The Ultimate Sleeping Beauty Problem

She responds "I'm sorry, but while I am a highly skilled mathematician, I'm actually from an alternate universe which is identical to yours except that in mine 'subjective probability' is the name of a particularly delicious ice cream flavor. Please precisely define what you mean by 'subjective probability', preferably by describing in detail a payoff structure such that my winnings will be maximized by selecting the correct answer to your query."

If you answered 1/2 to the original Sleeping Beauty Problem, the answer to this one should be reasonable to calculate. The probability of exactly n flips is (1/2)^n, so the probability of an even number of flips is (1/2)^2+(1/2)^4+(1/2)^6...=1/3.

If you answered 1/3 to the original Sleeping Beauty Problem, I do not think that there is any sensible answer to this one. I do not however consider this strong evidence that the answer of 1/3 is incorrect for the original problem. This could be an example of evidence for infinite set atheism. Analyzing this problem requires taking as given that the experiment can actually be repeated an arbitrarily large number of times, and we have thus far seen mostly evidence that this is not possible in our universe.

The expected value for "number of days lived by Sleeping Beauty" is an infinite series that diverges to infinity. If you think this is okay, then the Ultimate Sleeping Beauty problem isn't badly formed. Otherwise...

Coscott’s original problem is unsolvable by standard means because the expected number of wakings is infinite, so you can’t determine a frequency. That doesn’t mean it is unanswerable – we just need an isomorphism. After informing SB of the procedure and putting her to sleep the first time:

1) Set M=0. 2) Select a number N (I’ll discuss how later). 3) Flip a coin. a. If this (odd-numbered) flip lands heads, wake SB N times and end the experiment. b. If this flip (odd-numbered) lands tails, continue to step 4. 4) Flip the coin again. a. If this (even-numbered) flip lands heads, wake SB 3*N times and end the experiment; b. If this (even-numbered)flip lands tails, set M=M+1 go to step #2.

In Coscott’s version, we start with N=1 and multiply it by 9 each time we choose a new one; that is, N=9^M. But does the answer depend on N in any way? Halfers don’t think the answer depends on the number of wakings at all, and thirders think it depends only on the ratio of wakings in step 3a to those in step 4a, not the specific values.

So I maintain that my problem is the same as coscott’s, except in scale, no matter how we choose N. We can answer the original question by choosing N=1 every time.

There is a 2/3 chance of ending after an odd number of flips, and a 1/3 chance of ending after an even number. A halfer should claim SB gains no new knowledge by being awake, so P(odd|awake)=2/3 and P(even|awake)=1/3. A thirder should say there are four possible situations that awake SB could be in , and she cannot differentiate between them. Since 3 of them correspond to an even number of flips, P(odd|awake)=1/4 and P(even|awake)=3/4.

But like coscott’s, this variation, by itself, sheds no light on the original problem. We can even change numbers to something else:

1) Flip a coin. a. If this (odd-numbered) flip lands heads, wake SB N times and end the experiment. b. If this flip (odd-numbered) lands tails, continue to step 2. 2) Flip the coin again. a. If this (even-numbered) flip lands heads, wake SB M times and end the experiment; b. If this (even-numbered)flip lands tails, go to step #1.

You can decide for yourself whether you think the answer should depend on M and N, but I suspect most people will decide that based on whether they are halfers (“it can’t depend on M and N!”) or thirders (“it must depend on M and N!”), rather than what makes mathematical sense. (I’m not saying they will ignore mathematical, I’m saying they will define it by getting the answer they prefer.)

But as long as we are accepting the possibility of infinite wakings, what happens if we hold N constant and let M approach infinity? Halfers will still say the answers don’t change, thirders will say P(odd)=N/(M+N) and P(even)=M/(M+N).

But is it, or is it not, the same if we hold M constant, at a very large number, and let N approach 0? Because at N=0, P(odd)=0, which it can’t be if the halfers are right.

(1/2 3^0 + 1/8 3^2 + ...) / (1/2 3^0 + 1/4 3^1 + 1/8 * 3^2 + ...)

... which can be transformed into an infinite series with a Cesàro sum of 0.5, so that's my answer.

You might want to read some papers by Jacob Ross for similar examples, and discussion.

Note that if she has no way to test whether her calculation is correct, the notion of probability does not make sense in her situation. In other words, what would she do differently if she estimates the probability to be, say, 1/2 instead of, say, 1/3?

50%. Upon finding that the expected physical situation is undefined (a limit that does not converge), sensible agents should default to using a more limited set of information.

EDIT: All right then, if you downvoters are so smart, what would you bet if you were in sleeping beauty's place?