### The sleeping beauty problem: how some statisticians calculate probability

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This problem was drawn to my attention by calculations of physicists and philosophers (see my last post), but I see that some eminent statisticians (Jeffrey S. Rosenthal) are also convinced that the answer is $1/3$.

What I think is the real problem is that statisticians immediately start calculating with conditional probability, Bayes formula etc without modelling the problem mathematically. It is also important to generalize a problem to avoid being confused by particular details.

In any case, I want to clarify what I said in the last post by considering a simple example.

The essential point of the sleeping beauty problem (see my last post) is that the beauty is asked to estimate a probability without knowing how many steps of the process have occurred.

Consider the following Markov chain with $100$ states meant to represent the birthdays of my life:

$0\xrightarrow{1}1\xrightarrow{1}2 \xrightarrow{1}\cdots \xrightarrow{1}99\xrightarrow{1}100 $,

with also a loop of probability $1$ at $100$ when I am in my tomb.

Now it is clear that the probability of my being $1$ after $1$ year if $1$, the probability of being $2$ after $2$ years is $1$, the probability of being $n$ after $n$ years is $1$ ($n\leq 99$).

Moreover if you ask me what is the probability of being $5$ say during my life (that is, after $0$ or $1$ or $\cdots$ or $99$ years) I might be tempted to answer $1$ since it is certain I will be $5$ during my life.

However the sum of the probabilities that I am $n$ ($n\leq 99$) during my life must add to total probability $1$, and hence I must answer that the probability of being $5$ in my life is $1/100$. That is, rarely am I five.

Notice that if I ask what is the probability of being alive (that is, in one of the states $n$ (0$\leq n \leq 99$) in a period of $\leq 99,999$ years, I must answer $1/(1000)$; that is, mostly you will find me dead.

Applying the same reasoning to the sleeping beauty Markov chain in the last post, sleeping beauty must calculate the probabilities in the case of tails not knowing whether one or two days have passed, and so she must assign $1/4$ to the probability of Monday after tails, and $1/4$ to the probability of Tuesday after tails, so that the probability of Monday or Tuesday after tails is $1/2$.

There is a complication I didn't mention explicitly in the last post. In the case of heads the beauty knows that only one step has occurred. Hence she should calculate the probability of the heads case using the probabilities in one step, not one or two steps. Hence she should estimate the probability of Monday as $1/2$ after heads.

Notice that she has calculated probabilities in two different probability spaces. The fact that the sum of these two probabilities is $1$ means that knowing that she is being interviewed gives her no new information, or said in another way, conditioning on the fact that an interview is in process does not require normalization of the probabilities.

Let us see how one statistician claims to get a contradiction from this estimate. I will describe an argument from a paper. (The article begins sloppily because in stating the problem it does not specify that the beauty knows the protocol of the procedure to be carried out.)

The sample space is not described, and only partial inference is made about probabilities but sufficient to convince the author that $P(Heads)< \frac{1}{2}$. We will see that the error in the argument is that the beauty must make calculations in two different probability spaces (one step, and one or two step).

The argument begins as follows: $P(Heads|Monday)=1/2$, $P(Heads|Tuesday)=0$, $P(Tuesday)\geq 0$, $P(Monday)=1-P(Tuesday)$ and hence $P(Monday)< 1$. All these probabilities are conditioned on the fact that an interview is taking place.

Aside from the fact that he is really talking about two probability spaces (one step, and one or two steps) I could agree with these probabilities. However the real error comes in his next step, when he seeks to apply the law of total probability:

$P(Heads)=P(Heads|Monday)P(Monday)+P(Heads|Tueday)P(Tuesday)$

$P((Heads)=(1/2)P(Monday)+0\times P(Tuesday)< (1/2)$

contradicting $P(Heads)=1/2$.

The error comes from applying the law of total probability to two different probability spaces where it is not valid. Part of the equation refers to probabilities for one step, the other part to one or two steps.

This problem was drawn to my attention by calculations of physicists and philosophers (see my last post), but I see that some eminent statisticians (Jeffrey S. Rosenthal) are also convinced that the answer is $1/3$.

What I think is the real problem is that statisticians immediately start calculating with conditional probability, Bayes formula etc without modelling the problem mathematically. It is also important to generalize a problem to avoid being confused by particular details.

In any case, I want to clarify what I said in the last post by considering a simple example.

The essential point of the sleeping beauty problem (see my last post) is that the beauty is asked to estimate a probability without knowing how many steps of the process have occurred.

**A simpler problem**Consider the following Markov chain with $100$ states meant to represent the birthdays of my life:

$0\xrightarrow{1}1\xrightarrow{1}2 \xrightarrow{1}\cdots \xrightarrow{1}99\xrightarrow{1}100 $,

with also a loop of probability $1$ at $100$ when I am in my tomb.

Now it is clear that the probability of my being $1$ after $1$ year if $1$, the probability of being $2$ after $2$ years is $1$, the probability of being $n$ after $n$ years is $1$ ($n\leq 99$).

Moreover if you ask me what is the probability of being $5$ say during my life (that is, after $0$ or $1$ or $\cdots$ or $99$ years) I might be tempted to answer $1$ since it is certain I will be $5$ during my life.

However the sum of the probabilities that I am $n$ ($n\leq 99$) during my life must add to total probability $1$, and hence I must answer that the probability of being $5$ in my life is $1/100$. That is, rarely am I five.

Notice that if I ask what is the probability of being alive (that is, in one of the states $n$ (0$\leq n \leq 99$) in a period of $\leq 99,999$ years, I must answer $1/(1000)$; that is, mostly you will find me dead.

**The sleeping beauty problem**Applying the same reasoning to the sleeping beauty Markov chain in the last post, sleeping beauty must calculate the probabilities in the case of tails not knowing whether one or two days have passed, and so she must assign $1/4$ to the probability of Monday after tails, and $1/4$ to the probability of Tuesday after tails, so that the probability of Monday or Tuesday after tails is $1/2$.

There is a complication I didn't mention explicitly in the last post. In the case of heads the beauty knows that only one step has occurred. Hence she should calculate the probability of the heads case using the probabilities in one step, not one or two steps. Hence she should estimate the probability of Monday as $1/2$ after heads.

Notice that she has calculated probabilities in two different probability spaces. The fact that the sum of these two probabilities is $1$ means that knowing that she is being interviewed gives her no new information, or said in another way, conditioning on the fact that an interview is in process does not require normalization of the probabilities.

**A statistician's argument against**$1/2$Let us see how one statistician claims to get a contradiction from this estimate. I will describe an argument from a paper. (The article begins sloppily because in stating the problem it does not specify that the beauty knows the protocol of the procedure to be carried out.)

The sample space is not described, and only partial inference is made about probabilities but sufficient to convince the author that $P(Heads)< \frac{1}{2}$. We will see that the error in the argument is that the beauty must make calculations in two different probability spaces (one step, and one or two step).

The argument begins as follows: $P(Heads|Monday)=1/2$, $P(Heads|Tuesday)=0$, $P(Tuesday)\geq 0$, $P(Monday)=1-P(Tuesday)$ and hence $P(Monday)< 1$. All these probabilities are conditioned on the fact that an interview is taking place.

Aside from the fact that he is really talking about two probability spaces (one step, and one or two steps) I could agree with these probabilities. However the real error comes in his next step, when he seeks to apply the law of total probability:

$P(Heads)=P(Heads|Monday)P(Monday)+P(Heads|Tueday)P(Tuesday)$

$P((Heads)=(1/2)P(Monday)+0\times P(Tuesday)< (1/2)$

contradicting $P(Heads)=1/2$.

The error comes from applying the law of total probability to two different probability spaces where it is not valid. Part of the equation refers to probabilities for one step, the other part to one or two steps.

Labels: probability

## 6 Comments:

The given Markov chain reasoning would certainly be appropriate if we were to consider one of the following two tweaked situations:

The SB is awakened only once after tails, the day to be decided by random. (Then we trivially get a 1/2, 1/4, 1/4 distribution). Or...

The original SB experiment, except we arrive after the experiment and request a random interview from the experiment. Then again, we trivially get a 1/2, 1/4, 1/4 distribution.

However, the SB in the original SB experiment is simply in another situation (due to the protocol of repeated drugged awakenings) and the math is simple, but different.

In your previous post, you gave the "no new information" reasoning, and in this post, you argue against a "statistician's argument", saying:

"The error comes from applying the law of total probability to two different probability spaces where it is not valid. Part of the equation refers to probabilities for one step, the other part to one or two steps."

I disagree with both, but I think it better to start with a question on a tweaked version:

1. If heads, SB is awakened on Monday. (HM)

2. If tails, SB is awakened on Tuesday. (TT)

3. There may be an additional interview on Monday. This interview will take place 50% of the time (we can think of another coin to decide for this interview).

SB knows this, and she is awakened. Question: Is being in HM or TT equally likely from SB point of view? If yes, what is the chance she is in HM?

We now play again, but this time, the additional interview in 3 will take place after tails or heads of the original coin (whichever it is is withheld from the SB). Same question, is being in HM or TT equally likely from SB point of view?

Now play again, but tell SB the exact condition for the additional interview to take place (say, after tails). Why would this influence the HM and TT odds?

The answer (which I hope you will arrive at too) is that the HM and TT odds are unchanged all the way and that they are 1/3 (due to it being a fair coin etc.). But notice: This is due to us NOT requiring that precisely one and only one interview is regarded for each coin toss outcome (as in the "request" example in the top of this post).

In other words: As long as the 3rd additional interview occurs 50% pr. experiment, the specific condition for the 3rd additional interview to take place (heads, tails, another coin, partity of a dice, random.org, or whatever) does not matter for the chance of the SB to be in HM or TT.

I will conclude with a straightforward way to calculate the SB credence that fits with a frequentist as well as a betting subjectivist probablity interpretation ("the price at which you would buy or sell a bet that pays 1 unit of utility if E, 0 if not E."):

Number of heads interviews pr. experiment: 0.5

Number of tails interviews pr. experiment: 0.5 + 0.5 = 1

Number of interviews pr. experiment: 0.5 + 1 = 1.5

Probability of heads interview: #heads / #all = 0.5 / 1.5 = 1/3

You can use this fundamental "favorable outcomes out of all outcomes" method in all kinds of tweaks of the original problem. Like, SB is not being interviewed after heads at all, the coin is not fair but 70/30, the SB is only interviewed with chance c after tails with c in [0;0], SB being interviewed 17 times after heads, 3 times after tails, etc. etc. In each case, you are going to match the correct betting odds, as is easy to verify case by case until you eventually realize that this is much ado about nothing.

If interested, I have tried to explain in depth why the SB problem is deceptive and leads many people to conflate the question being asked with another question not being asked (IMHO of course):

http://www.preposterousuniverse.com/blog/2014/07/28/quantum-sleeping-beauty-and-the-multiverse/#comment-7295910552604295996

She hss the new information that she is awake. The probability is .25 for each of four outcomes. But she is asleep for one of the four. So she has three choices that total .75 of the probability. So 1/3.

I'm neither statisticion nor physicist. I think the reasoning simpletons like me use is as follows. Do the experiment (tossing the coin) 1000 times. 500 times the coin comes up tails, 500 times heads. If it comes up tails, you know for certain you're going to interview Sleeping Beauty another 500 times (on tuesday). Sleeping Beauty knows that too. So you have a universe of 1500 outcomes, and the probability of being interviewed monday after heads is 500/1500; of being interviewed on monday after tails is 500/1500; and on tuesday after tails is also 500/1500.

So, where "tossing a coin" has 1000 outcomes, "being interviewed" has 1500 outcomes. If you ask the question "what is the probability of being interviewed on monday after the coin came up tails", wouldn't the answer be 1/3 ?

If Sleeping beauty were presented with an opportunity to bet on heads or tails upon waking, what would be fair odd?

Suppose that we change the SB puzzle, but only in the protocol for Tuesday after tails (T/t). The new protocol is that she is awakened, and, after a pause of a minute, is put back to sleep (but only on T/t). Again, she knows the full protocol in advance. In the first minute after awakening, she assigns equal probabilities for M/h, T/h, M/t, T/t. After a minute, if she is still awake, she excludes T/t and is left with equal probabilities for M/h, M/t, T/h. So in this case the 1/3 is correct. Do you agree with this result for the modified problem? If so, how can the changed protocol for T/t affect her relative weightings of the other three cases? What if she's only awake for a second?

I totally agree with Bob that "What I think is the real problem is that statisticians immediately start calculating with conditional probability, Bayes formula etc without modelling the problem mathematically."

I believe that you should explicitly define the random experiment whose outcomes form the sample space you are using to calculate probabilities. Otherwise it is very easy to mix probabilities assigned to events of different sample spaces. This is the case in Elga's arguments in favor of 1/3. I define the involved random experiments for the SB problem and demonstrate why the double halfers' position is consistent in the following paper:

http://arxiv.org/ftp/arxiv/papers/1409/1409.3803.pdf

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