# Wager calculation for the thirder position (Sleeping Beauty problem)

Here is the problem for those unfamiliar with it:

Sleeping Beauty volunteers to undergo the following experiment and is told all of the following details: On Sunday she will be put to sleep. Once or twice, during the experiment, Sleeping Beauty will be awakened, interviewed, and put back to sleep with an amnesia-inducing drug that makes her forget that awakening. A fair coin will be tossed to determine which experimental procedure to undertake:

If the coin comes up heads, Sleeping Beauty will be awakened and interviewed on Monday only.

If the coin comes up tails, she will be awakened and interviewed on Monday and Tuesday.

In either case, she will be awakened on Wednesday without interview and the experiment ends.

Any time Sleeping Beauty is awakened and interviewed she will not be able to tell which day it is or whether she has been awakened before. During the interview Sleeping Beauty is asked: "What is your credence now for the proposition that the coin landed heads?"

Now I understand what the thirder position is and how they arrived there and I still think it is wrong. Though in my quest to understand the reasoning they came up with an example:

An alternative argument is as follows: Credence can be viewed as the amount a rational risk-neutral bettor would wager if the payoff for being correct is 1 unit (the wager itself being lost either way). In the heads scenario, Sleeping Beauty would spend her wager amount one time, and receive 1 money for being correct. In the tails scenario, she would spend her wager amount twice, and receive nothing. Her expected value is therefore to gain 0.5 but also lose 1.5 times her wager, thus she should break even if her wager is 1/3.

And quite frankly I don't understand how they ended up with 0.5 gain and 1.5 loss. Like what are they doing and why?

Like if she were to place a bet each time and lose a token for being incorrect and gain one for being correct, then there are 4 possible scenarios:

• She believes it's heads and it is heads. So she places one bet and wins 1 tokens.
• She believes it's heads but it is tails. She gambles 2 times and loses 2 tokens.
• She believes it's tails and it is tails. She gambles 2 times and wins 2 tokens.
• She believes it's tails but it is heads. She gambles 1 time and loses 1 token.

This is under the assumption that she always places the same bet because there's no rational reason why her credence should change over the course of the experiment. In case her credence is just random like a coin flip. Well that's essentially 1 coin flip of what she believes for heads and 2 for tails.

• So for heads is lose 1 token or win 1 token
• And for tails it's lose 25% lose 2 tokens, 50% win one lose one = gain 0 tokens, 25% win 2 tokens.

So there would be a difference between the outcomes of the two examples. Like in the steady guessing case it's:

• She believes it's heads 50% * 1 + 50% * -2 = -0.5 on average
• She believes it's tails 50% *2 + 50% *-1 = 0.5 on average

While in the random credence they are technically both just coin flips with an expected value of 0 but if you were to be loss avoidant then the tails case has a 75% to avoid losses and the heads case has only a 50% chance. But in either case that is not your choice to make so it doesn't matter.

However none of that yields a 0.5 gain and 1.5 loss. Am I missing something?

I don't understand how they ended up with 0.5 gain and 1.5 loss. Like what are they doing and why?

Well, they are not using ratios consistently, and "in the Heads scenario" means what you meant by "believes it is Heads." So look at only your first two cases.

If she plays the game 2N times, and bets on (i.e., "believes it is") Heads each time, she will wager a total of 3N tokens. She will get N/P tokens back, where P is the probability of Heads. Thus her average gain per game is the (N/P)/(2N)=0.5/P tokens. That's their 0.5. By "her loss", they mean the 3N tokens wagered, which averages (3N)/(2N)=1.5 tokens. To break even, we need 0.5/P=1.5, so P=1/3.

The problem is that not everybody believes that this is set up correctly. If you look at individual wagers, instead of individual games, you can use similar techniques to get P=1/2. But imo you have to ignore that some wagers are not independent, so that is wrong. A better conclusion is that wagering is not a good way to approach this problem.

Not many people recall it, but you digress from the original experimental procedure. (You also left out the amnesia drug, but you stated its result). All the original actually says is that SB will be wakened once if the coin lands Heads, and twice if it lands Tails. ("Self-locating belief and the Sleeping Beauty problem", Adam Elga). The two days, and the specifically the order of the wakings on Monday and Tuesday, was how Elga formulated his solution. There is a better way.

• Flip two coins, C1 and C2. C2 will have no bearing on what SB ever knows, it is just used to create the circumstances Elga asks for.
• If either coin shows Tails: wake SB, interview her, and put her back to sleep with amnesia.
• Turn C2 over to show its other face, and repeat that same process; that is, wake SB if either coin is showing Tails, and interview her.
• In the interview, regardless of whether C2 has been turned over, ask SB for her probability/confidence/credence/whatever that C1 is showing Heads.

To formulate her answer, SB knows that the state of the coins was evaluated before waking her. Regardless of previous or subsequent evaluations, there were four equally-likely states for the two coins: {HH, HT, TH, TT}. But since she is awake, she knows that HH is eliminated. In only one of the three equally-likely cases is C1 showing Heads, so the answer is 1/3.

The only issue is whether Elga's solution, which amounts to placing C2 on the table showing Tails instead of flipping it, changes the problem.

• Thanks for the answer it makes a little more sense now. But isn't using N/P already using the probability of heads to be 1/2 because otherwise it would need to be 2N/P? And how exactly does your algorithm for your game work? Commented Aug 26, 2022 at 11:19
• It uses correct probability theory, not the rationalization uses to explain wagers. Try it without sleep: In experiment A, I flip C1 and C2 out of your sight. If either shows Tails, I ask for your confidence that C1 landed on Heads; if both are tails, I say "oops." You'd answer 1/3 if you are asked, right? Experiment B works the same way except I turn C2 over after I flip it. You will still say 1/3. This is exactly what I described for SB above, except that she doesn't know if I used experiment A or experiment B. Just that I used one of them, and if I said "oops" she slept through it. Commented Aug 28, 2022 at 21:11
• Thanks again for the answer. Would you mind updating your answer to better explain the setup for your game, because it's a little confusing to be honest, especially the part where you "repeat the same process". So you're asking her whether it's head on c1 and are only left with 3 options because in the 4th case you wouldn't ask her? And what do you mean by turn over and is she aware of that, is it relevant to the interview/don't interview question? Commented Aug 30, 2022 at 13:22
• The process you repeat starts with looking at the coins, not flipping. The point of the procedure is to wake SB once, or twice, depending on C1. Not to start an infinite loop. The point of doing it this way is to make Monday and/or Tuesday irrelevant. The "prior" is the state of the coins when they are examined. Commented Sep 10, 2022 at 21:53
• Monday and Tuesday may (thirders) or may not (halfers) require a different probability distribution to describe that day. Even though SB does not know which day it is, in the thirder solution she applies another probability distribution to account for that unknown information (halfers deny that this is valid). In my version, only one probability space is needed, hence the ordering of the awakenings is irrelevant. Commented Sep 13, 2022 at 14:51