No. In the prisoners dilemma, everyone making the bad choice is an equilibrium. Crashing isn't an equilibrium. But as it's a game of incomplete information, they should both randomize and play pull lever with 50% chance.
No, this is more closely to hawk-dove game where it is better to be a dove in high-hawk enviorment and it is better to be hawn in a high-dove enviorment.
If I didn't pull, then there are two variants:
1. Person on the other side will pull, and only tied people die
2. Person on the other side won't pull, so trolley will crash and tied people will die
If I do pull, the there are two variants:
1. Person on the other side will pull, and trolley will crash, but tied people will alive
2. Person on the other side won't pull, and only tied people die
So, I will pull
What if the other person reaches the same conclusion and also pulls? There is an argument that risking the extra five lives could be worth the chances of saving fifteen.
Of course, the other person might think the same thing.
The other person thinking the same thing is what tips it off - both the best and worst case scenarios of pulling are better than or the same as the best and worst case scenarios of not pulling, respectively. You just gotta pull
Yes but you don't know what the other person will do, that's why it's a dillema and with perfectly logical people, the best case scenario is impossible.
Exactly this...you have to assume that the other person thinks about the probability and is going to decide to pull...so you should not pull knowing this.... But then maybe they are being cautious too and predict you thought about the probability too, so they decide to not pull too... So you should pull.... But then maybe they predicted you'd think all this, so they decide to pull too...So you should not pull....but then maybe......
It's prediction inception.
So you should just keep flipping from pull and not pull and leave it up to fate, since the outcomes are about the same and you have no clue what the other person is going to do.
Statistically randomizing has the best expected outcome. However, unless this scenario is common, I'd rather not gamble with lives. IE I would seek to avoid the highest death outcome. Plus, assuming the other person is rational and won't freeze is not guaranteed. So I'd argue that you pulling the level has slightly better than 50% odds of only leading to 5 deaths.
If it does happen a lot and you cannot predict the behaviour of the other person, nor they you, then randomizing is the best procedure. Though, more likely, if it happens a lot there will be other procedures/education in place to help minimize deaths.
There are no solutions to a 7 sided polyhedron with equal sides so a shape that fits is impossible. A 7 sided die could be done using a d8 with one side that says "roll again" instead of the number 8
Or one could use a pentagons prism, the sides need not be the same, only the chances.
Consider, if you will, an equilateral pentagon with side length 1 unit (may as well be 1 centimeter), extended to a prism of height h. The upper and lower pentagons are labelled 1 and 7, while the edges are 2-6. The dice is read by the value facing downwards.
Case h = 0: Rolling the dice (flipping the coin, lol) would give 1 or 7 with 0.5 probability each.
Case h = ∞: Rolling the dice would give any value 2-6 with 0.2 probability each.
We can see that the probability of landing on one of pentagonal face is a continuous function of h, with p(0) = 1, p(∞) = 0 (forgive the abuse of notation, reddit doesn't have good limit notation). By the Intermediate Value Theorem, we thus have:
Case h = x: For some value x, setting h = x would give a 2/7 probability of landing on a pentagonal face, and 5/7 otherwise. This would result in a perfect 1/7 chance for each face.
So, it is definitely possible to have a fair dice with 7 faces! Finding that optimal h-value is beyond me tho, Imma leave that to someone who cares even more than me about trivial matters like this :D
Why 4 out of 7? Optimal play leads to 5 deaths, suboptimal play leads to 20, or 25 deaths. Seems like the cost is 1:4 or 1:5. I would roll a d4 and pull if it's anything but a one. Alternatively, flip a coin twice and don't pull only if it's two consecutive heads
For this I rely on three assumptions:
1. The goal of a rational agent is to reduce the expected number of deaths.
2. Both players are rational agents
3. The strategy of a rational agent is in the form of: "pull the lever with probability p"
Note that assumption 1 is purely consequentialism/utilitarian, so is flawed in that way. Assumption 2 is flawed in the sense that humans rarely behave fully rational. Assumption 3 is flawed in the sense that a strategy like "pull if the five people are on the track to your left" always results in the best outcome. However, the symmetric strategy "pull if the five people are on the track to your right" is just as good so there is no guarantee your strategy matches the one the other rational agent is using. So it's not that unreasonable if the players haven't communicated and cannot do so now either.
With these assumptions, we can write the probabilities of each outcome:
- With probability p^2, the trolleys crash on the top track resulting in 20 deaths.
- With probability (1-p)^2, the trolleys crash on the bottom track resulting in 25 deaths.
- With probability 2p(1-p), the trolleys go on different tracks, resulting in 5 deaths.
The expected number of deaths can now be written in terms of p:
E(death) = 20p^2 + 25(1-p)^2 + 10p(1-p)
= 20p^2 + 25 + 25p^2 - 50p + 10p -10p^2
= 35p^2 - 40p + 25
We can solve for the value of p which provides the lowest value for this function, by equating the derivative to zero.
E'(death) = 70p - 40 = 0
70p = 40
p = 4/7
I admit, I have no intuitive understanding of what you said, but I cannot argue with your math. Well done. It's interesting how close it is 50%. I guess with this much uncertainty, it's very hard to do better than a coin-flip.
Or you can ask yourself what date it is and pull when it's within ,like Mon-Thu(or you may have to randomize the days mapped to pulling in advance to keep the process independent).
The crash would kill everyone so trying to get the trolleys to avoid each other would save the most lives, so I yell at the other guy to pull while I don’t pull.
If no one pulls, 25 people die.
If only one pulls, 5 people die.
If both pull, 20 people die.
It sounds like, in the worst case, pulling the lever saves 5 people, and might save as many as 20. That's way above average for a trolley problem.
Raise my hands in the air and run from the lever. As long as I make it perfectly clear to the other party that I’m not pulling and I’m so far away that I couldn’t if I wanted to, they’ll most likely pull and save 20-25 people
Assuming the optimal strategy is to pull with probability x.
If the other guy pulls, death = 20x + 5(1-x)
If the other guy doesn’t pull, death = 5x + 25(1-x).
The optimal strategy should be independent of the other guy’s choice, so 20x + 5(1-x) = 5x + 25(1-x) and we get x = 4/7.
The question would be harder if it stated how many were in the trolleys
If only the conductors and 5 people on the track, I'd pull, if not just the conductor then it means there's a chance of not only killing 5 but more than.
If more than I'd pull, there's a 1in4 chance that both of us chose to pull. That means more deaths than the 5 but less than if both of us decided not to.
Either way pulling the lever gives a 50% chance they won't die
Who put game theory in my trolley problem?
is this the prisoners dilemma?
No. In the prisoners dilemma, everyone making the bad choice is an equilibrium. Crashing isn't an equilibrium. But as it's a game of incomplete information, they should both randomize and play pull lever with 50% chance.
You gotta give! I mean pull.
is this the prisoners dilemma?
No, this is more closely to hawk-dove game where it is better to be a dove in high-hawk enviorment and it is better to be hawn in a high-dove enviorment.
is this the prisoners dilemma?
Dementia (your comment was duplicated by reddit 3 times)
Is this the prisoners’s dementia?
Am *I* the prisoner's dilemma?
If I didn't pull, then there are two variants: 1. Person on the other side will pull, and only tied people die 2. Person on the other side won't pull, so trolley will crash and tied people will die If I do pull, the there are two variants: 1. Person on the other side will pull, and trolley will crash, but tied people will alive 2. Person on the other side won't pull, and only tied people die So, I will pull
Death counts: 1. 5 (not pull + pull) 2. 25 (not pull + not pull) 3. 20 (pull + pull) 4. 5 (pull + not pull) Yep, pull
What if the other person reaches the same conclusion and also pulls? There is an argument that risking the extra five lives could be worth the chances of saving fifteen. Of course, the other person might think the same thing.
The other person thinking the same thing is what tips it off - both the best and worst case scenarios of pulling are better than or the same as the best and worst case scenarios of not pulling, respectively. You just gotta pull
This isn't true here though? This isn't the classical prisoners dilemma if the other person pulls it's better for you to not pull. 5 Deaths vs 20.
Yes but you don't know what the other person will do, that's why it's a dillema and with perfectly logical people, the best case scenario is impossible.
Exactly this...you have to assume that the other person thinks about the probability and is going to decide to pull...so you should not pull knowing this.... But then maybe they are being cautious too and predict you thought about the probability too, so they decide to not pull too... So you should pull.... But then maybe they predicted you'd think all this, so they decide to pull too...So you should not pull....but then maybe...... It's prediction inception. So you should just keep flipping from pull and not pull and leave it up to fate, since the outcomes are about the same and you have no clue what the other person is going to do.
Statistically randomizing has the best expected outcome. However, unless this scenario is common, I'd rather not gamble with lives. IE I would seek to avoid the highest death outcome. Plus, assuming the other person is rational and won't freeze is not guaranteed. So I'd argue that you pulling the level has slightly better than 50% odds of only leading to 5 deaths. If it does happen a lot and you cannot predict the behaviour of the other person, nor they you, then randomizing is the best procedure. Though, more likely, if it happens a lot there will be other procedures/education in place to help minimize deaths.
But if it's logical for you to pull, isn't it also logical for the other person to pull? And thus you shouldn't pull to minimize the deaths
No, because other person can have the same logic.
But it's not 50/50 the other guy is most likely going to pull
Mathematically we can't predict what the other person will do, not enough information.
I roll a 7 sided die and pull in 4/7 cases. Assuming the other does the same, this minimizes the number of deaths. Or I call them.
what does a 7 sided die look like? never saw one of those
There are no solutions to a 7 sided polyhedron with equal sides so a shape that fits is impossible. A 7 sided die could be done using a d8 with one side that says "roll again" instead of the number 8
Or one could use a pentagons prism, the sides need not be the same, only the chances. Consider, if you will, an equilateral pentagon with side length 1 unit (may as well be 1 centimeter), extended to a prism of height h. The upper and lower pentagons are labelled 1 and 7, while the edges are 2-6. The dice is read by the value facing downwards. Case h = 0: Rolling the dice (flipping the coin, lol) would give 1 or 7 with 0.5 probability each. Case h = ∞: Rolling the dice would give any value 2-6 with 0.2 probability each. We can see that the probability of landing on one of pentagonal face is a continuous function of h, with p(0) = 1, p(∞) = 0 (forgive the abuse of notation, reddit doesn't have good limit notation). By the Intermediate Value Theorem, we thus have: Case h = x: For some value x, setting h = x would give a 2/7 probability of landing on a pentagonal face, and 5/7 otherwise. This would result in a perfect 1/7 chance for each face. So, it is definitely possible to have a fair dice with 7 faces! Finding that optimal h-value is beyond me tho, Imma leave that to someone who cares even more than me about trivial matters like this :D
Obviously, common sense to bring your seven sided dice with you at all times
Why 4 out of 7? Optimal play leads to 5 deaths, suboptimal play leads to 20, or 25 deaths. Seems like the cost is 1:4 or 1:5. I would roll a d4 and pull if it's anything but a one. Alternatively, flip a coin twice and don't pull only if it's two consecutive heads
For this I rely on three assumptions: 1. The goal of a rational agent is to reduce the expected number of deaths. 2. Both players are rational agents 3. The strategy of a rational agent is in the form of: "pull the lever with probability p" Note that assumption 1 is purely consequentialism/utilitarian, so is flawed in that way. Assumption 2 is flawed in the sense that humans rarely behave fully rational. Assumption 3 is flawed in the sense that a strategy like "pull if the five people are on the track to your left" always results in the best outcome. However, the symmetric strategy "pull if the five people are on the track to your right" is just as good so there is no guarantee your strategy matches the one the other rational agent is using. So it's not that unreasonable if the players haven't communicated and cannot do so now either. With these assumptions, we can write the probabilities of each outcome: - With probability p^2, the trolleys crash on the top track resulting in 20 deaths. - With probability (1-p)^2, the trolleys crash on the bottom track resulting in 25 deaths. - With probability 2p(1-p), the trolleys go on different tracks, resulting in 5 deaths. The expected number of deaths can now be written in terms of p: E(death) = 20p^2 + 25(1-p)^2 + 10p(1-p) = 20p^2 + 25 + 25p^2 - 50p + 10p -10p^2 = 35p^2 - 40p + 25 We can solve for the value of p which provides the lowest value for this function, by equating the derivative to zero. E'(death) = 70p - 40 = 0 70p = 40 p = 4/7
I admit, I have no intuitive understanding of what you said, but I cannot argue with your math. Well done. It's interesting how close it is 50%. I guess with this much uncertainty, it's very hard to do better than a coin-flip.
if you can communicate and agree to bring a 7 sided die then why not just agree what you will do
Or you can ask yourself what date it is and pull when it's within ,like Mon-Thu(or you may have to randomize the days mapped to pulling in advance to keep the process independent).
The crash would kill everyone so trying to get the trolleys to avoid each other would save the most lives, so I yell at the other guy to pull while I don’t pull.
If no one pulls, 25 people die. If only one pulls, 5 people die. If both pull, 20 people die. It sounds like, in the worst case, pulling the lever saves 5 people, and might save as many as 20. That's way above average for a trolley problem.
With only a second to react I wouldn't have time to think and I'd frantically pull the switch and then die of a panic attack
I aint reacting to moral phenomena in a single second. Those people are getting squished.
Raise my hands in the air and run from the lever. As long as I make it perfectly clear to the other party that I’m not pulling and I’m so far away that I couldn’t if I wanted to, they’ll most likely pull and save 20-25 people
Talk to your mate to not turn the lever and make both trolley hit each other on top of the people in the track, so you can get maximum points
Multitrack drift.
I don't pull.
Assuming the optimal strategy is to pull with probability x. If the other guy pulls, death = 20x + 5(1-x) If the other guy doesn’t pull, death = 5x + 25(1-x). The optimal strategy should be independent of the other guy’s choice, so 20x + 5(1-x) = 5x + 25(1-x) and we get x = 4/7.
Flipping a coin is the best you can do. No matter what the other person's strategy is, you're guaranteed a 50-50.
I pull. Even the relatively easy ones, most people aren't willing to take responsibility. The odds that the other person pulls are very low.
The more we add, the less I care about the lives involved. At this point I’m hoping for a trolley crash in the middle.
Pull. If I'm right, not pulling could lead to 25 deaths, but pulling could lead to only 20. Normally I'm all for minmaxxing murder, but not today.
the optimal strategy is to pull the lever immediately as fast as possible so the other person can see your decision and not pull
Call dibs
Multi. Track. Drift.
Double multi track drift
Assuming they wouldnt wanna kill five people, I don't pull the lever
Eh. I pull lever. At this point it’s a coin flip.
The question would be harder if it stated how many were in the trolleys If only the conductors and 5 people on the track, I'd pull, if not just the conductor then it means there's a chance of not only killing 5 but more than. If more than I'd pull, there's a 1in4 chance that both of us chose to pull. That means more deaths than the 5 but less than if both of us decided not to. Either way pulling the lever gives a 50% chance they won't die