You're ignoring that with probability 1/4 agent ends up in room B.n that case you don't get to decide but you get to collect reward. Which is 3 for (the other agent) guessing T, or 0 for (the other agent) guessing H.

So basically guessing H is increasing your own expected reward at the expense of the other agent's expected reward (before you actually went to a room you didn't know if you'll be an agent which gets to decide or not so your expected reward also included part of expected reward for agent which doesn't get an opportunity to make a guess).

See also Psy-Kosh's non-anthropic problem.

I don't see the problem, and I suspect you're ignoring that the anthropic probability (2/3 probability for A1) is _ONLY_ in those worlds where you get to bet. Your payout is reduced whether you're in B or you're in A and bet wrong.

I _think_ that "always bet T" maximizes each agent's payout.

There are only 3 coins, so 8 possibilities (some of which collapse into the same case, but that doesn't matter). Payouts are for agent A (meaning 1+1 is 1 for A's bet and 1 for B's bet, paid to A). Agent B is symmetrical, so not shown. PayoutHiffA1 is if the strategy is to bet H in room A1 and T in A2. Total is for the rules as given, Agent pay is ONLY the agent's contribution, with no cross-agent payments.

COINS ROOMS PayoutH PayoutT PayoutHiffA1
 HHH  A1,A2  1+1     0+0      1+0
 HHT  A2,A1  1+1     0+0      0+1
 HTH  A1,A2  1+1     0+0      1+0
 HTT  A2,A1  1+1     0+0      0+1
 THH  A1,B   0,0     3+0      0+0
 THT  B,A1   0,0     0+3      0+0
 TTH  A2,B   0,0     3+0      3+0
 TTT  B,A2   0,0     0+3      0+3

 TOTAL PAY     8      12       10
 AGENT PAY     4       6        5

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.

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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.