### The situation

Some researchers would like to put you to sleep. Depending on the secret toss of a fair coin, they will briefly awaken you either once (Heads) or twice (Tails). After each waking, they will put you back to sleep with a drug that makes you forget that awakening. When you are awakened, to what degree should you believe that the outcome of the coin toss was Heads?

(OK, maybe you don’t want to be the subject of this experiment! Suppose instead that Sleeping Beauty (SB) agrees to it (with the full approval of the Magic Kingdom’s Institutional Review Board, of course). She’s about to go to sleep for one hundred years, so what are one or two more days, anyway?)

[Detail of a Maxfield Parrish illustration.]

### Are you a Halfer or a Thirder?

The Halfer position. Simple! The coin is fair--and SB knows it--so she should believe there's a one-half chance of heads.

The Thirder position. Were this experiment to be repeated many times, then the coin will be heads only one third of the time SB is awakened. Her probability for heads will be one third.

### Thirders have a problem

• On Sunday evening, just before SB falls asleep, she must believe the chance of heads is one-half: that’s what it means to be a fair coin.

• Whenever SB awakens, she has learned absolutely nothing she did not know Sunday night. What rational argument can she give, then, for stating that her belief in heads is now one-third and not one-half?

### Some attempted explanations

• SB would necessarily lose money if she were to bet on heads with any odds other than 1/3. (Vineberg, inter alios)

• One-half really is correct: just use the Everettian “many-worlds” interpretation of Quantum Mechanics! (Lewis).

• SB updates her belief based on self-perception of her “temporal location” in the world. (Elga, i.a.)

• SB is confused: “[It] seems more plausible to say that her epistemic state upon waking up should not include a definite degree of belief in heads. … The real issue is how one deals with known, unavoidable, cognitive malfunction.” [Arntzenius]

### The question

Accounting for what has already been written on this subject (see the references as well as a previous post), how can this paradox be resolved in a statistically rigorous way? Is this even possible?

### References

Arntzenius, Frank (2002). Reflections on Sleeping Beauty Analysis 62.1 pp 53-62.

Bradley, DJ (2010). Confirmation in a Branching World: The Everett Interpretation and Sleeping Beauty. Brit. J. Phil. Sci. 0 (2010), 1–21.

Elga, Adam (2000). Self-locating belief and the Sleeping Beauty Problem. Analysis 60 pp 143-7.

Franceschi, Paul (2005). Sleeping Beauty and the Problem of World Reduction. Preprint.

Groisman, Berry (2007). The end of Sleeping Beauty’s nightmare. Preprint.

Lewis, D (2001). Sleeping Beauty: reply to Elga. Analysis 61.3 pp 171-6.

Papineau, David and Victor Dura-Vila (2008). A Thirder and an Everettian: a reply to Lewis’s ‘Quantum Sleeping Beauty’.

Pust, Joel (2008). Horgan on Sleeping Beauty. Synthese 160 pp 97-101.

Vineberg, Susan (undated, perhaps 2003). Beauty’s Cautionary Tale.

• I was moved to post this as a separate question based on comments at stats.stackexchange.com/questions/23779. – whuber Oct 25 '12 at 20:10
• It would be good if you could describe the experiment a bit clearer. Without reading the original post, it is really hard to understand what the paradox is about. – sebhofer Oct 25 '12 at 22:51
• My comment wasn't meant to be rude btw. I realized later it might have come across a bit harsh. Hope you didn't take it the wrong way. – sebhofer Oct 28 '12 at 0:42
• You might be interested in the (now large) literature in philosophy on this paradox. Here is a fairly complete bibliography (with links): philpapers.org/browse/sleeping-beauty – user16414 Oct 31 '12 at 14:13
• The references to "Sunday" in the problem statement are a bit confusing, since there is no other mention of weekday. Judging from some answers, I suppose the original problem statement made reference to Monday and Tuesday, but they were edited out. I recommend editing out Sunday as well. – starchild Oct 9 '17 at 22:20

### Strategy

I would like to apply rational decision theory to the analysis, because that is one well-established way to attain rigor in solving a statistical decision problem. In trying to do so, one difficulty emerges as special: the alteration of SB’s consciousness.

• Rational decision theory has no mechanism to handle altered mental states.

• In asking SB for her credence in the coin flip, we are simultaneously treating her in a somewhat self-referential manner both as subject (of the SB experiment) and experimenter (concerning the coin flip).

Let’s alter the experiment in an inessential way: instead of administering the memory-erasure drug, prepare a stable of Sleeping Beauty clones just before the experiment begins. (This is the key idea, because it helps us resist distracting--but ultimately irrelevant and misleading--philosophical issues.)

• The clones are like her in all respects, including memory and thought.

• SB is fully aware this will happen.

We can clone, in principle. E. T. Jaynes replaces the question "how can we build a mathematical model of human common sense"--something we need in order to think through the Sleeping Beauty problem--by "How could we build a machine which would carry out useful plausible reasoning, following clearly defined principles expressing an idealized common sense?" Thus, if you like, replace SB by Jaynes' thinking robot, and clone that.

(There have been, and still are, controversies about "thinking" machines.

"They will never make a machine to replace the human mind—it does many things which no machine could ever do."

You insist that there is something a machine cannot do. If you will tell me precisely what it is that a machine cannot do, then I can always make a machine which will do just that!”

--J. von Neumann, 1948. Quoted by E. T. Jaynes in Probability Theory: The Logic of Science, p. 4.)

--Rube Goldberg

### The Sleeping Beauty experiment restated

Prepare $n \ge 2$ identical copies of SB (including SB herself) on Sunday evening. They all go to sleep at the same time, potentially for 100 years. Whenever you need to awaken SB during the experiment, randomly select a clone who has not yet been awakened. Any awakenings will occur on Monday and, if needed, on Tuesday.

I claim that this version of the experiment creates exactly the same set of possible results, right down to SB's mental states and awareness, with exactly the same probabilities. This potentially is one key point where philosophers might choose to attack my solution. I claim it's the last point at which they can attack it, because the remaining analysis is routine and rigorous.

Now we apply the usual statistical machinery. Let's begin with the sample space (of possible experimental outcomes). Let $M$ mean "awakens Monday" and $T$ mean "awakens Tuesday." Similarly, let $h$ mean "heads" and "t" mean tails. Subscript the clones with integers $1, 2, \ldots, n$. Then the possible experimental outcomes can be written (in what I hope is a transparent, self-evident notation) as the set

\eqalign{ \{&hM_1, hM_2, \ldots, hM_n, \\ &(tM_1, tT_2), (tM_1, tT_3), \ldots, (tM_1, tT_n), \\ &(tM_2, tT_2), (tM_2, tT_3), \ldots, (tM_2, tT_n), \\ &\cdots, \\ &(tM_{n-1}, tT_2), (tM_{n-1}, tT_3), \ldots, (tM_{n-1}, tT_n) & \}. }

### Monday probabilities

As one of the SB clones, you figure your chance of being awakened on Monday during a heads-up experiment is ($1/2$ chance of heads) times ($1/n$ chance I’m picked to be the clone who is awakened). In more technical terms:

• The set of heads outcomes is $h = \{hM_j, j=1,2, \ldots,n\}$. There are $n$ of them.

• The event where you are awakened with heads is $h(i) = \{hM_i\}$.

• The chance of any particular SB clone $i$ being awakened with the coin showing heads equals $$\Pr[h(i)] = \Pr[h] \times \Pr[h(i)|h] = \frac{1}{2} \times \frac{1}{n} = \frac{1}{2n}.$$

### Tuesday probabilities

• The set of tails outcomes is $t = \{(tM_j, tT_k): j \ne k\}$. There are $n(n-1)$ of them. All are equally likely, by design.

• You, clone $i$, are awakened in $(n-1) + (n-1) = 2(n-1)$ of these cases; namely, the $n-1$ ways you can be awakened on Monday (there are $n-1$ remaining clones to be awakened Tuesday) plus the $n-1$ ways you can be awakened on Tuesday (there are $n-1$ possible Monday clones). Call this event $t(i)$.

• Your chance of being awakened during a tails-up experiment equals $$\Pr[t(i)] = \Pr[t] \times P[t(i)|t] = \frac{1}{2} \times \frac{2(n-1}{n(n-1)} = \frac{1}{n}.$$

### Bayes' Theorem

Now that we have come this far, Bayes' Theorem--a mathematical tautology beyond dispute--finishes the work. Any clone's chance of heads is therefore $$\Pr[h | t(i) \cup h(i)] = \frac{\Pr[h]\Pr[h(i)|h]}{\Pr[h]\Pr[h(i)|h] + \Pr[t]\Pr[t(i)|t]} = \frac{1/(2n)}{1/n + 1/(2n)} = \frac{1}{3}.$$

Because SB is indistinguishable from her clones--even to herself!--this is the answer she should give when asked for her degree of belief in heads.

### Interpretations

The question "what is the probability of heads" has two reasonable interpretations for this experiment: it can ask for the chance a fair coin lands heads, which is $\Pr[h] = 1/2$ (the Halfer answer), or it can ask for the chance the coin lands heads, conditioned on the fact that you were the clone awakened. This is $\Pr[h|t(i) \cup h(i)] = 1/3$ (the Thirder answer).

In the situation in which SB (or rather any one of a set of identically prepared Jaynes thinking machines) finds herself, this analysis--which many others have performed (but I think less convincingly, because they did not so clearly remove the philosophical distractions in the experimental descriptions)--supports the Thirder answer.

The Halfer answer is correct, but uninteresting, because it is not relevant to the situation in which SB finds herself. This resolves the paradox.

This solution is developed within the context of a single well-defined experimental setup. Clarifying the experiment clarifies the question. A clear question leads to a clear answer.

I guess that, following Elga (2000), you could legitimately characterize our conditional answer as "count[ing] your own temporal location as relevant to the truth of h," but that characterization adds no insight to the problem: it only detracts from the mathematical facts in evidence. To me it appears to be just an obscure way of asserting that the "clones" interpretation of the probability question is the correct one.

This analysis suggests that the underlying philosophical issue is one of identity: What happens to the clones who are not awakened? What cognitive and noetic relationships hold among the clones?--but that discussion is not a matter of statistical analysis; it belongs on a different forum.

• This answer summarizes a talk I prepared in December 2008 and posted on the Web at that time in PowerPoint format. Its conclusion appears to be substantially similar to Groisman's, even though the justification may be different: "If we mean ‘This awakening is a Head-awakening under setup of wakening’, then her answer should be 1/3, but if we mean ‘The coin landed Heads under setup of coin tossing’, her answer should be 1/2." See philsci-archive.pitt.edu/3382/1/SB_PhilSci.pdf. – whuber Oct 25 '12 at 20:11
• I attack it at exactly your unscaled underbelly. See my detailed analysis below. – Dax Fohl Sep 1 '15 at 5:29
• You've made it far more complex than it needs to be, check my answer. – Kelvin Apr 13 '16 at 23:18

I just thought of the following situation which does not require fairies, miracles nor magic potions. Flip a fair coin on Monday noon. Upon 'Tails' send a mail to Alice and Bob (in a way that they don't know that the other has received a mail from you, and that they cannot communicate). Upon 'Heads', send a mail to one of them at random (with probability $1/2$).

When Alice receives a mail, what is the probability that the coin landed on 'Heads'? The probability that she receives a letter is $1/2 \times 1/2 + 1/2 = 3/4$, and the probability that the coin landed on 'Heads' is $1/3$.

Here there is no paradox because Alice does not receive a letter with probability $1/4$, in which case she knows the coin landed on 'Heads'. The fact that we don't ask her opinion in that case, does make this probability equal to 0.

So, what is the difference? Why would Alice gain information by receiving a mail, and SB would learn nothing being awakened?

Moving on to a more miraculous situation, we put 2 different SB to sleep. If the coin lands on 'Tails' we wake up both, if it lands on 'Heads' we wake up one of them at random. Here again, each of the SB should say that the probability of the coin landing on 'Heads' is $1/3$ and again there is no paradox because there is a $1/4$ chance that this SB would not be awakened.

But this situation is very close to the original paradox because erasing the memory (or cloning) is equivalent to having two different SB. So, I am with @Douglas Zare here (+1). SB has learned something by being awakened. The fact that she cannot express her opinion on Tuesday when the coin is 'Heads' up because she is sleeping does not erase the information she has by being awakened.

In my opinion the paradox lies in "she has learned absolutely nothing she did not know Sunday night" which is stated without justification. We have this impression because the situations when she is awakened are identical, but this is just like Alice receiving a mail: it is the fact that she is asked her opinion that gives her information.

MAJOR EDIT: After giving it a deep thought, I change my opinion: Sleeping Beauty has learned nothing and the example I give above is not a good analogue of her situation.

## The thirders win

Instead of a coin, lets a assume a fair dice:

on friday, the sleeping beauty will sleep:
if the dice == 1 , they will awake her on saturday;
if the dice == 2 , they will awake her on saturday and sunday;
if the dice == 3 , they will awake her on saturday, sunday and monday;
if the dice == 4 , they will awake her on saturday, sunday, monday and tuesday;
if the dice == 5 , they will awake her on saturday, sunday, monday, tuesday and wednesday;
if the dice == 6 , they will awake her on saturday, sunday, monday, tuesday, wednesday and thursday;


Every time they ask her ' to what degree should you believe that the outcome of the dice was 1?'

The halfers will say the probability of dice = 1 is 1/6 The thirders will say the probability of dice = 1 is 1/21

But simulation clearly solves the problem :

days <- c("saturday", "sunday", "monday", "tuesday", "wednesday", "thursday")

#she will answer the dice was 1 every time
#the trick here is that this is not absolutely random because every day implies the days before it.

number_of_days <- 0
for (i in 1:1000){
dice <- sample(1:6,1)
for (item in days[1:dice]){
number_of_days <- number_of_days + 1
}
}
#equals 1/21
#but if we divided by 1000 , which is incorrect because every experiment has more than one day we will get 1/6
#equals 1/6


Also we can simulate the toss problem

days <- c("monday", "tuesday")
number_of_tosses <- 0
for (i in 1:1000){
toss <- sample(1:2,1)
for (item in days[1:toss]){
number_of_tosses <- number_of_tosses + 1
}
}