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Let's say that there are three beauties; Michael, Jane, and Jill. They are put to sleep and assigned a random number from {1, 2, 3}.

If the coin lands heads then 1 is woken on Monday. If the coin lands tails then 2 is woken on Monday and 3 is woken on Tuesday.

If Michael is woken then what is his credence that the coin landed heads?

The halfer explanation given here states that:

Sleeping Beauty receives no new non-self-locating information throughout the experiment because she is told the details of the experiment. Since her credence before the experiment is P(Heads) = 1/2, she ought to continue to have a credence of P(Heads) = 1/2 since she gains no new relevant evidence when she wakes up during the experiment.

Michael's credence before the experiment is P(1) = 1/3, so if woken he ought to continue to have a credence of P(1) = 1/3 since he gains no new relevant evidence if he wakes up during the experiment.

However, given that if woken he is 1 iff the coin landed heads, he ought to either have a credence of P(Heads) = 1/3 or a credence of P(1) = 1/2.

So it appears that the priors P(1) = 1/3 and P(Heads) = 1/2 are in conflict?

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  • Why is your credence before the experiment 1/3? All three have a 1/2 chance of being woken up.
    – David Gudeman
    May 12 at 12:44
  • Because I'm randomly assigned a number between 1 and 3, therefore there's a 1/3 chance that I've been assigned 1, a 1/3 chance that I've been assigned 2, and a 1/3 chance that I've been assigned 3.
    – Michael
    May 12 at 12:46
  • Ah. I thought P(1) meant P(HEADS).
    – David Gudeman
    May 12 at 13:12
  • "Since her credence before the experiment is P(Heads) = 1/2, she ought to continue to have a credence of P(Heads) = 1/2 since she gains no new relevant evidence when she wakes up during the experiment." I'm confused as to why this is considered a relevant answer. The probability is P(Heads and awake), P(Heads) on it's own being 50% doesn't really matter, because for that to be the right probability she would also need to be awake, which heads on its own wouldnt guarantee.
    – JMac
    May 12 at 15:00
  • 1
    Your formulation of the problem is different than the traditional one, so I'm not sure if that "halfer" explanation is appropriate. In the typical problem, it's only one sleeper, and they always wake up at least once but they dont remember it, and dont know what day it is.
    – JMac
    May 12 at 21:01

4 Answers 4

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I think that Bayes' theorem can resolve this issue.

P(Heads|Awake) = P(Awake|Heads) * P(Heads)
                  -------------------------
                          P(Awake)

               = (1/3) * (1/2)
                 -------------
                      0.5

               = 1/3

Given that P(Heads|Awake) ≠ P(Heads), waking does in fact provide new relevant evidence. And the above shows that the correct answer is 1/3.

Applying this same formula to the original problem gives P(Heads|Awake) = P(Heads), and so is consistent with Lewis' reasoning and his conclusion that the correct answer is 1/2.

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When the halfer argument says "Sleeping Beauty receives no new non-self-locating information," it means that SB (their only subject) new she would be wakened, so that provides nothing that can change her assessment of the coin flip result. Yours do get new information, so your variation is comparing different things.

There are several ways to do what you are trying do, that don't have this problem. One is to use four volunteers, instead of their one or your three. Each will be assigned a random number, just like you do, but each will undergo an experiment that is functionally equivalent to the popular version of the problem. The same sleep and amnesia drugs will be used, and each will be awoken at least once, but maybe twice, based on the same fair coin toss. Only their schedules and the question they are asked will differ, but end up being equivalent to the popular problem. On Monday and Tuesday:

#1 Will be awoken unless it is Tuesday, after Heads.

#2 Will be awoken unless it is Tuesday, after Tails.

#3 Will be awoken unless it is Monday, after Heads.

#4 Will be awoken unless it is Monday, after Tails.

Each will be asked for their credence that this is the only time they will be awoken. For #1 and #3, that means credence in Heads. For #2 and #4, it is credence in Tails. For all four, the answer has to be the same as the correct answer to popular version of the Sleeping Beauty Problem.

One each day, we can bring the three awake volunteers together to discuss their answers. The only restriction is that none can know what numbers the others were assigned. Either because none know, or they can't share if they do know. Of these three, exactly one will not be, or was not, awakened on the other day of the experiment. But none of the three can have more, or less, credence that she is that one instead of one of the others.

So with three awake volunteers, one of whom will be awakened only once, the answer is 1/3.

Being awake does not provide any new information about the two-day experiment as a whole. But it does provide new information about the half of the experiment that is currently being conducted, that is isolated from the other half by the amnesia drug, and so is the only part of the experiment that a volunteer knows about. This isn't technically "add information," but it does "remove" some information from the knowledge that SB had on Sunday Night. It works the same way on their ability to update.

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  • I think yours might be a Monty Hall problem. There are 4 people in the experiment. I am the door I chose, the person known to be asleep is the door Monty opened. The principle of indifference doesn't apply to the remaining doors/people awake. The probability that I will only wake once is fixed at 1/2.
    – Michael
    May 18 at 16:51
  • You can't just assert a solution and claim that it is correct because you like the answer. In the MHP, it is the fact that the three doors each have different properties (one couldn't be opened, one could but wasn't, and one was) that makes the PoI inapplicable. Here, the three awake volunteers - including "you" - all have the same properties. To assert that the PoI doesn't apply, you need to identify a difference that makes one (or more) of them different from the rest. If you cannot, the PoI does apply ah\nd the answer is 1/3.
    – JeffJo
    May 18 at 23:52
  • On my waking day(s) I'm a door that can't be opened/left to sleep. Of the other three, only one that doesn't contain the prize (the prize being "is awake both days") is opened/left to sleep. That seems analogous to Monty Hall.
    – Michael
    May 19 at 9:43
  • On your "waking day," whatever it is that you think that means, you are a door that wasn't opened. Not one that could not have been, based on the unknown conditions that define that day. The same applies to the other two volunteers that share this particular "waking day." And the point is that your credence is based on the fact you have the same information about how those conditions apply to each of the three awake volunteers. That means the PoI applies.
    – JeffJo
    May 19 at 11:17
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I find questions about probabilities to be very confusing, as are some questions in physics, but a tactic I have found to be helpful is to consider extreme cases.

Let's take a simpler variant of your question. I perform an experiment in which I will toss a coin. If it is heads I will give you a ball with the number 1 printed on it. If it is tails I will extract one of two balls at random from a bag, one marked with a number 1, the other with 2. I perform the coin toss in secret then hand you a ball with the number 1 on it. Do you think the coin landed heads or tails? Now repeat the experiment, except this time if the coin lands tails I will randomly draw a numbered ball from a bag containing a million balls and give it to you. I toss the coin, pass you a ball and it is numbered 1. What now do you suppose is the probability of the coin being heads?

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  • This example is slightly different. In my example the coin toss and the number assignment are independent events, whereas in your example picking the random ball if the coin toss is tails is dependent on the coin toss.
    – Michael
    May 13 at 8:32
  • If my example is not sufficiently similar, try a more extreme variant of you own.
    – Marco Ocram
    May 13 at 12:59
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Well yes P(Heads) is different for and after the coin flip. While the coin is still in spin it's 50%:50% while after the coin has landed it's either 100%:0% or 0%:100%.

So after you've been woken up, you see the effect of the coin flip and you can deduce the probability of the status of the coin based on the probability of the effect that was cause by it (in that case that you are awake).

Like essentially your number assigning only has 2 relevant outcomes:

  1. You get 1 assigned.
  2. You don't get 1 assigned.

That's it. The probability for the 1st is 1/N and the probability for the 2nd is N-1/N.

Now in your case N is 3 but for the sake of argument let's make it 1 trillion. Then with near certainty you're not number 1. Now as not 1 and awake is only possible if the coin showed Tails, you're credence is now with near certainty that it must have shown tails. That has nothing to do with the credence in the fairness of the coin, but is deduced purely from what number you're most likely assigned and the fact that an effect has ALREADY HAPPENED (you are awake). Of course you being awake could also mean that you're 1 and that the coin landed Heads, but that's way less likely given that you were way less likely to be 1 to begin with.

However while waking up gives you a strong probability of Tails, not waking up tells you nothing, because you don't know if you had been awake prior to the end of the experiment or not, so it could have been Heads and you're never awake or it could have been Tails and you don't remember so you're still at 1/2.

So your new knowledge would change the credence in the state of the coin it would NOT change the credence in the state of the number.

So no the priors are not in conflict with each other.

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