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Selfish Trolley Car Debate
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So, I'm in disagreement with my boyfriend over the following scenario: given the trolley car problem (1 person on one track and 5 people on a second track, an out of control trolley car will kill the 5 unless you pull the lever to kill the 1), he knows that one of the people is me and the other 5 are strangers. Since I'm amazing, he selfishly wants to maximize the chance that I survive.
His solution: he says he would pull the lever to save the five people. Since he doesn't know which one I am, he should save the most people to maximize the chance of saving me.
My solution: I say I wish he would put more thought into saving me. Instead of assuming a uniform distribution on my position, he should assume a uniform distribution on all distributions. So, he should pull the lever with 5/6 probability and let it kill the 5 with 1/6 probability.
So, what's the right answer? Is there even a right answer?
probability
asked Jul 29 at 19:43
user144527
38619
 |Â
show 5 more comments
up vote
3
down vote
favorite
So, I'm in disagreement with my boyfriend over the following scenario: given the trolley car problem (1 person on one track and 5 people on a second track, an out of control trolley car will kill the 5 unless you pull the lever to kill the 1), he knows that one of the people is me and the other 5 are strangers. Since I'm amazing, he selfishly wants to maximize the chance that I survive.
His solution: he says he would pull the lever to save the five people. Since he doesn't know which one I am, he should save the most people to maximize the chance of saving me.
My solution: I say I wish he would put more thought into saving me. Instead of assuming a uniform distribution on my position, he should assume a uniform distribution on all distributions. So, he should pull the lever with 5/6 probability and let it kill the 5 with 1/6 probability.
So, what's the right answer? Is there even a right answer?
probability
asked Jul 29 at 19:43
user144527
38619
1
Your answer doesn't make sense unless you are going to repeat the experiment (which will not be possible if he got it wrong).
– Rob Arthan
Jul 29 at 19:52
@RobArthan: It would still be the wrong answer even if the experiment were repeatable (e.g. you'd lose a dollar instead of your life).
– joriki
Jul 29 at 19:53
1
@joriki: the scenario gives the boyfriend one binary choice in one experiment. The suggestion that the choice is made one way 5/6 of the time and the other 1/6 of the time doesn't make sense.
– Rob Arthan
Jul 29 at 20:01
1
@joriki: so the OP's suboptimal solution is that the boyfriend should equip himself with a high-probability random number generator and use it to make his decision.
– Rob Arthan
Jul 29 at 20:54
1
I said he should roll a d6.
– user144527
Jul 29 at 21:22
 |Â
show 5 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
So, I'm in disagreement with my boyfriend over the following scenario: given the trolley car problem (1 person on one track and 5 people on a second track, an out of control trolley car will kill the 5 unless you pull the lever to kill the 1), he knows that one of the people is me and the other 5 are strangers. Since I'm amazing, he selfishly wants to maximize the chance that I survive.
His solution: he says he would pull the lever to save the five people. Since he doesn't know which one I am, he should save the most people to maximize the chance of saving me.
My solution: I say I wish he would put more thought into saving me. Instead of assuming a uniform distribution on my position, he should assume a uniform distribution on all distributions. So, he should pull the lever with 5/6 probability and let it kill the 5 with 1/6 probability.
So, what's the right answer? Is there even a right answer?
probability
asked Jul 29 at 19:43
user144527
38619
So, I'm in disagreement with my boyfriend over the following scenario: given the trolley car problem (1 person on one track and 5 people on a second track, an out of control trolley car will kill the 5 unless you pull the lever to kill the 1), he knows that one of the people is me and the other 5 are strangers. Since I'm amazing, he selfishly wants to maximize the chance that I survive.
His solution: he says he would pull the lever to save the five people. Since he doesn't know which one I am, he should save the most people to maximize the chance of saving me.
My solution: I say I wish he would put more thought into saving me. Instead of assuming a uniform distribution on my position, he should assume a uniform distribution on all distributions. So, he should pull the lever with 5/6 probability and let it kill the 5 with 1/6 probability.
So, what's the right answer? Is there even a right answer?
probability
asked Jul 29 at 19:43
user144527
38619
asked Jul 29 at 19:43
user144527
38619
asked Jul 29 at 19:43
user144527
38619
asked Jul 29 at 19:43
user144527
38619
38619
1
Your answer doesn't make sense unless you are going to repeat the experiment (which will not be possible if he got it wrong).
– Rob Arthan
Jul 29 at 19:52
@RobArthan: It would still be the wrong answer even if the experiment were repeatable (e.g. you'd lose a dollar instead of your life).
– joriki
Jul 29 at 19:53
1
@joriki: the scenario gives the boyfriend one binary choice in one experiment. The suggestion that the choice is made one way 5/6 of the time and the other 1/6 of the time doesn't make sense.
– Rob Arthan
Jul 29 at 20:01
1
@joriki: so the OP's suboptimal solution is that the boyfriend should equip himself with a high-probability random number generator and use it to make his decision.
– Rob Arthan
Jul 29 at 20:54
1
I said he should roll a d6.
– user144527
Jul 29 at 21:22
 |Â
show 5 more comments
1
Your answer doesn't make sense unless you are going to repeat the experiment (which will not be possible if he got it wrong).
– Rob Arthan
Jul 29 at 19:52
@RobArthan: It would still be the wrong answer even if the experiment were repeatable (e.g. you'd lose a dollar instead of your life).
– joriki
Jul 29 at 19:53
1
@joriki: the scenario gives the boyfriend one binary choice in one experiment. The suggestion that the choice is made one way 5/6 of the time and the other 1/6 of the time doesn't make sense.
– Rob Arthan
Jul 29 at 20:01
1
@joriki: so the OP's suboptimal solution is that the boyfriend should equip himself with a high-probability random number generator and use it to make his decision.
– Rob Arthan
Jul 29 at 20:54
1
I said he should roll a d6.
– user144527
Jul 29 at 21:22
1
1
Your answer doesn't make sense unless you are going to repeat the experiment (which will not be possible if he got it wrong).
– Rob Arthan
Jul 29 at 19:52
Your answer doesn't make sense unless you are going to repeat the experiment (which will not be possible if he got it wrong).
– Rob Arthan
Jul 29 at 19:52
@RobArthan: It would still be the wrong answer even if the experiment were repeatable (e.g. you'd lose a dollar instead of your life).
– joriki
Jul 29 at 19:53
@RobArthan: It would still be the wrong answer even if the experiment were repeatable (e.g. you'd lose a dollar instead of your life).
– joriki
Jul 29 at 19:53
1
1
@joriki: the scenario gives the boyfriend one binary choice in one experiment. The suggestion that the choice is made one way 5/6 of the time and the other 1/6 of the time doesn't make sense.
– Rob Arthan
Jul 29 at 20:01
@joriki: the scenario gives the boyfriend one binary choice in one experiment. The suggestion that the choice is made one way 5/6 of the time and the other 1/6 of the time doesn't make sense.
– Rob Arthan
Jul 29 at 20:01
1
1
@joriki: so the OP's suboptimal solution is that the boyfriend should equip himself with a high-probability random number generator and use it to make his decision.
– Rob Arthan
Jul 29 at 20:54
@joriki: so the OP's suboptimal solution is that the boyfriend should equip himself with a high-probability random number generator and use it to make his decision.
– Rob Arthan
Jul 29 at 20:54
1
1
I said he should roll a d6.
– user144527
Jul 29 at 21:22
I said he should roll a d6.
– user144527
Jul 29 at 21:22
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
3
down vote
His answer is the right answer. If he acts according to his answer, you survive with probability $frac56$. If he acts according to your answer, you survive with probability
$$
frac56cdotfrac56+frac16cdotfrac16=frac25+136=frac1318ltfrac1518=frac56;.
$$
answered Jul 29 at 19:50
joriki
164k10179328
1
I don't understand how you are interpreting the OP's suggested answer: the boyfriend has just one opportunity to make a decision.
– Rob Arthan
Jul 29 at 19:53
@RobArthan: Yes, that's right, I don't think we're interpreting the answer any differently. Only one probability in each term is for the boyfriend's decision; the other is for the girlfriend's identity.
– joriki
Jul 29 at 19:55
You cannot say I survive with probability 5/6 because that is under the assumption that I am placed according to a uniform distribution. In fact, a malevolent entity could have placed me as the lone person to try to trick my boyfriend into killing me, etc. You have to consider all possible distributions. Can you justify using a uniform distribution to compute my survival odds? How?
– user144527
Jul 29 at 20:06
@user144527 Boo. I liked your amusing presentation of this problem, but now I'm going to have to vote to close, for reasons that this is not about mathematics. :-(
– Theo Bendit
Jul 29 at 20:09
@user144527: I don't quite understand. You wrote in your justification for your proposed answer that "he should assume a uniform distribution on all distributions". I took that to mean that he should assume that it's equally likely that you're any one of the $6$ people (which seems like a reasonable assumption, absent any additional knowledge). If that's not what you meant by that formulation, then what did you mean?
– joriki
Jul 29 at 20:12
 |Â
show 6 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
His answer is the right answer. If he acts according to his answer, you survive with probability $frac56$. If he acts according to your answer, you survive with probability
$$
frac56cdotfrac56+frac16cdotfrac16=frac25+136=frac1318ltfrac1518=frac56;.
$$
answered Jul 29 at 19:50
joriki
164k10179328
1
I don't understand how you are interpreting the OP's suggested answer: the boyfriend has just one opportunity to make a decision.
– Rob Arthan
Jul 29 at 19:53
@RobArthan: Yes, that's right, I don't think we're interpreting the answer any differently. Only one probability in each term is for the boyfriend's decision; the other is for the girlfriend's identity.
– joriki
Jul 29 at 19:55
You cannot say I survive with probability 5/6 because that is under the assumption that I am placed according to a uniform distribution. In fact, a malevolent entity could have placed me as the lone person to try to trick my boyfriend into killing me, etc. You have to consider all possible distributions. Can you justify using a uniform distribution to compute my survival odds? How?
– user144527
Jul 29 at 20:06
@user144527 Boo. I liked your amusing presentation of this problem, but now I'm going to have to vote to close, for reasons that this is not about mathematics. :-(
– Theo Bendit
Jul 29 at 20:09
@user144527: I don't quite understand. You wrote in your justification for your proposed answer that "he should assume a uniform distribution on all distributions". I took that to mean that he should assume that it's equally likely that you're any one of the $6$ people (which seems like a reasonable assumption, absent any additional knowledge). If that's not what you meant by that formulation, then what did you mean?
– joriki
Jul 29 at 20:12
 |Â
show 6 more comments
up vote
3
down vote
His answer is the right answer. If he acts according to his answer, you survive with probability $frac56$. If he acts according to your answer, you survive with probability
$$
frac56cdotfrac56+frac16cdotfrac16=frac25+136=frac1318ltfrac1518=frac56;.
$$
answered Jul 29 at 19:50
joriki
164k10179328
1
I don't understand how you are interpreting the OP's suggested answer: the boyfriend has just one opportunity to make a decision.
– Rob Arthan
Jul 29 at 19:53
@RobArthan: Yes, that's right, I don't think we're interpreting the answer any differently. Only one probability in each term is for the boyfriend's decision; the other is for the girlfriend's identity.
– joriki
Jul 29 at 19:55
You cannot say I survive with probability 5/6 because that is under the assumption that I am placed according to a uniform distribution. In fact, a malevolent entity could have placed me as the lone person to try to trick my boyfriend into killing me, etc. You have to consider all possible distributions. Can you justify using a uniform distribution to compute my survival odds? How?
– user144527
Jul 29 at 20:06
@user144527 Boo. I liked your amusing presentation of this problem, but now I'm going to have to vote to close, for reasons that this is not about mathematics. :-(
– Theo Bendit
Jul 29 at 20:09
@user144527: I don't quite understand. You wrote in your justification for your proposed answer that "he should assume a uniform distribution on all distributions". I took that to mean that he should assume that it's equally likely that you're any one of the $6$ people (which seems like a reasonable assumption, absent any additional knowledge). If that's not what you meant by that formulation, then what did you mean?
– joriki
Jul 29 at 20:12
 |Â
show 6 more comments
up vote
3
down vote
up vote
3
down vote
His answer is the right answer. If he acts according to his answer, you survive with probability $frac56$. If he acts according to your answer, you survive with probability
$$
frac56cdotfrac56+frac16cdotfrac16=frac25+136=frac1318ltfrac1518=frac56;.
$$
answered Jul 29 at 19:50
joriki
164k10179328
His answer is the right answer. If he acts according to his answer, you survive with probability $frac56$. If he acts according to your answer, you survive with probability
$$
frac56cdotfrac56+frac16cdotfrac16=frac25+136=frac1318ltfrac1518=frac56;.
$$
answered Jul 29 at 19:50
joriki
164k10179328
answered Jul 29 at 19:50
joriki
164k10179328
answered Jul 29 at 19:50
joriki
164k10179328
answered Jul 29 at 19:50
joriki
164k10179328
164k10179328
1
I don't understand how you are interpreting the OP's suggested answer: the boyfriend has just one opportunity to make a decision.
– Rob Arthan
Jul 29 at 19:53
@RobArthan: Yes, that's right, I don't think we're interpreting the answer any differently. Only one probability in each term is for the boyfriend's decision; the other is for the girlfriend's identity.
– joriki
Jul 29 at 19:55
You cannot say I survive with probability 5/6 because that is under the assumption that I am placed according to a uniform distribution. In fact, a malevolent entity could have placed me as the lone person to try to trick my boyfriend into killing me, etc. You have to consider all possible distributions. Can you justify using a uniform distribution to compute my survival odds? How?
– user144527
Jul 29 at 20:06
@user144527 Boo. I liked your amusing presentation of this problem, but now I'm going to have to vote to close, for reasons that this is not about mathematics. :-(
– Theo Bendit
Jul 29 at 20:09
@user144527: I don't quite understand. You wrote in your justification for your proposed answer that "he should assume a uniform distribution on all distributions". I took that to mean that he should assume that it's equally likely that you're any one of the $6$ people (which seems like a reasonable assumption, absent any additional knowledge). If that's not what you meant by that formulation, then what did you mean?
– joriki
Jul 29 at 20:12
 |Â
show 6 more comments
1
I don't understand how you are interpreting the OP's suggested answer: the boyfriend has just one opportunity to make a decision.
– Rob Arthan
Jul 29 at 19:53
@RobArthan: Yes, that's right, I don't think we're interpreting the answer any differently. Only one probability in each term is for the boyfriend's decision; the other is for the girlfriend's identity.
– joriki
Jul 29 at 19:55
You cannot say I survive with probability 5/6 because that is under the assumption that I am placed according to a uniform distribution. In fact, a malevolent entity could have placed me as the lone person to try to trick my boyfriend into killing me, etc. You have to consider all possible distributions. Can you justify using a uniform distribution to compute my survival odds? How?
– user144527
Jul 29 at 20:06
@user144527 Boo. I liked your amusing presentation of this problem, but now I'm going to have to vote to close, for reasons that this is not about mathematics. :-(
– Theo Bendit
Jul 29 at 20:09
@user144527: I don't quite understand. You wrote in your justification for your proposed answer that "he should assume a uniform distribution on all distributions". I took that to mean that he should assume that it's equally likely that you're any one of the $6$ people (which seems like a reasonable assumption, absent any additional knowledge). If that's not what you meant by that formulation, then what did you mean?
– joriki
Jul 29 at 20:12
1
1
I don't understand how you are interpreting the OP's suggested answer: the boyfriend has just one opportunity to make a decision.
– Rob Arthan
Jul 29 at 19:53
I don't understand how you are interpreting the OP's suggested answer: the boyfriend has just one opportunity to make a decision.
– Rob Arthan
Jul 29 at 19:53
@RobArthan: Yes, that's right, I don't think we're interpreting the answer any differently. Only one probability in each term is for the boyfriend's decision; the other is for the girlfriend's identity.
– joriki
Jul 29 at 19:55
@RobArthan: Yes, that's right, I don't think we're interpreting the answer any differently. Only one probability in each term is for the boyfriend's decision; the other is for the girlfriend's identity.
– joriki
Jul 29 at 19:55
You cannot say I survive with probability 5/6 because that is under the assumption that I am placed according to a uniform distribution. In fact, a malevolent entity could have placed me as the lone person to try to trick my boyfriend into killing me, etc. You have to consider all possible distributions. Can you justify using a uniform distribution to compute my survival odds? How?
– user144527
Jul 29 at 20:06
You cannot say I survive with probability 5/6 because that is under the assumption that I am placed according to a uniform distribution. In fact, a malevolent entity could have placed me as the lone person to try to trick my boyfriend into killing me, etc. You have to consider all possible distributions. Can you justify using a uniform distribution to compute my survival odds? How?
– user144527
Jul 29 at 20:06
@user144527 Boo. I liked your amusing presentation of this problem, but now I'm going to have to vote to close, for reasons that this is not about mathematics. :-(
– Theo Bendit
Jul 29 at 20:09
@user144527 Boo. I liked your amusing presentation of this problem, but now I'm going to have to vote to close, for reasons that this is not about mathematics. :-(
– Theo Bendit
Jul 29 at 20:09
@user144527: I don't quite understand. You wrote in your justification for your proposed answer that "he should assume a uniform distribution on all distributions". I took that to mean that he should assume that it's equally likely that you're any one of the $6$ people (which seems like a reasonable assumption, absent any additional knowledge). If that's not what you meant by that formulation, then what did you mean?
– joriki
Jul 29 at 20:12
@user144527: I don't quite understand. You wrote in your justification for your proposed answer that "he should assume a uniform distribution on all distributions". I took that to mean that he should assume that it's equally likely that you're any one of the $6$ people (which seems like a reasonable assumption, absent any additional knowledge). If that's not what you meant by that formulation, then what did you mean?
– joriki
Jul 29 at 20:12
 |Â
show 6 more comments
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1
Your answer doesn't make sense unless you are going to repeat the experiment (which will not be possible if he got it wrong).
– Rob Arthan
Jul 29 at 19:52
@RobArthan: It would still be the wrong answer even if the experiment were repeatable (e.g. you'd lose a dollar instead of your life).
– joriki
Jul 29 at 19:53
1
@joriki: the scenario gives the boyfriend one binary choice in one experiment. The suggestion that the choice is made one way 5/6 of the time and the other 1/6 of the time doesn't make sense.
– Rob Arthan
Jul 29 at 20:01
1
@joriki: so the OP's suboptimal solution is that the boyfriend should equip himself with a high-probability random number generator and use it to make his decision.
– Rob Arthan
Jul 29 at 20:54
1
I said he should roll a d6.
– user144527
Jul 29 at 21:22