Sleeping Beauty Problem - A Generic Solution

---

One Generalized Variation of the Sleeping Beauty Problem

Background

The Sleeping Beauty Problem is highly controversial. No consensus has ever been reached among scholars and/or amateurs. Why is that? Some possible reasons are:

• some people complain that the problem statement itself is vaguely defined and hence is subject to different interpretations
• some people said it is due to Antropic Bias. As discussed in the book , Self-Sampling Assumption (SSA) vs Self-indication Assumption (SIA) etc are causing different results
• some people might find it confusing to handle the fact that: (a) Head-events and Tail-events are mutually exclusive and (b) inside the Tail-event space, there are more than one must-happen events
• some people might have found the problem parameters simple, and hence have resorted to intuition instead of painstaking basic probability calculations

Sleeping Beauty Problem - Simplest Solution

Indeed, there is a straightforward proof that supports the Halfer position:

• Probability that the coin landed head, given Sleeping Beauty is being interviewed

P(Head | at least one interview)

= P(Head ∩ "at least one interview“) / P(at least one interview)

= P(Head ∩ TRUE) / P(TRUE)

= P(Head)

= 1/2

IMPORTANT: P(at least one interview) = P(TRUE) = 100% because the problem statement has made it clear that there is always at least one interview no matter the coin landed Head or Tail.

One "Generalized" Variation of the Sleeping Beauty problem

The approach being taken to generalize the problem is as follows:

• Generalize the problem with a biased M-face die, instead of a fair coin
• Generalize the problem with N days, instead of simply Monday and Tuesday
• Generalize the problem with a probabilistic function V(d,n) that decides whether an interview is to be conducted on Day-n if the die roll result is "d", instead of the simple rules stated in the original problem.

Rules of the "Generalized" Experiment

• Rule 0 - Any time Sleeping Beauty is awakened and interviewed, she will not be able to tell which day it is or whether she has been awakened before.
• Rule 1 - A biased M-face die is rolled once.

$D(d)=$ probability of die roll result being Die = d

$\sum _{d=1}^{M}D(d)=1$

• Rule 2 - Based on the die roll result “d”, each morning for the next N days, we will make an independent probabilistic decision on whether to conduct an interview that day.

$V(n|d)=$ probability of conducting an interview on Day = n, given the die roll result being Die = d

Note: V(n1|d1) and V(n2|d2) are independent of each other, unless d1 = d2 AND n1 = n2.

• Rule 3 - During the interview, Sleeping Beauty is asked: "What is your credence now for the proposition that the die roll result is Die = d ?"

Step by Step Calculations

What exactly is the question that Sleeping Beauty is trying to answer?

• The question that Sleeping Beauty is being asked is: "What is your credence now for the proposition that the coin landed heads?".

• Many people interpret this question as:

What is P(Head | being in an interview)?

This will involve the calculation of 1/P(being in an interview). However, there are very different understandings of how to calculate this P(being in an interview), resulting in several logical but conflicting results. In particular, the fact that (a) Head-events and Tail-events are mutually exclusive and (b) there are more than one must-happen events inside the Tail-event space, are prone to misunderstanding and erroneous treatments.

• A more precise interpretation of the question should be:

What is P(Head | at least one interview)?

This is exactly the only thing that Sleeping Beauty is aware of. With this interpretation, people will have a unified approach to deal with the two error-prone facts mentioned above.

Warming Up

• Probability of Die = d and then conducting an interview, on Day = n

P(Die=d ∩ Interview | Day=n)

= D(d) ∗ V(n|d)

• Probability of conducting an interview for all possible Die = d, on Day = n

P(Interview | Day=n)

=$\sum _{d=1}^M[D(d)*V(n|d)]$

IMPORTANT: The different Die=d event spaces are mutually exclusive. Hence, we use P(A∪B) = P(A) + P(B).

Actual Calculations

• Probability of having no interviews, given Die = d

P(no interview | Die=d)

= P(no interview | Day=1 ∩ Die=d) * P(no interview | Day=2 ∩ Die=d) * ... * P(no interview | Day=M ∩ Die=d)

= [1 - P(Interview | Day=1 ∩ Die=d)] * [1 - P(Interview | Day=2 ∩ Die=d)] * ... * [1 - P(Interview | Day=M ∩ Die=d)]

= [1 - V(Day=1 | Die=d)] * [1 - V(Day=2 | Die=d)] * ... * [1 - V(Day=M | Die=d)]

=$\prod _{n=1}^N[1-V(n|d)]$

IMPORTANT: The different Day=n event spaces are mutually independent. Hence, we use P(A∩B) = P(A) * P(B).

• Probability of having at least one interview, given Die = d

P(at least one interview | Die=d)

= 1 - P(no interview | Die=d)

= 1 - $\prod _{n=1}^N[1-V(n|d)]$

• Probability of having Die = d, and then (given Die = d) having at least one interview

P(Die=d ∩ "at least one interview")

= D(d) * P(at least one interview | Die=d)

=$D(d)*[1-\prod _{n=1}^N[1-V(n|d)]]$

• Probability of having at least one interview, for all possible Die = d where d = 1..M

P(at least one interview)

=$\sum _{d=1}^M$[D(d) * P(at least one interview | Die=d)], because various Die = d are mutually exclusive

=$\sum _{d=1}^M[D(d)*[1-\prod _{n=1}^N[1-V(n|d)]]]$

= 1 - $\sum _{d=1}^M[D(d)*\prod _{n=1}^N[1-V(n|d)]]$

• Probability of die roll result being Die = d, given at least one interview is conducted

P(Die=d | at least one interview)

= P(Die=d ∩ "at least one interview") / P(at least one interview)

=$\frac{D(d)*[1-\prod _{n=1}^N[1-V(n|d)]]}{1-\sum _{d=1}^M[D(d)*\prod _{n=1}^N[1-V(n|d)]]}$

Sleeping Beauty Problem as a Special Case

Now, we could use the generic solution developed above to solve the Sleeping Beauty Problem, by plugging appropriate parameters into the formula above.

• Fiar Coin vs Biased Die:

a fair coin => M=2 and D(d=1) = D(d=2) =1/2

• Interview:

Monday and Tuesday => N=2

Inteview on Monday only for Head => V(n=1 | d=1) = 1 and V(n=2 | d=1) = 0

Inteviews on Monday+Tuesday for Tail => V(n=1 | d=2) = 1 and V(n=2 | d=2) = 1

• Sleeping Beauty being asked during an interview:

the only knowledge Sleeping Beauty has is "at least one interview" is conducted

P(Head ∩ "at least one interview")

=$D(d=1)*[1-\prod _{n=1}^2[1-V(n|d=1)]]$

= D(d=1) * { 1 - (1 - V(n=1|d=1)) * (1 - V(n=2|d=1)) }

= 1/2 * { 1 - (1-1)*(1-0))

= 1/2

P(at least one interview)

= 1 - $\sum _{d=1}^2[D(d)*\prod _{n=1}^2[1-V(n|d)]]$

= 1 - $\sum _{d=1}^2[D(d)*[1-V(n=1|d)]*[1-V(n=2|d)]]$

= 1 - { [ D(d=1) * [1 - V(n=1|d=1)] * [1 - V(n=2|d=1)] ] + [ D(d=2) * [1 - V(n=1|d=2)] * [1 - V(n=2|d=2)] ] }

= 1 - { [ 1/2 * [1 - 1]*[1 - 0] ] + [ 1/2 * [1 - 1]*[1 - 1] ] }

= 1

P(Head | at least one interview)

= P(Head ∩ "at least one interview") / P(at least one interview)

= 1/2

References

This article "Sleeping Beauty Problem - A Generic Solution" is from Wikipedia. The list of its authors can be seen in its historical and/or the page Edithistory:Sleeping Beauty Problem - A Generic Solution. Articles copied from Draft Namespace on Wikipedia could be seen on the Draft Namespace of Wikipedia and not main one.

---