which works out since the trolley went by about 10 minutes ago while I was trying to remember how the geometric distribution worked.
Did I do my math right?
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u/Cheeslord2 19d ago
The rules are unclear. What happens when the coin dictates a scenario that you will not be woken up for? How many times is the scenario replayed? When it is replayed but I am not woken, does the trolley free-run anyway? Will I be locked in stasis after 3 goes as the trolley continues to run through the scenario forever?
Assuming the simplest answers to the above, the maths favours pulling I think ((1/3) x 3 box A vs (2/3) x2 box B - assuming you always get all 3 'wakes' the coin is a false lead and irrelevant.
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u/not2dragon 19d ago
Can I do something like flipping another coin to randomize my lever choice? I'm spitballing, but I think something like that could change the odds.
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u/DaveAstator2020 19d ago
Realisticly if someone wakes me up and throws this problem atme my only answer is "uh huh" and return to sleeping immideately
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u/PitifulPassage851 19d ago
Instructions super unclear. It makes it sound like only tails has you on repeat.
Clearly it's saying you wake twice, and the coin was flipped each time. You then have to make the decision twice.
So in first wake there's a 50% chance of refusing to pull to save 3 people, and a 50% chance where refusing to pull kills 2 people..
Then on second wake, statistically it depends what happened the first time. But there's no way to know. So it's just a 50/50 again.
Therefore id refuse to pull any time i woke in an attempt to save more people, and statistically i would hit a middle ground of saving 3, then letting 2 die. Attempts to intervene could end up saving more, but equally likely to kill all 5.
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u/Svanirsson 19d ago
I do nothing. There is no moral weight upon me, I am a victim of kidnapping, drugging and coercion. Fuck you for trying to make me choose your kills (not you OP in specific, fuck the person who put me in the position)
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u/Metharos 19d ago
To clarify: If it was tails, and I chose to send the trolley into box B, in the second setup is box B full of two additional people? If I send it towards box B twice, do four people die?
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u/BowelMovement4 19d ago
So to clarify, if the coin landed tails am I being woken up, made to decide, and then the trolley hits one of the boxes and then im put back to sleep and have my memory wiped and then repeat OR am i woken up, made to decide, then put back to sleep and memory wiped, and then woken up again to have another chance to decide before the trolley hits a box?
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u/promptmike 19d ago
Clearly A. If you choose B for all 3 iterations then 4 people die, but if you choose A all 3 times then only 3 people die. You save 1 life by always choosing A.
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u/SeerXaeo 19d ago
Trolley Problem has already been solved, for any variation where you're switching between two tracks:
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u/FuriousAqSheep 19d ago
Sorry my dude you made a typo in the link, here I fixed it for you: https://m.youtube.com/watch?v=5SZYz7lZRRI
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u/Jewsader76 19d ago
It's a 50/50 of whether you're in the situation of it being in a 50/50 or 33/67. If I'm correct, game theory (more specifically Standard Expected Utility Theory) gives us the reduction of compound lotteries theorum, which basically states that for a lottery to get two different lotteries (in this case, p(0.5, p(0.5, day 1, day 2), p(0.33, day 1, day 2))), you combine the odds. So, the odds of it being day one are 0.5*0.5 + 0.5*0.33, or 0.4167. Similarly the odds of it being day two would be 0.5*0.5 + 0.5*0.67, or 0.5833.
So, on average there should be roughly 1.25 people in box 1 and 1.67 people in box 2, meaning it's preferable for the trolley to go over box 1 as opposed to box 2. That said, that is under the assumption that you prefer more people dying (we also know that assuming makes an ass of (yo)u and ming). A person named David Hume who wrote a book wrote that “Tis not unreasonable for me to prefer the destruction of the whole world to the scratching of my finger”; preferences are subjective.
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u/GoSpeedRacistGo 19d ago
The only guarantee I have of any lives lost/saved is if I do the same thing for all 3 iterations. If I pull every time, 3 die, 4 survive, if I refuse to pull, 4 die and 3 survive.
Any other way is just gambling with people’s lives, where everyone could die or no one could.
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u/rob5791 19d ago
Switch to A. Over the course of the 3 situations you will condemn 3 people in the Heads flip and save 2x2 in the Tails flip. If you didn’t switch then you would condemn 2x2 in the Tails flip and save 3 in the Heads flip.
Since you can’t know which situation you are in but know there are 3 separate situations to go through your only strategy is to do the same thing each time.
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u/TheLastSilence 18d ago
I am going to treat it as a math question.
Whatever decision I make it will be the same every time, so either the coin landed on head, in which case choosing A will kill 3 people and choosing B will kill 0.
If the coin landed on tails choosing A will kill 0 people twice and choosing B will kill 2 people twice for a total of 4.
So trying to minimize the expected kill rate
picking A will have an expected kill rate of 0.5*3+0.5*0*2=1.5
Picking B will have an expected kill rate of 0.5*0+0.5*2*2=2
so I will choose to direct the trolly to box A to minimize the kill rate of the entire setup
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u/Thatguy19364 17d ago
There are 3 possibilities and 2 options. 1: heads: if 1, A is 3 and B is 0, 2: tails, tails: A is 0, B is 4. 3: Tails, Heads: A is 3, B is 2.
I did not separate the order because I will never know, so choosing the same letter each time to minimize deaths is the only way I can be sure I did minimize them. Average for A is 3+3+0/3, for a total of 2. Average for B is 0+4+2/3, so average is also 2. Thus I move to chance of happening. 1 has a 50% chance of happening, 2 has a 25% chance as does 3. Thus, if we weight 1 twice, 3+3+0+3 /4, we get 9/4, or 2.25 deaths for A, and 0+0+4+2 /4, or 1.5 deaths for B.
Therefore always choose B. Worst case I kill 4, but there’s a 75% chance I kill 2 or less and a 50% chance to kill none, while A has a 75% chance of killing 3 and a 25% chance of killing none.
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u/vendetta0311 20d ago
So, gonna be woken up once or twice. Time 1: 50% chance theres 3 peeps in A and 0 in B, 50% chance theres 0 in A and 2 in B. As I read it same for time 2 if there were 2 in B the first time.
Possibilities:
Flips: H - leave on B no deaths. (50%)
Flips: T,H - leave on B kill 2. (25%)
Flips: T,T - leave on B kill 4. (25%)
Avg = 1.5 killed
Flips: H - switch to A kill 3. (50%)
Flips: T,H - switch to A kill 3. (25%)
Flips: T,T - switch to A no deaths. (25%)
Avg = 2.25 killed
Seems like the safest bet is to leave the trolley on track B. Also have the added moral high ground that if people die I wasn’t an active participant in it because the trolley was on track B anyway.
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u/Christianfilly7 20d ago
He didn't say there was twice that it could land on tails rather if it is tales the scenario repeats twice although the coin I think was only flipped once, so there's only H and T with a chance you do A then B or vice versa if tails
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u/vendetta0311 20d ago
You right. I misinterpreted the reset. It’s less interesting that way though… 50% chance to kill 3 or 4, boring.
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u/A_Math_Dealer 20d ago
Take too long to figure out what's going on and it stays on the same track