# The fairness of the Sleeping Beauty

This post will attempt a (yet another) analysis of the problem of the Sleeping Beauty, in terms of Jaynes’ framework “probability as extended logic” (aka objective Bayesianism).

**TL,DR: ***The problem of the sleeping beauty reduces to interpreting the sentence “a fair coin is tossed”: it can mean either that no results of the toss is favourite, or that the coin toss is not influenced by anthropic information, but not both at the same time. Fairness is a property in the mind of the observer that must be further clarified: the two meanings cannot be confused.*

What I hope to show is that the two standard solutions, ^{1}⁄_{3} and ^{1}⁄_{2} (the ‘thirder’ and the ‘halfer’ solutions), are both consistent and correct, and the confusion lies only in the incorrect specification of the sentence “a fair coin is tossed”.

The setup is given both in the Lesswrong’s wiki and in Wikipedia, so I will not repeat it here.

I’m going to symbolize the events in the following way:

- It’s Monday = Mon

- It’s Tuesday = Tue

- The coin landed head = H

- The coin landed tail = T

- statement “A and B” = A & B

- statement “not A” = ~A

The problem setup leads to an uncontroversial attributions of logical structure:

1) H = ~T (the coin can land only on head or tail)

2) Mon = ~Tue (if it’s Tuesday, it cannot be Monday, and viceversa)

And of probability:

3) P(Mon|H) = 1 (upon learning that the coin landed head, the sleeping beauty knows that it’s Monday)

4) P(T|Tue) = 1 (upon learning that it’s Tuesday, the sleeping beauty knows that the coin landed tail)

Using the indifference principle, we can also derive another equation.

Let’s say that the Sleeping Beauty is awaken and told that the coin landed tail, but nothing else. Since she has no information useful to distinguish between Monday and Tuesday, she should assign both events equal probability. That is:

5) P(Mon|T) = P(Tue|T)

Which gives

6) P(Mon & T) = P(Mon|T)P(T) = P(Tue|T)P(T) = P(Tue & T)

It’s here that the analysis between “thirder” and “halfer” starts to diverge.

The wikipedia article says “Guided by the objective chance of heads landing being equal to the chance of tails landing, it should therefore hold that”. We know however that there’s no such thing as ‘the objective chance’.

Thus, “a fair coin will be tossed”, in this context, will mean different things for different people.

The thirders interpret the sentence to mean that beauty learns no new facts about the coin upon learning that it is Monday.

They thus make the assumption:

(TA) P(T|Mon) = P(H|Mon)

So:

7) P(Mon & H) = P(H|Mon)P(Mon) = P(T|Mon)P(Mon) = P(Mon & T)

From 6) and 7) we have:

8) P(Mon & H) = P(Mon & T) = P(Tue & T)

And since those events are a partition of unity, P(Mon & H) = ^{1}⁄_{3}.

And indeed from 8) and 3):

9) ^{1}⁄_{3} = P(Mon & H) = P(Mon|H)P(H) = P(H)

So that, under TA, P(H) = ^{1}⁄_{3} and P(T) = ^{2}⁄_{3}.

Notice that also, since if it’s Monday the coin landed either on head or tail, P(H|Mon) = ^{1}⁄_{2}.

The thirder analysis of the Sleeping Beauty problem is thus one in which “a fair coin is tossed” means “Sleeping Beauty receives no information about the coin from anthropic information”.

There is however another way to interpret the sentence, that is the halfer analysis:

(HA) P(T) = P(H)

Here, a fair coin is tossed means simply that we assign no preference to either side of the coin.

Obviously from 1:

10) P(T) + P(H) = 1

So that, from 10) and HA)

11) P(H) = ^{1}⁄_{2}, P(T) = ^{1}⁄_{2}

But let’s not stop here, let’s calculate P(H|Mon).

First of all, from 3) and 11)

12) P(H & Mon) = P(H|Mon)P(Mon) = P(Mon|H)P(H) = ^{1}⁄_{2}

From 5) and 11) also

13) P(Mon & T) = ^{1}⁄_{4}

But from 12) and 13) we get

14) P(Mon) = P(Mon & T) + P(Mon & H) = ^{1}⁄_{2} + ^{1}⁄_{4} = ^{3}⁄_{4}

So that, from 12) and 14)

15) P(H|Mon) = P(H & Mon) / P(Mon) = ^{1}⁄_{2} / ^{3}⁄_{4} = ^{2}⁄_{3}

We have seen that either P(H) = ^{1}⁄_{2} and P(H|Mon) = ^{2}⁄_{3}, or P(H) = ^{2}⁄_{3} and P(H|Mon) = ^{1}⁄_{2}.

Nick Bostrom is correct in saying that self-locating information changes the probability distribution, but this is true in both interpretations.

The problem of the sleeping beauty reduces to interpreting the sentence “a fair coin is tossed”: it can mean either that no results of the toss is favourite, or that the coin toss is not influenced by anthropic information, that is, you can attribute the fairness of the coin to prior or posterior distribution.

Either P(H)=P(T) or P(H|Mon)=P(T|Mon), but **both at the same time is not possible**.

If probability were a physical property of the coin, then so would be its fairness. But since the causal interactions of the coin possess both kind of indifference (balance and independency from the future), that would make the two probability equivalent.

That such is not the case just means that fairness is a property in the mind of the observer that must be further clarified, since the two meanings cannot be confused.

- 17 Sep 2015 7:41 UTC; 0 points) 's comment on Open thread, Sep. 14 - Sep. 20, 2015 by (

This is technically accurate, but irrelevant, since everyone understands the phrase “a fair coin is tossed” to refer to prior distribution.

The problem with the Sleeping Beauty Problem, is that probability can be thought of as a rate: #successes per #trials. But this problem makes #trials a function of #successes, introducing what could be called a feedback loop into this rate calculation, and fracturing our concepts of what the terms mean. All of the analyses I’ve seen struggle to put these fractured meanings back together, without fully acknowledging that they are broken. MrMind comes closer to acknowledging it than most, when he says “‘A fair coin will be tossed,’ in this context, will mean different things for different people.”

But this fractured terminology can be overcome quite simply. Instead of one volunteer, use four.

Each will go through a similar experience where they will be woken at least once and maybe twice, on Monday and/or Tuesday, depending on the result of the same fair coin flip.

All four will be wakened both days with the following exceptions: SB1 will be left asleep on Monday if Heads is flipped. SB2 will be left asleep on Monday if Tails is flipped. SB3 will be left asleep on Tuesday if Heads is flipped. And SB4 will be left asleep on Tuesday if Tails is flipped. Note that SB3′s schedule corresponds to the original version of the problem.

This way, three of the volunteers will be wakened on Monday. Two of those will be wakened on again Tuesday, while the third will be left asleep and be replaced by the one who slept through Monday. And each has the same chance to be wakened just once.

Put the three in a room together, and allow them to discuss anything EXCEPT the coin result and day that they would sleep through. Ask each for their confidence in the assertion that she will be wakened just once during the experiment.

No matter what day it is, or how the coin landed, the assertion will be true for one of the three awake volunteers, and false for the other two. So their confidences should sum to 1. No matter what combination of day and result each was assigned to sleep through, each has the same information upon which to base her confidence. So their confidences should be the same.

The only possible solution is that the confidences should all be

^{1}⁄_{3}. If, instead, SB3 is just told about the other three volunteers, but never meets them, she can still reason the same way and get the answer^{1}⁄_{3}. And since “I, SB3, will be wakened only once” is equivalent to “the fair coin landed Heads,” our original volunteer can give the same answer.