A friend of mine recently brought a logic problem to my attention. It’s called the Sleeping Beauty Problem, and it’s considered something of a paradox. From the Wikipedia page describing it <Sleeping Beauty Problem>:

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, 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, 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 Beauty is asked: “What is your credence now for the proposition that the coin landed heads?”

First of all, you might wonder what credence is. It’s just probability. In this example, it is how sure Sleeping Beauty is that the coin landed heads. Of course she could be wrong. She could be 100% sure the coin landed on heads. But for this problem, we assume that she is perfectly logical, and that her credence is exactly the same as the correct probability. In short, credence is different from probability in that the people who coined the first term were apparently unaware that the second term already meant exactly the same thing.

But whatever.

So what’s the answer? Apparently there are two, which the same Wikipedia page names the Thirder Position and the Halfer Position. (There is a third position given but it’s wrong and even the guy who argued it knew it was wrong.) These arguments are paraphrased from the Wikipedia page:

The Thirder Position – If the coin comes up tails, there will be two interviews which SB could not distinguish between, so the probability that is both tails and Monday, P(tails and Monday), must be equal to the probability that it is both tails and Tuesday, P(tails and Tuesday). Similarly, if SB assumed the day of the interview was Monday, she would have no reason to believe heads or tails was more likely than the other. The procedure, she notes, does not require the coin to actually be thrown until Tuesday. So, Monday, the probability of the result being heads is equal to the probability of it being tails. Therefore P(tails and Monday) = P(tails and Tuesday).

P(tails and Monday) = P(tails and Tuesday) = P(heads and Monday)

As these are the only three possibilities, they add to 1, and so are each equal to 1/3. Only P(heads and Monday) has the coin flip landing heads, and so P(heads) equals 1/3.

The Halfer Position – The probability of a fair coin landing heads is 1/2, by definition. SB is awakened and interviewed. She knew at the outset of the experiment that this would happen, so she has gained no new information, so her credence cannot change. It must still be 1/2.

Let’s consider the Halfer Position. If we suppose P(heads) = 1/2 as it argues, we can deduce some odd findings. If P(heads) = 1/2, then P(heads and Monday) = 1/2, because if the coin lands on heads, there is no interview Tuesday.

We also find P(tails and Monday) = P(tails and Tuesday) for the exact same reason as it was true in the Thirder Position. So, they each equal 1/4

The term P(heads|Monday) means the probability that the coin flip landed on heads *given* that the day is Monday. It’s called conditional probability. It means, in this case, if SB somehow learns the day is Monday, what is her credence then that the coin landed heads?

We can calculate P(heads|Monday) and several other values as follows. By the definition of conditional probability:

P(heads and Monday) = P(heads|Monday)*P(Monday)

P(heads and Monday) = P(Monday|heads)*P(heads)

P(heads and Tuesday) = P(heads|Tuesday)*P(Tuesday)

P(heads and Tuesday) = P(Tuesday|heads)*P(heads)

P(tails and Tuesday) = P(tails|Tuesday)*P(Tuesday)

P(tails and Tuesday) = P(Tuesday|tails)*P(tails)

By solving these six equations simultaneously – which turns out to be a simpler algebra problem than it appears – We find the following:

P(heads|Monday) = 2/3.

This means if SB somehow learned during an interview that the day were Monday, she would have to answer that her credence that the coin landed on heads was 2/3. But, what would she have learned that would have changed her credence?

The Halfer Position is based on the assertion that waking and being interviewed doesn’t give SB any new information. After all, she knew she would be interviewed regardless of the coin flip. Being interviewed tells her only that there is an interview conducted, which is not new information, and therefore she cannot change her credence. That’s the argument.

However, the same argument can be applied to her being interviewed and knowing the day is Monday. This scenario would tell her only that there is an interview conducted *on Monday.* Again, she knew this would happen from the outset, so learning it here doesn’t give her any new information, and therefore she cannot change her credence. But she must, because for P(heads) to equal 1/2, P(heads|Monday) must equal 2/3.

Compare this experiment to a similar one, in which there is one interview performed on Monday if the coin flip lands on heads, but there are three interviews, performed Monday, Tuesday, and Wednesday if the flip lands on tails. Either way, she is released on Thursday.

In this experiment, SB’s credence that the coin flip is heads must be 3/4 if she somehow learned the day were Monday. Or, P(heads|Monday) = 3/4. The only difference between the two experiments, however is the number of interviews that have yet to be performed at the time this interview takes place.

What if she didn’t know which of these two experiments she was in? Consider a third experiment. Heads, one interview on Monday. Tails, one interview on Monday, followed by a random number of interviews on subsequent days.

In this experiment, the following chain of events might occur. SB is put to sleep. A fair coin is flipped. She is awakened, somehow learns that the day is Monday, and is interviewed. When asked for her credence, she can only reasonably answer that she does not know, that she would need to know how many interviews the experimenters are planning to perform in subsequent days.

Strangely, if the experiment isn’t disrupted, if she is awakened on Monday but did not know it was Monday, she could confidently answer that her credence was 1/2. If she *then* learned that it was Monday, she would have to quickly change her answer. She previously thought it was 1/2, but upon learning that it was Monday, she no longer has enough information to answer the question.

So the Halfer Position is wrong. But why is it wrong?

The Halfer Position asserts SB does not learn anything from being awakened and interviewed. Is that true?

Maybe. But her credence could change for another reason: the amnesia drug. Her change of credence stems not from something she knows but from something she doesn’t know.

Notice how trivial the interview becomes if there is a calendar in the room. Monday, she correctly answers 1/2. Tuesday, if there is an interview, she correctly answers 0.

Without the calendar, she knows what the probability is in the case that the day is Monday, and she knows what the probability is in the case that the day is Tuesday. What she does not know is which case she is in.

She doesn’t *know*.* *That is true. However – the point that the Halfer Position neglects – she does know the *probability* that the day is Monday. And she can use that information to adjust her credence.

It isn’t trivial to calculate, as P(Monday) is known only in terms of P(heads). There are numerous correct approaches to take to solve the problem this way, but the most straightforward and elegant is the one presented in the Thirder Position, above.

Here are a few of the key results:

P(heads) = 1/3

P(heads|Monday) = 1/2

P(heads|Tuesday) = 0

If it is Monday, the odds are 1/2, as expected. If it is Tuesday, the flip is guaranteed to be tails, also as expected.

Some have called this problem a paradox. A paradox occurs when the intuitive answer to a question is different from the correct one.

Strike the notion from your mind. Do not believe there is any such thing. If your intuition differs from reality, your intuition is deficient. If you allow ideas to persist in your mind under the heading of Paradoxes, you are allowing this deficiency to continue to be a part of your logical process.

Retrain yourself to intuitively see the world as it actually is.

In this case, a tool to keep in the toolbox is this: Information helps you make better judgments. Right?

Notice how in the Thirder Position, SB answers that her credence is 2/3, which is not the precise probability in either case. But upon *learning* the day, she can adjust her answer to either 1/2 or 0 (depending on what day she learns it is), which is the correct probability in each case.

However, in the Halfer Position, SB answers 1/2, but when she learns the day is Monday, her credence gets *worse*, moving to 2/3.

And remember the Halfer Position applied to the experiment with a random number of interviews. If SB doesn’t know the day, she answers 1/2, but upon *learning* that the day is Monday, no longer knows enough to answer the question.

Of course, we could contrive examples where a subject answers some question correctly just by dumb luck, and additional information causes the answer to deviate, but generally speaking, knowledge is helpful, and any case where the opposite appears true should make you suspicious.