Sleeping Beauty

For over 15 years, some people—particularly philosophers—continue to be confused by the so-called “Sleeping Beauty Problem.” This is a rather straight-forward exercise in conditional probability that should be accessible to a student in an undergraduate course on probability and statistics. Nevertheless, there are people who have managed to arrive at the wrong answer to this problem.

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The Problem

The Sleeping Beauty Problem is usually described as follows:

Beauty is going to be the subject of an experiment that will take place over three days. On Sunday, Beauty will be told the plan for the experiment, and then given a drug and put to sleep. The drug will cause her to sleep until Wednesday, but the experimenters plan to wake her up some number of times. To decide how many times to wake her up, they toss a coin. If the coin lands heads, they will wake her on Monday briefly, and then return her to sleep. If the coin lands tails, they will instead wake her briefly on both Monday and Tuesday. After they put her back to sleep on Monday, they will erase her memory, so that upon waking on Tuesday, her last memories will be having gone to sleep on Sunday. The circumstances of her wakings are identical, so she cannot tell from her environment what day it is or which way the coin landed. What should Beauty think about the probability of heads and tails upon awakening?

Alternatively, the question that is often asked is, “What is the probability that the coin came up heads?”

The Wrong Answer

A popular, incorrect answer to this problem is that Beauty should believe that, because she woke up, the fair coin is twice as likely to have come up tails as to come up heads. Since this results in a probability of heads of 1/3, the people who argue this point of view are often called “thirders.”

In the most naive argument of this type that is put forward, the proponent points out that Beauty can wake in one of three states:

(1) The coin toss was heads and it’s Monday
(2) The coin toss was tails and it’s Monday
(3) The coin toss was tails and it’s Tuesday

The proponent then (implicitly or explicitly) assumes that all three states are equally likely to occur—i.e., each occurs with the same probability as the others. Since the coin came up heads in only one of the three states, he or she concludes that the probability of heads is 1/3.

More complicated, but equally wrong, arguments are addressed below.

The Right Answer

Here we will not only provide the correct answer to the problem (which is trivial) but also rigorously explore the estimates of probabilities that Beauty should assign to other parts of the experiment based on what information she has available.

In the interest of conserving space, let’s introduce the following notation for conditions of the experiment:\[
H &{}= \text{The coin came up heads}\\
T &{}= \text{The coin came up tails}\\
D_1 &{}= \text{It is the first day (Monday)}\\
D_2 &{}= \text{It is the second day (Tuesday)}
\]The notation used above for Monday and Tuesday readily generalizes to the scenario proposed by Nick Bostrom whereby Beauty is wakes for many, many (e.g., a million) days if the coin comes up tails (more below).

Expression of the Problem as Probabilities

The assumption that the coin is fair implies\[
P(H) &{}= 1/2\\
P(T) &{}= 1/2
\]If the coin comes up heads, Beauty will be woken on Monday, but not on Tuesday. This means\[
P(D_1|H) &{}= 1\\
P(D_2|H) &{}= 0
\]If the coin comes up tails, Beauty will be woken on both Monday and Tuesday. She does not know which day it is when she wakes, but she knows that if tails shows up, she will be woken on both days, so in this situation the frequency of Monday and Tuesday awakenings will be the same. Therefore, they have an equal probability:\[
P(D_1|T) &{}= 1/2\\
P(D_2|T) &{}= 1/2

From that, the grid of all four possibilities falls out:\[
P(H \cap D_1) &{}= P(H) \cdot P(D_1|H) & & {}= 1/2\\
P(H \cap D_2) &{}= P(H) \cdot P(D_2|H) & & {}= 0\\
P(T \cap D_1) &{}= P(T) \cdot P(D_1|T) & & {}= 1/4\\
P(T \cap D_2) &{}= P(T) \cdot P(D_2|T) & & {}= 1/4

On Waking Up

Let’s get one thing out of the way, Beauty is always going to wake up:\[
P(\text{wake}) = P(\text{wake}|H) = P(\text{wake}|T) =
P(\text{wake}|D_1) = P(\text{wake}|D_2) = 1
\]So the fact that she wakes and is “in the moment” doesn’t give her or us any more information than she started with. Since Beauty has not been provided with any new information, the probability of the coin coming up heads remains \(P(H) = 1/2\), which is the right answer to the problem.

Additional Probabilities

With that out of the way, we can use the information above to determine the probability of Beauty waking on a Monday,\[
P(D_1) = P(H) \cdot P(D_1|H) + P(T) \cdot P(D_1|T) = 3/4
\]and the probability of it being a Tuesday,\[
P(D_2) = P(H) \cdot P(D_2|H) + P(T) \cdot P(D_2|T) = 1/4
\]If Beauty is informed that she has risen on a Monday, then she can reevaluate the probability of the coin landing on heads or tails by applying Bayes’s Theorem:\[
P(H|D_1) &{}= \frac{P(D_1|H) \cdot P(H)}{P(D_1)} & &{}= 2/3\\
P(T|D_1) &{}= \frac{P(D_1|T) \cdot P(T)}{P(D_1)} & &{}= 1/3
\]Similarly, if she is told that it is a Tuesday, then she knows for certain that the coin toss was tails:\[
P(H|D_2) &{}= \frac{P(D_2|H) \cdot P(H)}{P(D_2)} & &{}= 0\\
P(T|D_2) &{}= \frac{P(D_2|T) \cdot P(T)}{P(D_2)} & &{}= 1

Common Mistakes

Although the flaws in logic employed by the “thirders” encompass a wide range of fallacies, a couple of mistakes appear quite frequently. Here we’ll examine why they are wrong.

What Happens on Monday

The mistake that most “thirders,” including the original guy who published the first article on this, make is to reason that, since she must wake up on Monday—regardless of the coin flip—we must consider her situation at that time. The usual explanation that is given is that the experimenters might not have flipped the coin until after Beauty had been woken on Monday, because the coin doesn’t affect that part of the experiment. The coin toss determines only whether Beauty has her memory erased and is put back to bed. Nothing in the experiment requires that the coin is flipped before Beauty is woken and asked about the result of the coin toss.

They then argue that it is ridiculous, in the situation where Beauty is asked on Monday about the probability of heads on a coin toss that has not yet happened, for her to say anything but 1/2. In one sense, they have a valid point. If Beauty is woken and told that it is Monday and that the coin toss hasn’t yet occurred, then she should reasonably conclude that the probability of the (future) coin toss coming up heads is 1/2. However, this is not the question that is being asked in the problem.

By focusing on the result of a coin toss on Monday—regardless of whether the coin has actually been tossed or not—the “thirders” have eliminated the possibility that the coin toss could have resulted in Beauty waking on Tuesday. In their line of reasoning, it’s simply not possible. Therefore, the coin toss that they are considering is not the coin toss that Beauty is asked about in the problem.

Following up on this mistake by applying it to probability calculations, they then (erroneously) reason that\[
P(H|D_1) &{}= 1/2 \\
P(T|D_1) &{}= 1/2
\]and work backward through Bayes’s Theorem to conclude the wrong answer:\[
P(H) = 1/3 \rlap{\qquad\text{(wrong)}}

The source of their error is that they have changed the definition of the problem to a different problem in which the day is always Monday:\[
P(D_1) &{}= 1 \\
P(D_2) &{}= 0
\]Naturally, the coin toss can have no effect on this:\[
P(D_1|H) &{}= P(D_1|T) & &{}= 1 \\
P(D_2|H) &{}= P(D_2|T) & &{}= 0
\]It is important to note that the claim made above about the probability of heads and tails on Monday, which is wrong in the original problem, is correct in this new, different problem,\[
P(H|D_1) &{}= \frac{P(D_1|H) \cdot P(H)}{P(D_1)} & &{}= 1/2\\
P(T|D_1) &{}= \frac{P(D_1|T) \cdot P(T)}{P(D_1)} & &{}= 1/2
\]but only if \(P(H) = P(T) = 1/2\), so even their modified problem doesn’t demonstrate their claim that the probability of heads is 1/3. When the assumptions are explicitly stated and the math is done correctly, it merely reinforces the proof that the probability is 1/2.

You Wanna Bet?

The other mistake that is commonly encountered in “thirder” reasoning is to treat the Sleeping Beauty problem as a betting proposition. There are several variants as to how the wager is presented—whether only one bet or multiple bets are considered, whether odds are given on the bet, the amount of the bet, etc.—but they all suffer from the same critical error. There is even confusion over what constitutes a “bet.” While it is generally agreed that Beauty bets on the result of the toss of the coin every time she wakes up, some people count the number of bets as the number of times the coin is tossed \(n_{\text{t}}\), which can be controlled, since it’s the number of times Beauty goes to sleep on Sunday. Meanwhile, other people consider every waking to be a separate bet, even though this number is randomly determined by the coin tosses. The best that can be said is that the number of times Beauty wakes \(N_{\text{w}}\) is expected to be\[
E(N_{\text{w}}) = n_{\text{t}} P(H) + 2 n_{\text{t}} P(T)
= 3 n_{\text{t}} / 2
\]That is, Beauty can be expected to wake about 50% more than the number of times that she was put to sleep on Sunday, but the exact number in each series of experiments depends on the outcome of the coin tosses.

To avoid this ambiguity, let’s define “toss” to be every time the coin is tossed (and Beauty is put to sleep on Sunday) and “bet” as every time she has a chance to win or lose money (which is every time that Beauty wakes and is interviewed). Beauty has two options. She either can decide what she will bet when she wakes up each morning, or she can decide what to bet when she goes to sleep on Sunday and consistently place the same bet each time she wakes. Either way, her expected winnings will be the same. She can decide always to bet heads, always to bet tails, or to bet heads randomly at a specified fraction of the time \(f_{\text{H}}\).

Her expected winnings are\[
E(W) = f_{\text{H}}\cdot\bigl(P(H) – 2P(T)\bigr) +
(1 – f_{\text{H}})\cdot\bigl(2P(T) – P(H)\bigr)
\]which becomes\[
E(W) = (2f_{\text{H}} – 1)\cdot\bigl(P(H) – 2P(T)\bigr)
\]So if Beauty always bets heads (\(f_{\text{H}} = 1\)),\[
E(W) = -1/2
\] and if Beauty always bets tails (\(f_{\text{H}} = 0\)),\[
E(W) = 1/2
\]If she bets heads only part of the time (\(0 < f_{\text{H}} < 1\)) then the expected winnings vary linearly with \(f_{\text{H}}\) between these two extremes.

So obviously, Beauty’s best strategy that maximizes her winnings is to always bet tails. For every, two dollars she offers to wager, she is expected to win one, which leads the “thirders” to mistakenly conclude that the odds of the coin coming up heads is 1 to 2 or that heads has a probability of \(P(H) = 1/3\).

However, this is not the case. If she seals in her bet (or bets, in case of tails) on Sunday by always betting either heads or tails, she is making a wager with equally likely results on the toss, but an uneven payoff on the result of the toss. The payoff table looks like the following:

Bet Heads Bet Tails
Result Heads Win 1 Lose 1
Result Tails Lose 2 Win 2

The “thirders” have confused betting odds with statistical odds. They are not the same thing. Therefore, a wise Beauty can make easy money by always betting tails and taking the two-to-one payoff, but she should not delude herself that the probabilities of the coin toss are anything but even.

The Groundhog Day Fallacy

Philosopher Nick Bostrom has proposed a scenario—possibly inspired by the movie Groundhog Day—in which Beauty, instead of being woken only on Monday and Tuesday when the coin comes up tails, is woken many, many times—perhaps as many as a million times. (Never mind that a person who lives to be 100 years old will have less than 37,000 days on which to wake up. This is a philosopher talking.)

The idea behind this argument is that the set of possible days on which Beauty can wake up is overwhelmed by all of these additional days making it “more absurd” for someone to argue that the one day with heads out of all of these possible days should be so frequent. This is nonsense, of course.

The coin will come up heads 1/2 of the time, and the experiment will end on Monday every time this happens. The other half of the time, Beauty will be woken on one of the days, each of which is no more likely than the rest without additional information. Thus, if there are \(n_{\text{T}}\) such days, then\[
P(D_i|T) = 1/n_{\text{T}}, \quad i = 1, 2, \ldots, n_{\text{T}}
\] So the problem generalizes as follows\[
P(D_1|H) &{}= 1\\
P(D_i|H) &{}= 0, \quad & i &{}= 2, 3, \dots, n_{\text{T}} \\
P(D_i|T) &{}= 1/n_{\text{T}}, \quad & i &{}= 1, 2, \dots, n_{\text{T}} \\
P(H \cap D_1) &{}= P(H) \cdot P(D_1|H) & & {}= 1/2 & & \\
P(H \cap D_i) &{}= P(H) \cdot P(D_i|H) & & {}= 0, & \quad
i &{}= 2, 3, \dots, n_{\text{T}} \\
P(T \cap D_i) &{}= P(T) \cdot P(D_i|T) & & {}= \frac{1}{2n_{\text{T}}}, & \quad
i &{}= 1, 2, \dots, n_{\text{T}}

There is nothing in this more general problem to imply anything other than \(P(H) = 1/2\).

The Punters

Finally, there are a few people who claim that the framing of the problem is too ambiguous to evaluate the probability of the coin landing on heads. I don’t quite understand how they have confused themselves enough to reach this conclusion, but they are clearly wrong, as has been demonstrated above.


The correct answer to the Sleeping Beauty Problem is that Beauty has received no additional information upon waking. Therefore the probability that the coin has or will come up heads is 1/2, which is exactly the same as it was when she went to sleep on Sunday. A fair coin is, after all, a fair coin.

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