Consider the following classic problem:

Some researchers would like to put Sleeping Beauty to sleep on Sunday. Depending on the secret toss of a fair coin, they will briefly awaken her either once on Monday (Heads) or twice (first on Monday then again on Tuesday) (Tails). After each waking, they will put her back to sleep with a drug that makes her forget that awakening, and finally she will be awakened on Wednesday without being asked any questions and the experiment will end. When she is awakened (before Wednesday—and she will be told it is before Wednesday, but not whether it is Monday or Tuesday), to what degree should Sleeping Beauty believe that the outcome of the coin toss was Heads?

In a previous thread (where I borrowed and slightly modified the quotation), whuber convincingly argues that the problem as stated above is ambiguous and gives interpretations under which the answer is either $$13\frac{1}{3}$$ or $$12\frac{1}{2}$$, with $$13\frac{1}{3}$$ being the more interesting answer. I recommend reading whuber’s response before attempting to respond to this post.

Now consider the following modification, borrowed from a 2015 blog post:

Before going to sleep on Sunday, Sleeping Beauty makes a bet at odds of 3:2 that the coin will come down heads. (This is favourable for her when the probability of heads is 1/2, and unfavourable when the probability of heads is 1/3). She is told that whenever she is woken up, she will be offered the opportunity to cancel any outstanding bets. Later she finds herself woken up, and asked whether she wants to cancel any outstanding bets. Should she say yes or no? (Let’s say she doesn’t have access to any external randomness to help her choose). Is her best answer compatible with a “belief of 1/3 that the coin is showing heads”?

The issue in the modified version is that because the coin is fair, the expected value of the bet should be $$3⋅12−2⋅12>03 \cdot \frac{1}{2} - 2 \cdot \frac{1}{2} > 0$$. But when Sleeping Beauty is awoken, by whbuer’s reasoning, she assigns a probability of $$13\frac{1}{3}$$ to the coin coming coming up Heads given that she is the one awakened. In this case, the expected value is $$3⋅13−2⋅23<03 \cdot \frac{1}{3} - 2 \cdot \frac{2}{3} < 0$$, so she should cancel the bet. Yet nothing about the bet seems intuitively to have changed.

In the blog post linked above, Sleeping Beauty reasons that while the probability she assigns to the coin coming up Heads based on being awakened before Wednesday is $$13\frac{1}{3}$$, on Wednesday she will experience the event of waking up on a Wednesday, at which point the probability of Heads given waking up on Wednesday will be $$12\frac{1}{2}$$, so she defers the bet to later.

However, since Sleeping Beauty already knows earlier that she will eventually wake up on Wednesday, doesn't that argument mean that the answer to the original Sleeping Beauty paradox should "morally" be $$12\frac{1}{2}$$ rather than $$13\frac{1}{3}$$? How do you resolve the clash between the intuitive feeling that Sleeping Beauty should not cancel the bet and whuber's reasoning for the probability of Heads given that she was the one awoken being $$13\frac{1}{3}$$? Should Sleeping Beauty cancel her bet?

What this line of thinking fails to account for is the fact that, if it's tails and she cancelled the bet on Monday, then cancelling the bet again on Tuesday does nothing. The benefit of the action of cancelling the bet on Tuesday depends on what she did on Monday. From the sleeping beauty's perspective, when she's awakened, she should be thinking "what is my expected gain, from the baseline scenario of the bet, if I cancel the bet?" In other words, what is the expected difference between cancelling the bet, and not cancelling the bet? (This, of course, is just the negative of the EV of the bet, because the value of no bet is simply zero.)

Well let's examine the three cases:

Case 1 - Monday Heads: If she cancels the bet, she loses 3

Case 2 - Monday Tails: If she cancels the bet, she gains 2

Case 3 - Tuesday Tails: If she cancels the bet, and she didn't cancel the bet on Monday, she gains 2. If she cancels the bet, but she already cancelled the bet on Monday, then nothing changes.

To ascertain the expected gain of the bet-cancelling action, we have to know the probability that she cancelled the bet on Monday, given that she cancels on Tuesday, because her gain from the action of "cancel bet" on Tuesday depends on that. However, the problem isn't specified enough to tell us that.

We can make a simplifying assumption. Let's assume she does the same thing every time, that is, she picks a strategy "do this if I'm awakened". So if she cancels the bet on Tuesday, then she necessarily cancelled the bet on Monday. Thus, her gain from the baseline, for cancelling the bet on Tuesday, is zero. Then her expected gain from cancelling the bet is $$−3⋅13+2⋅13+0⋅13=−13.-3 \cdot \frac{1}{3} + 2 \cdot \frac{1}{3} + 0 \cdot \frac{1}{3} = -\frac{1}{3}.$$

This is in contrast to the naive approach, where the expected gain from cancelling the bet on Tuesday is equal to that of Monday on a tails roll, in which case, the calculation is (falsely) $$−3⋅13+2⋅13+2⋅13=13.-3 \cdot \frac{1}{3} + 2 \cdot \frac{1}{3} + 2 \cdot \frac{1}{3} = \frac{1}{3}.$$ Notice that this is simply the negative of what you wrote as the EV of the bet in the thirder's perspective. The expected gain from cancelling a bet is the negative of the expected value of the bet. But that expected gain naively ignores the fact that cancelling on Tuesday does nothing if you cancelled on Monday. When you don't ignore that, as I showed above, the expected gain of cancelling the bet is negative.

Thus, even from the Thirder's perspective, she should not cancel the bet.