I’m going to rephrase this using as many integers as possible because humans are better at reasoning about those. I know I personally am.
Instead of randomness, we have four teams that perform this experiment. Teams 1 and 2 represent the first flip landing on heads. Team 3 is tails then heads, and team 4 is tails then tails. No one knows which team they’ve been assigned to.
Also, instead of earning $1 or $3 for both participants, a correct guess earns that same amount once. They still share finances so this shouldn’t affect anyone’s reasoning; I just don’t want to have to double it.
Team 1 makes 2 guesses. Each “heads” guess earns $1, each “tails” guess earns nothing.
Team 2 makes 2 guesses. Each “heads” guess earns $1, each “tails” guess earns nothing.
Team 3 makes 1 guess. Guessing “heads” earns nothing, guessing “tails” earns $3.
Team 4 makes 1 guess. Guessing “heads” earns nothing, guessing “tails” earns $3.
If absolutely everyone guesses “heads,” teams 1 and 2 will earn $4 between them. If absolutely everyone guesses “tails,” teams 3 and 4 will earn $6 between them. So far, this matches up.
Now let’s look at how many people were sent to each room.
Three people visit room A1: one from team 1, one from team 2, and one from team 3. 2⁄3 of them are there because the first “flip” was heads.
Three people visit room A2: one from team 1, one from team 2, and one from team 4. 2⁄3 of them are there because the first “flip” was heads.
Two people visit room B: one from team 3 and one from team 4. They don’t matter.
The three visitors to A1 know they aren’t on team 4, thus they can subtract that team’s entire winnings from their calculations, leaving $4 vs. $3.
The three visitors to A2 know they aren’t on team 3, thus they can subtract that team’s entire winnings from their calculations, leaving $4 vs. $3.
Do you see the error? Took me a bit.
If you’re in room A1, you need to subtract more than just team 4′s winnings. You need to subtract half of team 1 and team 2′s winnings. Teams 1 and 2 each have someone in room A2, and you can’t control their vote. Thus:
Three people visit room A1: one from team 1, one from team 2, and one from team 3. If all three guess “heads” they earn $2 in all. If all three guess “tails” they earn $3 in all.
Three people visit room A1: one from team 1, one from team 2, and one from team 4. If all three guess “heads” they earn $2 in all. If all three guess “tails” they earn $3 in all.
Guessing “tails” remains the best way to maximize expected value.
---
The lesson here isn’t so much to do with EDT agents, it’s to do with humans and probabilities. I didn’t write this post because I’m amazing and you’re a bad math student, I wrote this post because without it, I wouldn’t have been able to figure it out either.
Whenever this sort of thing comes up, try to rephrase the problem. Instead of 85%, imagine 100 people in a room, with 85 on the left and 15 on the right. Instead of truly random experiments, imagine the many-worlds interpretation, where each outcome is guaranteed to come up in a different branch. (And try to have an integer number of branches, each representing an equal fraction.) Or use multiple teams like I did above.
I don’t think this is right. A superrational agent exploits the symmetry between A1 and A2, correct? So it must reason that an identical agent in A2 will reason the same way as it does, and if it bets heads, so will the other agent. That’s the point of bringing up EDT.
Oh right, I see where you’re coming from. When I said “you can’t control their vote” I was missing the point, because as far superrational agents are concerned, they do control each other’s votes. And in that case, it sure seems like they’ll go for the $2, earning less money overall.
It occurs to me that if team 4 didn’t exist, but teams 1-3 were still equally likely, then “heads” actually would be the better option. If everyone guesses “heads,” two teams are right, and they take home $4. If everyone guesses “tails,” team 3 takes home $3 and that’s it. On average, this maximizes winnings.
Except this isn’t the same situation at all. With group 4 eliminated from the get go, the remaining teams can do even better than $4 or $3. Teammates in room A2 knows for a fact that the coin landed heads, and they automatically earn $1. Teammates in room A1 are no longer responsible for their teammates’ decisions, so they go for the $3. Thus teams 1 and 2 both take home $1 while team 3 takes home $3, for a total of $5.
Maybe that’s the difference. Even if you know for a fact that you aren’t on team 4, you also aren’t in a world where team 4 was eliminated from the start. The team still needs to factor into your calculations… somehow. Maybe it means your teammate isn’t really making the same decision you are? But it’s perfectly symmetrical information. Maybe you don’t get to eliminate team 4 unless your teammate does? But the proof is right in front of you. Maybe the information isn’t symmetrical because your teammate could be in room B?
I don’t know. I feel like there’s an answer in here somewhere, but I’ve spent several hours on this post and I have other things to do today.
I do want to add—separately—that superrational agents (not sure about EDT) can solve this problem in a roundabout way.
Imagine if some prankster erased the “1” and “2″ from the signs in rooms A1 and A2, leaving just “A” in both cases. Now everyone has less information and makes better decisions. And in the real contest, (super)rational agents could achieve the same effect by keeping their eyes closed. Simply say “tails,” maximize expected value, and leave the room never knowing which one it was.
None of which should be necessary. (Super)rational agents should win even after looking at the sign. They should be able to eliminate a possibility and still guess “tails.” A flaw must exist somewhere in the argument for “heads,” and even if I haven’t found that flaw, a perfect logician would spot it no problem.
I’m going to rephrase this using as many integers as possible because humans are better at reasoning about those. I know I personally am.
Instead of randomness, we have four teams that perform this experiment. Teams 1 and 2 represent the first flip landing on heads. Team 3 is tails then heads, and team 4 is tails then tails. No one knows which team they’ve been assigned to.
Also, instead of earning $1 or $3 for both participants, a correct guess earns that same amount once. They still share finances so this shouldn’t affect anyone’s reasoning; I just don’t want to have to double it.
If absolutely everyone guesses “heads,” teams 1 and 2 will earn $4 between them. If absolutely everyone guesses “tails,” teams 3 and 4 will earn $6 between them. So far, this matches up.
Now let’s look at how many people were sent to each room.
Do you see the error? Took me a bit.
If you’re in room A1, you need to subtract more than just team 4′s winnings. You need to subtract half of team 1 and team 2′s winnings. Teams 1 and 2 each have someone in room A2, and you can’t control their vote. Thus:
Guessing “tails” remains the best way to maximize expected value.
---
The lesson here isn’t so much to do with EDT agents, it’s to do with humans and probabilities. I didn’t write this post because I’m amazing and you’re a bad math student, I wrote this post because without it, I wouldn’t have been able to figure it out either.
Whenever this sort of thing comes up, try to rephrase the problem. Instead of 85%, imagine 100 people in a room, with 85 on the left and 15 on the right. Instead of truly random experiments, imagine the many-worlds interpretation, where each outcome is guaranteed to come up in a different branch. (And try to have an integer number of branches, each representing an equal fraction.) Or use multiple teams like I did above.
I don’t think this is right. A superrational agent exploits the symmetry between A1 and A2, correct? So it must reason that an identical agent in A2 will reason the same way as it does, and if it bets heads, so will the other agent. That’s the point of bringing up EDT.
Oh right, I see where you’re coming from. When I said “you can’t control their vote” I was missing the point, because as far superrational agents are concerned, they do control each other’s votes. And in that case, it sure seems like they’ll go for the $2, earning less money overall.
It occurs to me that if team 4 didn’t exist, but teams 1-3 were still equally likely, then “heads” actually would be the better option. If everyone guesses “heads,” two teams are right, and they take home $4. If everyone guesses “tails,” team 3 takes home $3 and that’s it. On average, this maximizes winnings.
Except this isn’t the same situation at all. With group 4 eliminated from the get go, the remaining teams can do even better than $4 or $3. Teammates in room A2 knows for a fact that the coin landed heads, and they automatically earn $1. Teammates in room A1 are no longer responsible for their teammates’ decisions, so they go for the $3. Thus teams 1 and 2 both take home $1 while team 3 takes home $3, for a total of $5.
Maybe that’s the difference. Even if you know for a fact that you aren’t on team 4, you also aren’t in a world where team 4 was eliminated from the start. The team still needs to factor into your calculations… somehow. Maybe it means your teammate isn’t really making the same decision you are? But it’s perfectly symmetrical information. Maybe you don’t get to eliminate team 4 unless your teammate does? But the proof is right in front of you. Maybe the information isn’t symmetrical because your teammate could be in room B?
I don’t know. I feel like there’s an answer in here somewhere, but I’ve spent several hours on this post and I have other things to do today.
I do want to add—separately—that superrational agents (not sure about EDT) can solve this problem in a roundabout way.
Imagine if some prankster erased the “1” and “2″ from the signs in rooms A1 and A2, leaving just “A” in both cases. Now everyone has less information and makes better decisions. And in the real contest, (super)rational agents could achieve the same effect by keeping their eyes closed. Simply say “tails,” maximize expected value, and leave the room never knowing which one it was.
None of which should be necessary. (Super)rational agents should win even after looking at the sign. They should be able to eliminate a possibility and still guess “tails.” A flaw must exist somewhere in the argument for “heads,” and even if I haven’t found that flaw, a perfect logician would spot it no problem.