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.
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.