Breaking the SIA with an exponentially-Sleeping Beauty

post by 0naut (kellerjordan) · 2022-02-20T08:03:32.324Z · LW · GW · 9 comments

Contents

  The original Sleeping Beauty problem
  A new construction: Exponentially-Sleeping Beauty problem
  Notes
None
9 comments

In this post I'd like to construct an extension to the Sleeping Beauty problem which is unsolvable from an thirder/SIA perspective. The core of the construction is having the number of wakings Beauty will experience follow a distribution with an infinite mean. This leads to Beauty being unable to assign normalized probability values when reasoning under SIA. Given this forum's fascination with anthropic reasoning and this problem, this may be of interest.

The original Sleeping Beauty problem

The following is a summary of the thirder/SIA approach to the original Sleeping Beauty problem. From what I can tell, this forum is very well versed here, so I won't go into too much background detail.

In the original problem, we have the following two steps for the experimenter:

1. Flip a coin. If heads, record the value $N=0$. If tails, record the value $N=1$.
2. Put Beauty to sleep [on Sunday]. She will be woken once on each of the following $N+1$ days, with memory erased between wakings. (i.e. once if heads, twice if tails)

Let $M_k$ be the proposition "I am awake and today is the $k$th day since the experiment began". (I.e. $M_1$ is "I am awake and it is Monday"; $M_2$ is "I am awake and it is Tuesday".) Also let $M$ be the proposition "I am awake", in the sense that $M=M_1\vee M_2$.

The problem is then to give Beauty a value to assign to the conditional probability of heads given that she has been awoken, i.e. to

$P(N=0\mid M)$.

One version of the thirder/SIA calculation for this value then proceeds as follows.

$\begin{array}{cc}\frac{P(N=0\mid M)}{P(N=1\mid M)}& =\frac{P(M\mid N=0)}{P(M\mid N=1)}\frac{P(N=0)}{P(N=1)}\\ & =\frac{P(M_1\mid N=0)+P(M_2\mid N=0)}{P(M_1\mid N=1)+P(M_2\mid N=1)}\frac{P(N=0)}{P(N=1)}\\ & =\frac{P(M_1\mid N=0)+0}{2\ast P(M_1\mid N=1)}\frac{P(N=0)}{P(N=1)}\\ & =\frac{1}{2}\frac{P(N=0\mid M_1)}{P(N=1\mid M_1)}\\ & =\frac{1}{2}\frac{1/2}{1/2}\\ & =\frac{1}{2}.\end{array}$

Where the steps are respectively due to:

1. Bayes' law.
2. That $M=M_1\vee M_2$ and that $M_1$, $M_2$ are disjoint propositions.
3. For the numerator, that $M_2$ is impossible given $N=0$ (i.e. we cannot wake up on Tuesday if the coin came up heads). For the denominator, this is the part of the SIA assumption: given that the coin came up tails, and we cannot distinguish a waking on Monday or Tuesday, we must put equal weight on both possibilities.
4. Bayes' law.
5. Following SIA/thirder reasoning, we put equal weight on heads or tails given that we are waking on Monday. I leave the subtler point of justifying this assignment to the many expositions of thirder/SIA reasoning.

Combining this determined ratio with the fact that $P(N=0\mid M)+P(N=1\mid M)=1$, we must have $P(N=0\mid M)=1/3$. This of course is what gives proponents of the above calculation the name "thirders".

A new construction: Exponentially-Sleeping Beauty problem

Now suppose we extend the experiment as follows. The experimenter will:

1. Sample an integer $N\ge 0$ according to the distribution $P_0(N=k)=1/2^{k+1}$ for $k\ge 0$. Perhaps the experimenter implements this by flipping a coin until the first time it comes up tails, and then recording the number of heads.
2. Wake up Beauty $2^N$ times.

Once again, let $M$ be the proposition "I am awake" and $M_k$ be the proposition "I am awake and today is the $k$th day/waking-time since the start of the experiment". What value should a waking Beauty now assign to e.g. the probability $P(N=5\mid M)$? 

We will see that in this new problem, assigning such a value will be problematic. Following the thirder/SIA approach to the original problem, we calculate the ratio

$\begin{array}{cc}\frac{P(N=k_1\mid M)}{P(N=k_2\mid M)}& =\frac{P(M\mid N=k_1)}{P(M\mid N=k_2)}\frac{P(N=k_1)}{P(N=k_2)}\\ & =\left(\frac{{\sum }_{i=1}^{i=2^{k_1}}P(M_i\mid N=k_1)}{{\sum }_{i=1}^{i=2^{k_2}}P(M_i\mid N=k_2)}\right)\frac{P(N=k_1)}{P(N=k_2)}\\ & =\frac{2^{k_1}}{2^{k_2}}\frac{P(M_1\mid N=k_1)}{P(M_1\mid N=k_2)}\frac{P(N=k_1)}{P(N=k_2)}\\ & =2^{k_1-k_2}\frac{P(N=k_1\mid M_1)}{P(N=k_2\mid M_1)}\frac{P\left(M_1\right)}{P\left(M_1\right)}\\ & =2^{k_1-k_2}\frac{1/2^{k_1+1}}{1/2^{k_2+1}}\\ & =1.\end{array}$

Where the steps are resp. due to:

1. Bayes' law/defn. of conditional probability
2. We have $M=(M_1\vee M_2\vee \dots )$ and  $P(M_{2^k+1}\vee M_{2^k+2}\vee \cdots \mid N=k)=0$, so that $P(M\mid N=k)=P(M_1\vee M_2\vee \dots M_{2^k}\mid N=k)$. These are mutually exclusive/disjoint propositions, so we can transform into a sum.
3. Following thirder/SIA reasoning, given that $N=k$ for some $k$, we must assign equal probability for each of the possible days that we can awake, i.e. $P(M_i\mid N=k)=P(M_1\mid N=k)$ for all $1\le i\le 2^k$.
4. (Bayes' law.)
5. Following thirder/SIA reasoning, given only that it is "Monday", awoken Beauty's conditional probability on the outcome of the experimenter's coin tosses is the same as what is was before she went to sleep.

Hence we have shown that following thirder/SIA reasoning, exponentially-sleeping Beauty must assign the probability value ratio $\frac{P(N=k_1\mid M)}{P(N=k_2\mid M)}=1$ for every $0\le k_1,k_2$ pair, hence $P(N=k\mid M)=P(N=0\mid M)$ for all $k\ge 0$.

This is a problem. We have an infinite collection of mutually-exclusive probabilities $\left\{P\right(N=k\mid M){\}}_{k\ge 0}$ which we have proven under SIA reasoning to all be of equal value. It is impossible to satisfy this while having our probabilities sum to unity as in ${\sum }_{k=0}^{\infty }P(N=k\mid M)=1$.

Wat does dis mean; how should Beauty assign probabilities here; is this a problem for the SIA?

Notes

• This construction requires a universe which is infinite in either time or space, in order to have the volume to support a potentially unlimited number of Beauty awakenings. However, we should note that this assumption can be somewhat weakened: If even the expected value of Beauty's probability distribution over her number of universe-supported wakings is infinite, then the calculation can still go through to a point where we cannot assign probabilities under SIA reasoning.
• Alternate constructions with some interesting properties can be realized by having Beauty wake ${1.5}^N$, $\frac{2^N}{N}$ or $3^N$ times.

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