And here are examples that I don’t think that rephrasing as betting resolves:
Convinced by the Sleeping Beauty problem, you buy a lottery ticket and set up a robot to put you to sleep and then, if the lottery ticket wins, wake you up 1 billion times, and if not just wake you up once. You wake up. What is the expected value of the lottery ticket you’re holding? You knew ahead of time that you will wake up at least once, so did you just game the system? No, since I would argue that this system is better modeled by the Sleeping Beauty problem when you get only a single payout regardless of how many times you wake up.
Or: if the coin comes up heads, then you and your memories get cloned. When you wake up you’re offered the deal on the spot 1:1 bet on the coin. Is this a good bet for you? (Your wallet gets cloned too, let’s say.) That depends on how you value your clone receiving money. But why should P(H|awake) be different in this scenario than in Sleeping Beauty, or different between people who do value their clone versus people who do not?
Or: No sleeping beauty shenanigans. I just say “Let’s make a bet. I’ll flip a coin. If the coin was heads we’ll execute the bet twice. If tails, just once. What odds do you offer me?” Isn’t that all that you are saying in this Sleeping Beauty with Betting scenario? The expected value of a bet is a product of the payoff with the probability—the payoff is twice as high in the case of heads, so why should I think that the probability is also twice as high?
I argue that this is the very question of the problem: is being right twice worth twice as much?
Or: No sleeping beauty shenanigans. I just say “Let’s make a bet. I’ll flip a coin. If the coin was heads we’ll execute the bet twice. If tails, just once. What odds do you offer me?”
By executing the bet twice, do you mean I lose/win twice as much money as I’d otherwise lost/won?
Yes, exactly. You choose either heads or tails. I flip the coin. If it’s tails and matches what you chose, then you win $1 otherwise lose $1. If it’s heads and matches what you chose, you win $2 otherwise you lose $2. Clearly you will choose heads in this case, just like the Sleeping Beauty when betting every time you wake up. But you choose heads because we’ve increased the payout not the probabilities.
If it’s tails and matches what you chose, then you win $1 otherwise lose $1. If it’s heads and matches what you chose, you win $2 otherwise you lose $2.
Both options have the expected value equal to zero though? (0.5⋅1−0.5⋅1 versus 0.5⋅2−0.5⋅2.)
If you choose heads, you either win $2 (ie win $1 twice) or lose $1. If you choose tails then you either win $1 or lose $2. It’s exactly the same as the Sleeping Beauty problem with betting, just you have to precommit to a choice of heads/tail ahead of time. Sorry that this situation is weird to describe and unclear.
And here are examples that I don’t think that rephrasing as betting resolves:
Convinced by the Sleeping Beauty problem, you buy a lottery ticket and set up a robot to put you to sleep and then, if the lottery ticket wins, wake you up 1 billion times, and if not just wake you up once. You wake up. What is the expected value of the lottery ticket you’re holding? You knew ahead of time that you will wake up at least once, so did you just game the system? No, since I would argue that this system is better modeled by the Sleeping Beauty problem when you get only a single payout regardless of how many times you wake up.
Or: if the coin comes up heads, then you and your memories get cloned. When you wake up you’re offered the deal on the spot 1:1 bet on the coin. Is this a good bet for you? (Your wallet gets cloned too, let’s say.) That depends on how you value your clone receiving money. But why should P(H|awake) be different in this scenario than in Sleeping Beauty, or different between people who do value their clone versus people who do not?
Or: No sleeping beauty shenanigans. I just say “Let’s make a bet. I’ll flip a coin. If the coin was heads we’ll execute the bet twice. If tails, just once. What odds do you offer me?” Isn’t that all that you are saying in this Sleeping Beauty with Betting scenario? The expected value of a bet is a product of the payoff with the probability—the payoff is twice as high in the case of heads, so why should I think that the probability is also twice as high?
I argue that this is the very question of the problem: is being right twice worth twice as much?
By executing the bet twice, do you mean I lose/win twice as much money as I’d otherwise lost/won?
Yes, exactly. You choose either heads or tails. I flip the coin. If it’s tails and matches what you chose, then you win $1 otherwise lose $1. If it’s heads and matches what you chose, you win $2 otherwise you lose $2. Clearly you will choose heads in this case, just like the Sleeping Beauty when betting every time you wake up. But you choose heads because we’ve increased the payout not the probabilities.
Both options have the expected value equal to zero though? (0.5⋅1−0.5⋅1 versus 0.5⋅2−0.5⋅2.)
If you choose heads, you either win $2 (ie win $1 twice) or lose $1. If you choose tails then you either win $1 or lose $2. It’s exactly the same as the Sleeping Beauty problem with betting, just you have to precommit to a choice of heads/tail ahead of time. Sorry that this situation is weird to describe and unclear.