Thoughts on the Sleeping Beauty Paradox
Introduction
In 2000, Adam Elga proposed the following problem:
Some researchers are going to put you to sleep. During the two days that your sleep will last, they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you are first awakened, to what degree ought you believe that the outcome of the coin toss is Heads?
First answer: 1/2, of course! Initially you were certain that the coin was fair, and so initially your credence in the coin’s landing Heads was 1/2. Upon being awakened, you receive no new information (you knew all along that you would be awakened). So your credence in the coin’s landing Heads ought to remain 1/2.
Second answer: 1/3, of course! Imagine the experiment repeated many times. Then in the long run, about 1/3 of the wakings would be Headswakings — wakings that happen on trials in which the coin lands Heads. So on any particular waking, you should have credence 1/3 that that waking is a Headswaking, and hence have credence 1/3 in the coin’s landing Heads on that trial. This consideration remains in force in the present circumstance, in which the experiment is performed just once.
I will argue that the correct answer is 1/3.
I, on the other hand, will argue that the arguments in favour of 1/3 are incorrect, and that the answer is, in fact, 1/2.
The longrun argument
I will start with the “long run” argument, which superficially resembles a probability as a frequentist would define it (the relative frequency of an outcome in infinite repetitions of a “random trial”), and may therefore hold some initial appeal in that way.
This position, however, requires that the “random” experiment be each awakening – but some of them are logically tied together by the same coin flip, and therefore arguably not random trials in the sense that frequentist probability requires.
Put another way: note that there is no basis for giving anything but the same answer in all awakenings. Further, that answer can be arrived at without actually conducting the experiment (indeed, as far as I know, that’s all that’s ever been done), since in that setup, waking up gives you no information, since you already knew it would happen with 100% certainty regardless of the outcome of the flip.^{1} We can therefore replace the setup with one where you write your answer on a piece of paper, we dispense with the sleeping and the waking up, and we decide that the piece of paper will be read on Monday if the coin flip comes up “heads”, or on Monday and Tuesday if it comes up “tails”. The thirder position, then, holds that the number of times the paper will be read should affect the answer. There is a case to be made that in a decisional context, the number of readings could potentially affect the loss function, but that’s distinct from the inferential problem.
A similar problem afflicts the argument based on betting (as found e.g. on Wikipedia). It is true that one should wager 1/3, but that’s not because it reflects one’s true credence; it’s because the bet is effectively executed twice in one case and only once in the other, which distorts the betting procedure and breaks the usual correspondence between wager and epistemic probability. Just because you take part in a rigged bet, should that change what you believe to be true?
Anyway, let’s now explore Adam Elga’s longer argument.
The temporal argument
The setup and the first part of the argument are as follows:
Suppose that the first waking happens on Monday, and that the second waking (if there is one) happens on Tuesday. Then when you wake up, you’re certain that you’re in one of three ‘predicaments’:
 $H_1$: HEADS and it is Monday.
 $T_1$: TAILS and it is Monday.
 $T_2$: TAILS and it is Tuesday.
[…]
Let $P$ be the credence function you ought to have upon first awakening. Upon first awakening, you are certain of the following: you are in predicament $H_1$ if and only if the outcome of the coin toss is Heads. Therefore, calculating $P(H_1)$ is sufficient to solve the Sleeping Beauty problem. I will argue first that $P(T_1) = P(T_2)$, and then that $P(H_1) = P(T_1)$.
If (upon first awakening) you were to learn that the toss outcome is Tails, that would amount to your learning that you are in either $T_1$ or $T_2$. Since being in $T_1$ is subjectively just like being in $T_2$, and since exactly the same propositions are true whether you are in $T_1$ or $T_2$, even a highly restricted principle of indifference yields that you ought then to have equal credence in each. But your credence that you are in $T_1$, after learning that the toss outcome is Tails, ought to be the same as the conditional credence $P(T_1 \mid T_1 \text{ or } T_2)$, and likewise for $T_2$. So $P(T_1 \mid T_1 \text{ or } T_2) = P(T_2 \mid T_1 \text{ or } T_2)$, and hence $P(T_1) = P(T_2)$.
Note the ambiguity as to what it might mean to learn that “it” is Monday. What is “it”? In which circumstances might we come to learn or not learn that? This is important to know to make sense of the information.^{2} As enunciated, the statement is illdefined; its truth value depending on the specific instant it is evaluated and therefore being of undefined probability.
If we try to make sense of it as a probabilistic statement from the numerical statements given ($P(T_1 \mid T_1 \text{ or } T_2) = P(T_2 \mid T_1 \text{ or } T_2)$, i.e. $P(\text{Monday} \mid \text{Tails}) = P(\text{Tuesday} \mid \text{Tails}) = 1/2$, and later in the text, $P(\text{Heads} \mid \text{Tuesday}) = 0$, implying that either $P(\text{Heads}) = 0$, which I don’t think the author believes, or $P(\text{Tuesday} \mid \text{Heads}) = 0$, and so $P(\text{Monday} \mid \text{Heads}) = 1$), “it” seems to refer to “one of the possible awakenings, with no reason to favour either Monday and Tuesday if both are possible”. Having “it” now more properly defined will allow us to proceed.
The researchers have the task of using a fair coin to determine whether to awaken you once or twice. They might accomplish their task by either
 first tossing the coin and then waking you up either once or twice depending on the outcome; or
 first waking you up once, and then tossing the coin to determine whether to wake you up a second time.
Your credence (upon awakening) in the coin’s landing Heads ought to be the same regardless of whether the researchers use method 1 or 2. So without loss of generality suppose that they use — and you know that they use — method 2.
Now: if (upon awakening) you were to learn that it is Monday, that would amount to your learning that you are in either $H_1$ or $T_1$. Your credence that you are in $H_1$ would then be your credence that a fair coin, soon to be tossed, will land Heads. It is irrelevant that you will be awakened on the following day if and only if the coin lands Tails — in this circumstance, your credence that the coin will land Heads ought to be 1/2. But your credence that the coin will land Heads (after learning that it is Monday) ought to be the same as the conditional credence $P(H_1 \mid H_1 \text{ or } T_1)$. So $P(H_1 \mid H_1 \text{ or } T_1) = 1/2$, and hence $P(H_1) = P(T_1)$.
No, it is not irrelevant; not with “it is Monday” as apparently defined. The outcome of the flip is relevant, because it affects the set of wakings that “it” could have been.
Indeed, from Bayes’ theorem, we find that $O(\text{Heads} \mid \text{Monday}) = 2 \cdot O(\text{Heads})$. The argument we have here is therefore one that begs the question that the posterior probability $P(\text{Heads} \mid \text{Monday}) = 1/2$ (odds of 1:1), and then works backwards to imply that the prior $P(\text{Heads})$ must have been 1/3 (odds of 1:2).
Arguably, what we have instead is $P(\text{Heads}) = 1/2$, and then $P(\text{Heads} \mid \text{“Monday”}) = 2/3$.
Combining results, we have that $P(H_1) = P(T_1) = P(T_2)$. Since these credences sum to 1, $P(H_1) = 1/3$.
Rather, $P(H_1) = 1/2$ and $P(T_1) = P(T_2) = 1/4$.
Let $H$ be the proposition that the outcome of the coin toss is Heads. Before being put to sleep, your credence in $H$ was 1/2. I’ve just argued that when you are awakened on Monday, that credence ought to change to 1/3.
No, what has actually been argued there is closer to the opposite: that before being put to sleep, the credence in $H$ should have been 1/3; and that it would have turned into 1/2 upon learning “one of the possible wakings turned out to be a Monday”.
Either that, or: illdefined information with changing truth value through time, but not incorporating time in the statement, and thereby violating the laws of probability by effectively changing on its own, was introduced to argue that a related probability should also change through time, which begs the question.
What 1/3 actually answers
Recall one of Adam Elga’s claims regarding the answer of 1/3:
This consideration remains in force in the present circumstance, in which the experiment is performed just once.
I believe this to be a cue as to what we can change in the question to make the answer 1/3, so that we can then educate our intuition by looking at what really makes it different from the problem as originally stated.
1/3 is the answer to the following question:
“In a series of infinite repetitions of the original experiment, from which we select an awakening at random, what is the probability that the last flip by that point came up heads?”
Or, equivalently:
“In a series of infinite repetitions of the original experiment, in which we also flip an additional coin between tailsMondays and tailsTuesdays (but discard the result), and from which we select an awakening at random, what is the probability that the last experimentstarting flip by that point came up heads?”
I propose that the fundamental difference here is that which coin flip we are talking about is now a moving target, and by weighting the flips by number of awakenings, we effectively increase the chance of picking a flip that came out tails – even though the unconditional probability that we would assign to any specific flip would be 1/2. The weighting by awakening just makes it more likely that “the last flip” is one of those that happened to turn out tails; a form of selection bias, in a sense.
I therefore posit that the following excerpt from the original article has things backwards:
On Monday, your unconditional credence in H differs from 1/2 because it is a weighted average of these two conditional credences — that is, a weighted average of 1/2 and 0.
It is not the unconditional credence on Monday that the one coin flip came up heads that is equal to 1/3 by being a weighted average of the probability that it’s heads if it’s Monday and the probability that it’s heads if it’s Tuesday (since that would be $\frac 3 4 \times \frac 2 3 + \frac 1 4 \times 0 = \frac 1 2$), it’s the unconditional credence of the last experimentstarting coin flip being heads that is 1/3 by being a weighted average of the probability that it’s Monday and that the last experimentstarting coin flip, which would have been last night, came up heads; and of the probability that it’s Tuesday and that the last experimentstarting coin flip was an independent one two nights ago that came up heads (which it cannot have, if indeed it’s Tuesday).
Conclusion
I must concede that this is quite a clever problem. It has an incorrect answer for which various subtly fallacious justifications can be given, such that one might even end up convinced. The main one, ultimately,
 sets up an alternative scenario in which you receive some additional information;
 omits to properly work out the consequences that this information has on your conclusions;
 from there, has you accept what should be your prior probability as the posterior one;
 through roundabout logic, arrives at the prior probability that would lead you to that posterior, and gaslights you into believing that that was your prior all along.
In other words, the “paradox” arises from misuse of probability theory. If you agree that, knowing only about the existence of the coin flip and not what will be done with it, the probability that its result would be “heads” is 1/2, then that’s also what you should say when awoken (once or twice) during the experiment, because nothing you can learn during the experiment can change that, since you know that the experience of waking up will happen either way. If you were to learn that one awakening out of all those possible within the experiment, chosen at random, happened to be a Monday, then that would update your probability of “heads” to 2/3, but that’s irrelevant to the experiment, in which you don’t learn that. There being a chance, in one case, of being asked unknowingly to repeat your answer by people who have no reason to expect anything different may affect your sense of shame but not your beliefs.
It is possible to describe a different setup, with a different question, to which the answer should indeed be 1/3, but that’s not the original problem.

Interestingly, this stops being true if, instead of waking up twice if tails and once if heads, we wake up once if tails and “half a time” if heads (for example by flipping another coin to decide whether there will be a waking), because then waking up at all is indeed more likely if tails, so observing it is evidence for tails. ↩︎

A similar ambiguity affects some formulations of the infamous Monty Hall problem: if you pick door 1 and then learn that Monty opens door 2 and there is a goat behind it, does that mean “he chose one of the unpicked doors that had a goat behind it, and it happened to be door 2”, or does it mean “he chose an unpicked door indifferently (door 2), and it happened to have a goat behind it”? In the former case, you should switch to door 3; in the latter case, it makes no difference. Which one it is affects your conclusion because it affects how likely that observation would be if door 3 wins vs. if it doesn’t, i.e. the likelihood ratio. ↩︎