# Anthropic Bias Part 1

## Sat Apr 29 2023E.W.Ayers

This is a post about the Sleeping Beauty Problem and the accompanying book Anthropic Bias by Nick Bostrom.

This book broke me. This is a multi-part blog series of offloading.

# 1. Back to basics

There is a classic teaser that you give to people learning probability theory called the The Two Children

A: You are walking down the street and you meet a woman called Alice and her son. She tells you she has two children. What is the probability that both her children are boys?

B: A company is holding a dinner for working mothers with at least on son. You meet Betty who tells you she has two children. What is the probability that both children are boys?

For Alice, the answer is ⅓. For Betty, the answer is ½. I remember this being surprising and counterintuitive when I learnt it in school, but at that point I had been beaten over the head with so many probability gotchas that I had given up on anything making sense.

Why does this result feel wierd? It seems on the face of it that A and B are the same scenario, in both cases we have the information that there is one boy and the other child is unknown. The reason for the different probabilities is because the wording of the questions imply that we are drawing from different distributions.

In A, we are drawing from the space {BB, BG, GB, GG} × {1, 2} where 1 is "the mother takes out youngest child" and 2 is ditto "eldest child". When we draw (bump into Alice), we are told the child is 'B', so the calculation to do is (BB1 + BB2) / (BB1 + BB2 + BG1 + GB2) = ½.

In B, we are drawing from {BB, BG, GB}, so it's ⅓.

Once you frame it like this it's ok. The lesson from this 'paradox' is that the space that you are drawing from is not often spelled out and can be an exercise in interpretation. Indeed there is not always an unambiguous interpretation.

## 1.1. Ambiguity

In the original statement of the problem in Scientific American, case B was worded as

Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

It was noted (see wikipedia page for citations etc) that this can give ½ or ⅓ depending on exactly how you interpret "at least one of them is a boy":

1. From all mothers with two children, draw from those with at least one boy → ⅓

2. From all mothers with two children, draw one of those children, and then constrain that the child is a boy. → ½

I avoided this ambiguity by using a variant that states it was a dinner for mothers with at least one boy.

So sometimes it's possible for people to pose maths problems where the question is ambiguous or doesn't give you enough infomation to answer. That's ok; we can ask the question-poser to clarify things. We've got a lid on probability.

And then I learnt about the Sleeping Beauty Problem.

# 2. The Sleeping Beauty Problem

Note, people say credence rather than probability. Credence means: how much does a person believe that a proposition is true. Afaict, it's just used as a more precise word than 'probability' when it is about a person's beliefs.

Let's write down the events:

• H = the coin flip came up heads

• T = tails

• Mon = it is Monday

• Tue = it is Tuesday

Let p(H) be the probability that Sleeping Beauty assigns to heads. The problem is similar to the two-children problem in that the trickiness comes from deciding how we are drawing from our sample space.

The Halfer Argument: You draw from the space {H, T}. In the case that H was drawn, you then set the day to Mon. In the T case - draw uniformly from {Mon, Tue}. Hence

The Thirder Argument: The space is {(H Mon), (T Mon), (T Tue)}. Hence the chance that the coin landed heads is p(H Mon) / (p(H Mon) + p(T Mon) + p(T Tue)) = ⅓.

I have heard people doggedly argue both for thirder and halfer position. I don't know which one is right.

Let's turn the dials and see what happens.

## 2.1. Big Tails

Let N := 1,000,000

Suppose that if T is thrown, we instead wake up Sleeping Beauty on N consecutive days. H is only woken on Monday still.

Now the Halfer's new calculation will still come out as ½, but the Thirder will now declare 1 / N. That is, p(H) → 0 as N → ∞.

## 2.2. The Reveal

Now go back to original setup, and suppose that you wake up and it is then revealed that it is Monday. What is your new updated probability of heads?

Note by Bayes' Rule we have p(H | Mon) = p(Mon | H) p(H) / p(Mon). And p(Mon) = p(Mon|H) p(H) + p(Mon|T) p(T) since H and T are mutex.

Halfer: We have p(H | Mon) = 1 × ½ / ¾ = ⅔

Thirder: We have p(H | Mon) = 1 × ⅓ / ⅔ = ½

A tempting position for the Halfer would be to say: "I don't get any information from knowing the day, the coin flip is independent of what day I got woken up, so it's still ½ after I learn that it is monday". Adopting this position means abandoning Bayes' rule: since p(Mon|H) = 1, we have p(H) = p(H|Mon) p(Mon), so believing p(H) = p(H|Mon) = ½ means you believe p(Mon) = 1. But this is absurd because you could plausibly learn that it is Tuesday.

## 2.3. Dutch books

Can we solve it with a Dutch Book argument?

In the two-children-problem. We can test what the correct probability is by doing a repeated experiment. In particular we can do a repeated experiment where you are asked to bet on whether both of the children are boys. If you have chosen your probability right, then there is no way that a bookie with the same knowledge as you can't exploit your probability.

In the case of Sleeping Beauty it is ambiguous how you should perform the betting process, let's say that Sleeping Beauty has the opportunity to bet whenever she is woken up. So in the T branch, Sleeping Beauty gets to bet on both Monday and Tuesday. In that case she should bet at 1:2 odds on H, since a double-payout is recieved on the T branch.

Note that this is consistent with both the Halfer and Thirder position. The Halfer says that even though it's 1:1 heads vs tails, we should put more weight on the T branch because payout is 2× the payout on H branch. The Thirder says no, there is the same payout on all three, equally likely, outcomes {(H Mon), (T Mon), (T Tue)}, so bet 1:2 odds on H.

Both Halfers and Thirders will behave the same decision-theoretically, the question is about what even is probability.

Positivists can get off the train here. It doesn't matter which one is right because they both lead to the same behaviour and there is no experiment to determine which is the 'right' probability assignment.

# 3. Rephrasings of the SBP

Here are two other formulations of the SBP that have the same structure.

## 3.1. The Sailor's Child

Now rather than being the same person at different times, you and your possible sibling are now different people.

I'll stipulate some labels to match this up with Sleeping Beauty: let's suppose the islands are called Mondagascar (Mon for short) and Tuesdegascar (Tue), but these are nautical names that the islanders don't know. Barry always has a child on Mondagascar and only a Tuesdegascar child if tails is thrown.

Revisiting the betting argument, let's suppose each child gets to bet. Should you be maximising total payout from you and your sibling (you should bet 1:2 for H)? Or just your own payout (bet 1:1)? Are you a Halfer or a Thirder?

## 3.2. The Incubator

This is the thought experiment proposed by Bostrom in Anthropic Bias. I'll copy verbatim.

I'd say this is more similar to the sailor case than the sleeping beauty case, but now the thought experiment mechanics use wierd sci-fi and beards instead. Black beard is Mon, white is Tue.

Bostrom then adds a second step to the experiment:

Stage (b): A little later, the lights are switched on, and you discover that you have a black beard. Question: What should your credence in heads be now?

We can add this stage (b) to SBP by telling Sleeping Beauty that it is Monday. Similarly, tell the child that they are on Mondagascar.

I think the Sailor Barry version of the problem does everything that The Incubator does, without being really wierd, so I'm going to stick to the Sleeping Beauty and Sailor Barry variants.

# 4. Wrap up

It's a strange problem. It has the feels of "does a tree that fall in the woods make a sound". If we find two people in the forest arguing about whether the tree made a sound, there is no experiment that we could do that would cause one to be right and the other to be wrong. They would both agree on what the noise meter said.

Similarly here, there is no way to bet or perform an experiment that would vindicate the Halfer over the Thirder.

However, whether you are a Halfer or a Thirder has some strange consequences.

.... part 2 coming soon.