Top YouTube channel Veritasium releases video on Sleeping Beauty Problem

Unless I'm missing something, for optimal betting to be isomorphic to a correct application of Bayes' theorem, you have to bet for every event in the set that you're being asked about. If you're asked "conditional on you waking up, what's the probability of the coin having landed heads," the isomorphic question is "if I bet every time I wake up on the coin landing heads, what odds should I bet at to cut even," for which the answer is 2:1, which makes the corresponding probability $p=\frac{1}{3}$.

That assumption literally changes the nature of the problem, because the offer to bet, is information that you are using to update your posterior probability.

You can repair that problem by always offering the bet and ignoring one of the bets on tails. But of course that feels like cheating - I think most people would agree that if the odds makers are consistently ignoring bets on one side, then the odds no longer reflect the underlying probability.

Maybe there's another formulation that gives 1:1 odds, but I can't think of it.

If a non-negligible amount of money is at stake only for Tuesday bets, then it's obvious that you should bet on Tails, since Tuesday bets exist only when the coin lands Tails.

One can come up with a variety of betting scenarios, but for all the ones I've seen, you get the correct decision regarding what odds offered would give good bets only if you think the probability of Heads is 1/3.  You do have to be sure you're doing the decision theory correctly, though, accounting for the mechanism of the bets.

Wouldn't P(you wake up | heads) be 1/2, since you wake up on Monday but not on Tuesday? That would give P(you wake up) = 3/4, which gives P(heads | you wake up) = 1/3. That is, "you wake up" is not taken to mean "you wake up at least once," but "you wake up today (where today is Monday or Tuesday)." P(heads | you woke up at least once during the two days) is obviously 1/2.

I think not waking up/getting asked on HeadsTuesday (i.e. the Heads was flipped and the date is Tuesday) is important. This is the information you "gain" when you wake up (oh hey, it's not HeadsTuesday). When you are answering on Sunday, you answer for all outcomes, including when you don't wake up, but "given you're awake" excludes HeadsTuesday.

Let A, B, C, D denote HeadsMonday, HeadsTuesday, TailsMonday, TailsTuesday, and W = {A, C, D} be the set of days you wake up. We need to assign values P(A), P(B), P(C), P(D) with sum 1, and are looking for P(H | W) = P(A) / (P(A) + P(C) + P(D)).  From the coin being fair, P(H) = P(A) + P(B) = P(T) = P(C) + P(D) = 1/2.

If you think the chosen date and the coin flip are independent then you must have P(A) = P(B) = P(C) = P(D) = 1/4, so P(H | W) = 1/3. If not, the only way to get P(H | W) = 1/2 is for P(A) = 1/2, P(B) = 0. So, if Tuesday simply doesn't exist after flipping heads, you could be able to get P(H | wake up) = 1/2.