Unstated in the problem (which is the main point of confusion IMO) is what decision Beauty is making, and what are the payouts/costs if she’s right or wrong. What difference does it make to her if she says 1/pi as the probability, because that’s prettier than 1⁄2 or 1/3?
If the payout is +1 utility for being correct and −1 for being incorrect, calculated each time the question’s asked, then 1⁄3 is the correct answer, because she’ll lose twice if wrong, but only win once if right. If the payout is calculated only once on Wednesday, with a 0 payout if she manages to answer differently Monday and Tuesday, then 1⁄2 is the right answer.
They are not simultaneously making that decision for all situations that they will ever find themselves in where the rational decision to make happens to be the same (even if the rationale for making that decision is also the same).
Aren’t they? If there is zero detectable change in cognition or evidence between the two decisions, how could the second one be different?
No, you get the wrong answer in your second scenario (with −1, 0, or +1 payoff) if you assign a probability of 1⁄2 to Heads, and you get the right answer if you assign a probability of 1⁄3.
In this scenario, guessing right is always better than guessing wrong. Being right rather than wrong either (A) gives a payoff of +1 rather than −1, if you guess only once, or (B) gives a payoff of +1 rather than 0, if you guess correctly another day, or (C) gives a payoff of 0 rather than −1, if you guess incorrectly another day. Since the change in payoff for (B) and (C) are the same, one can summarize this by saying that the advantage of guessing right is +2 if you guess only once (ie, the coin landed Heads), and +1 if you guess twice (ie, the coin landed Tails).
A Halfer will compute the difference in payoff from guessing Heads rather than Tails as (1/2)*(+2) + (1/2)*(-1) = 1⁄2, and so they will guess Heads (both days, presumably, if the coin lands Tails). A Thirder will compute the difference in payoff from guessing Heads rather than Tails as (1/3)*(+2) + (2/3)*(-1) = 0, so they will be indifferent between guessing Heads or Tails. If we change the problem slightly so that there is a small cost (say 1⁄100) to guessing Heads (regardless of whether this guess is right or wrong), then a Halfer will still prefer Heads, but a Thirder will now definitely prefer Tails.
What will actually happen without the small penalty is that both the Halfer and the Thirder will get an average payoff of zero, which is what the Thirder expects, but not what the Halfer expects. If we include the 1⁄100 penalty for guessing Heads, the Halfer has an expected payoff of −1/100, while the Thirder still has an expected payoff of zero, so the Thirder does better.
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If today you choose chocolate over vanilla ice cream today, and yesterday you did the same, and you’re pretty sure that you will always choose chocolate over vanilla, is your decision today really a decision not for one ice cream cone but for thousands of cones? Not by any normal idea of what it means to “decide”.
No, you get the wrong answer in your second scenario (with −1, 0, or +1 payoff) if you assign a probability of 1⁄2 to Heads, and you get the right answer if you assign a probability of 1⁄3.
Huh? Maybe I wasn’t clear in my second scenario. This is the situation where the bet is resolved only once, on Wednesday, with payouts being: a) +1 if it was heads, and she said “heads” on Tuesday (not being woken up on Monday); b) −1 if it was heads but she said “tails” on tuesday; c) +1 if it was tails and she said “tails” on both monday and tuesday; d) −1 if it was tails and she said heads on both monday and tuesday; e) (for completeness, I don’t believe it’s possible) 0 if it was tails and she gave different answers on monday and tuesday.
1⁄2 is the right answer here—it’s a literal coinflip and we’ve removed the double-counting of the mindwipe.
You’ve swapped Monday and Tuesday compared to the usual description of the problem, but other than that, your description is what I am working with. You just have a mistaken intuition regarding how the probabilities relate to decisions—it’s slightly non-obvious (but maybe not obvious that it’s non-obvious). Note that this is all using completely standard probability and decision theory—I’m not doing anything strange here.
In this situation, as explained in detail in my reply above, Beauty gets the right answer regarding how to bet only if she gives probability 1⁄3 to Heads whenever she is woken, in which case she is indifferent to guessing Heads versus Tails (as she should be—as you say, it’s just a coin flip), whereas if she gives probability 1⁄2 to Heads, she will have a definite preference for guessing Heads. If we give guessing Heads a small penalty (say on Monday only, to resolve how this works if her guesses differ on the two days), in order to tip the scales away from indifference, the Thirder Beauty correctly guesses Tails, which does indeed maximizes her expected reward, whereas the Halfer Beauty does the wrong thing by still guessing Heads.
Unstated in the problem (which is the main point of confusion IMO) is what decision Beauty is making, and what are the payouts/costs if she’s right or wrong. What difference does it make to her if she says 1/pi as the probability, because that’s prettier than 1⁄2 or 1/3?
If the payout is +1 utility for being correct and −1 for being incorrect, calculated each time the question’s asked, then 1⁄3 is the correct answer, because she’ll lose twice if wrong, but only win once if right. If the payout is calculated only once on Wednesday, with a 0 payout if she manages to answer differently Monday and Tuesday, then 1⁄2 is the right answer.
Aren’t they? If there is zero detectable change in cognition or evidence between the two decisions, how could the second one be different?
No, you get the wrong answer in your second scenario (with −1, 0, or +1 payoff) if you assign a probability of 1⁄2 to Heads, and you get the right answer if you assign a probability of 1⁄3.
In this scenario, guessing right is always better than guessing wrong. Being right rather than wrong either (A) gives a payoff of +1 rather than −1, if you guess only once, or (B) gives a payoff of +1 rather than 0, if you guess correctly another day, or (C) gives a payoff of 0 rather than −1, if you guess incorrectly another day. Since the change in payoff for (B) and (C) are the same, one can summarize this by saying that the advantage of guessing right is +2 if you guess only once (ie, the coin landed Heads), and +1 if you guess twice (ie, the coin landed Tails).
A Halfer will compute the difference in payoff from guessing Heads rather than Tails as (1/2)*(+2) + (1/2)*(-1) = 1⁄2, and so they will guess Heads (both days, presumably, if the coin lands Tails). A Thirder will compute the difference in payoff from guessing Heads rather than Tails as (1/3)*(+2) + (2/3)*(-1) = 0, so they will be indifferent between guessing Heads or Tails. If we change the problem slightly so that there is a small cost (say 1⁄100) to guessing Heads (regardless of whether this guess is right or wrong), then a Halfer will still prefer Heads, but a Thirder will now definitely prefer Tails.
What will actually happen without the small penalty is that both the Halfer and the Thirder will get an average payoff of zero, which is what the Thirder expects, but not what the Halfer expects. If we include the 1⁄100 penalty for guessing Heads, the Halfer has an expected payoff of −1/100, while the Thirder still has an expected payoff of zero, so the Thirder does better.
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If today you choose chocolate over vanilla ice cream today, and yesterday you did the same, and you’re pretty sure that you will always choose chocolate over vanilla, is your decision today really a decision not for one ice cream cone but for thousands of cones? Not by any normal idea of what it means to “decide”.
Huh? Maybe I wasn’t clear in my second scenario. This is the situation where the bet is resolved only once, on Wednesday, with payouts being: a) +1 if it was heads, and she said “heads” on Tuesday (not being woken up on Monday); b) −1 if it was heads but she said “tails” on tuesday; c) +1 if it was tails and she said “tails” on both monday and tuesday; d) −1 if it was tails and she said heads on both monday and tuesday; e) (for completeness, I don’t believe it’s possible) 0 if it was tails and she gave different answers on monday and tuesday.
1⁄2 is the right answer here—it’s a literal coinflip and we’ve removed the double-counting of the mindwipe.
You’ve swapped Monday and Tuesday compared to the usual description of the problem, but other than that, your description is what I am working with. You just have a mistaken intuition regarding how the probabilities relate to decisions—it’s slightly non-obvious (but maybe not obvious that it’s non-obvious). Note that this is all using completely standard probability and decision theory—I’m not doing anything strange here.
In this situation, as explained in detail in my reply above, Beauty gets the right answer regarding how to bet only if she gives probability 1⁄3 to Heads whenever she is woken, in which case she is indifferent to guessing Heads versus Tails (as she should be—as you say, it’s just a coin flip), whereas if she gives probability 1⁄2 to Heads, she will have a definite preference for guessing Heads. If we give guessing Heads a small penalty (say on Monday only, to resolve how this works if her guesses differ on the two days), in order to tip the scales away from indifference, the Thirder Beauty correctly guesses Tails, which does indeed maximizes her expected reward, whereas the Halfer Beauty does the wrong thing by still guessing Heads.