A more Rigorous Treatment of Sleeping Beauty

Posted on by Sean Paul Campbell

In this post I will define the state space, \sigma-algebra, and associated probability measure for the generalization of the sleeping beauty problem that I presented in my previous post on the topic. At the end, we will derive the same result as before, but the process will involve more jargon.

Let’s start by enumerating the possible states of the system the problem as stated has two main components, the result of the coin toss which we will denote by \mathbb{Z}_2=\{0,1\} (with zero for heads), and the day which we will denote by \mathbb{Z}_n where zero represents the first day sleeping beauty can be woken up. There is also an integer k\leq n such that if the coin lands on heads, she will not wake up if the day is not in \mathbb{Z}_k, but this will not be important until we are actually attempting to solve the problem. Hence, we will take the state space to be \Omega=\mathbb{Z}_2\times\mathbb{Z}_n.

Since \Omega is a finite set, we may take the \sigma-algebra on \Omega to simply be the power set \mathcal{P}(\Omega).

Now for the meat of the issue, defining the probability measure. We will start with low hanging fruit and see how much of the domain we may define the function on before problems start to show up:

First, \mathbb{P}(\{0\}\times\mathbb{Z}_n)=p to denote the probability that the coin comes up heads. Similarly, \mathbb{P}(\{1\}\times\mathbb{Z}_n)=1-p.

Next we will suppose a uniform distribution on the days of the experiment, that is \mathbb{P}(\mathbb{Z}_2\times\{a\})=\mathbb{P}(\mathbb{Z}_2\times\{b\})=\frac1n for any a,b\in\mathbb{Z}_n.

And finally, we will use the definition of a product measure to give the probability of any singleton in \mathcal{P}(\Omega). \mathbb{P}(a\times b)=\frac pn whenever a=0 and \mathbb{P}(a\times b)=\frac{1-p}n when a=1.

We may now solve the problem by computing the probability that the coin came up heads given that sleeping beauty is awake:

\displaystyle{\mathbb{P}\big(\{0\}\times\mathbb{Z}_n\big|(\{0\}\times\mathbb{Z}_k)\cup(\{1\}\times\mathbb{Z}_n)\big)=\frac{\mathbb{P}\Big(\big(\{0\}\times\mathbb{Z}_n\big)\cap\big((\{0\}\times\mathbb{Z}_k)\cup(\{1\}\times\mathbb{Z}_n)\big)\Big)}{\mathbb{P}\big((\{0\}\times\mathbb{Z}_k)\cup(\{1\}\times\mathbb{Z}_n)\big)}}

Recall that \{0\}\times\mathbb{Z}_n is nothing but the event that the coin came up heads on any day of the experiment. And since sleeping beauty knows that she will only wake up the first k days of the experiment if the coin lands on heads but she will wake up all n days if the coin lands on tails, so the event that she wakes up is given by the union of the first k days with the coin on heads and all n days with the coin on tails.

Now we will simplify the above quantity further by noting that \big(\{0\}\times\mathbb{Z}_n\big)\cap\big((\{0\}\times\mathbb{Z}_k)\cup(\{1\}\times\mathbb{Z}_n)\big)=\{0\}\times\mathbb{Z}_k.

Computing the probability in the numerator is quite simple at this point:

\displaystyle{\mathbb{P}(\{0\}\times\mathbb{Z}_k)=\sum\limits_{j=0}^{k-1}\mathbb{P}(0\times j)=\sum\limits_{j=0}^{k-1}\frac pn=\frac{kp}n}

And for the denominator we may use countable additivity of measures to reduce the computation to the sum of two probabilities that we have already established:

\displaystyle{\mathbb{P}\big((\{0\}\times\mathbb{Z}_k)\cup(\{1\}\times\mathbb{Z}_n)\big)=\mathbb{P}(\{0\}\times\mathbb{Z}_k)+\mathbb{P}(\{1\}\times\mathbb{Z}_n)=\frac{kp}n+1-p}

Finally we see that

\displaystyle{\mathbb{P}\big(\{0\}\times\mathbb{Z}_n\big|(\{0\}\times\mathbb{Z}_k)\cup(\{1\}\times\mathbb{Z}_n)\big)=\frac{\frac{kp}n}{\frac{kp}n+1-p}=\frac1{1+\frac nk\frac{1-p}p}}

Which is in fact, not the probability I previously computed! I was wrong on the internet! How wonderful it is to contradict oneself, it staves off boredom.

If you have an opinion as to which answer is correct or where the discrepancy lies in my arguments please let me know in the comments! I will do another update in October after my Ph.D. exam is over.

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Published by Sean Paul Campbell