I think you’re missing a term in your second calculation. And why are anthropism and copies of you necessary for this puzzle. I suspect the answer will indicate something I’m completely missing about this series.
Take this for straight-up probability:
I have two jars of marbles, one with 18 green and 2 red, the other with 18 red and two green. Pick one jar at random, then look at one marble from that jar at random.
If you pick green, what’s the chance that your jar is mostly green? I say 90%, by fairly straightforward application of bayes’ rule.
I offer a wager: you get $1 per green and lose $3 per red marble in the jar you chose.
After seeing a green marble, I think your EV is $5.60. After seeing a red marble, I think your EV is $0 (you decline the bet). If you are forced to make the wager before seeing anything, conditional on drawing green, I think your EV is $2.80. I calculate it thus: 50% to get mostly-green jar, and 90% of that will you see green and take the bet, which is worth +$1*18 - $3*2 in this case. 50% to get mostly-red, 10% of which will you draw green, worth +1*2 - $3*18.
0.5 * 0.9 (1 \ 18 − 3 * 2) + 0.5 * 0.1 (1 \ 2 − 3 * 18) = 2.80, which is consistent: half the time you pick green, with EV of 5.60.
I think you left out the probability that you’ll get green and take the bet in each of your 0.5 probabilities for the conditional strategy. Multiply a 0.9 to the first term and 0.1 into the second, and everything gets consistent.
The problem is that we aren’t asking one randomly selected person, we’re asking all of the green ones (they have to agree unanimously for the Yes vote to go through).
Ah, I see. You’re asking all the green ones, but only paying each pod once. This feels like reverse-weighting the payout, so it should still be -EV even after waking up, but I haven’t quite worked out a way to include that in the numbers...
“However, before the experiment, you calculate the general utility of the conditional strategy “Reply ‘Yes’ to the question if you wake up in a green room” as (50% ((18 +$1) + (2 -$3))) + (50% ((18 -$3) + (2 +$1))) = -$20. You want your future selves to reply ‘No’ under these conditions.”
The sum given is the one you would perform if you did not know which room you woke up in. Surely a different sum is appropriate with the additional evidence that you awoke in a green room.
Incidentally, this problem seems far too complicated! I feel like the programmer faced with a bug report which failed to find some simple code that nontheless manages to reproduce the problem. Simplify, simplify, simplify!
I think you’re missing a term in your second calculation. And why are anthropism and copies of you necessary for this puzzle. I suspect the answer will indicate something I’m completely missing about this series.
Take this for straight-up probability:
I have two jars of marbles, one with 18 green and 2 red, the other with 18 red and two green. Pick one jar at random, then look at one marble from that jar at random.
If you pick green, what’s the chance that your jar is mostly green? I say 90%, by fairly straightforward application of bayes’ rule.
I offer a wager: you get $1 per green and lose $3 per red marble in the jar you chose.
After seeing a green marble, I think your EV is $5.60. After seeing a red marble, I think your EV is $0 (you decline the bet). If you are forced to make the wager before seeing anything, conditional on drawing green, I think your EV is $2.80. I calculate it thus: 50% to get mostly-green jar, and 90% of that will you see green and take the bet, which is worth +$1*18 - $3*2 in this case. 50% to get mostly-red, 10% of which will you draw green, worth +1*2 - $3*18. 0.5 * 0.9 (1 \ 18 − 3 * 2) + 0.5 * 0.1 (1 \ 2 − 3 * 18) = 2.80, which is consistent: half the time you pick green, with EV of 5.60.
I think you left out the probability that you’ll get green and take the bet in each of your 0.5 probabilities for the conditional strategy. Multiply a 0.9 to the first term and 0.1 into the second, and everything gets consistent.
The problem is that we aren’t asking one randomly selected person, we’re asking all of the green ones (they have to agree unanimously for the Yes vote to go through).
Ah, I see. You’re asking all the green ones, but only paying each pod once. This feels like reverse-weighting the payout, so it should still be -EV even after waking up, but I haven’t quite worked out a way to include that in the numbers...
The second sum still seems wrong. Here it is:
“However, before the experiment, you calculate the general utility of the conditional strategy “Reply ‘Yes’ to the question if you wake up in a green room” as (50% ((18 +$1) + (2 -$3))) + (50% ((18 -$3) + (2 +$1))) = -$20. You want your future selves to reply ‘No’ under these conditions.”
The sum given is the one you would perform if you did not know which room you woke up in. Surely a different sum is appropriate with the additional evidence that you awoke in a green room.
Incidentally, this problem seems far too complicated! I feel like the programmer faced with a bug report which failed to find some simple code that nontheless manages to reproduce the problem. Simplify, simplify, simplify!