A Sleeping
beauty without indifference.
Doug
Clark, 22nd November 2003. [email protected] Last revised 8th December 2003.
In this account ‘OR’ is the ‘exclusive or’, ‘probability’ refers to objective probability and P(X) is the objective probability function, which is equal to the probability that the proposition X is true.
1. The Experiment.
Suppose you toss an unbiased coin, which has a probability of landing heads equal to a half. It would seem appropriate that your degree of belief (credence or subjective probability) that the coin will land heads should also be a half. Consider another type of uncertainty. Suppose you wake up one morning in the weekend. You are certain it is the weekend but uncertain as to whether it is Saturday or Sunday. The probability that Saturday will occur during the weekend is equal to the probability that Sunday will occur during the weekend is equal to 1. But these facts are not enough to give you equal credence that it is Saturday or Sunday. This requires a new premise, a Principle of Indifference stating that under such circumstances your credence in the two situations should be equal. The Sleeping Beauty faces a similar uncertainty and in some solutions of the problem can resolve it by means of a Principle of Indifference but in the story told below she uses a different strategy.
This is my version of the experiment performed on the Sleeping Beauty:
On Sunday evening the researchers explain the experimental protocol to Beauty and later put her to sleep. On Monday morning they awaken her, interview her and then give her a potion, which puts her to sleep and erases all memory of the Monday morning awakening. Depending on the toss of a fair coin the experiment then follows one of two routes. If heads; she remains asleep until Wednesday when the experiment ends. If tails; she is woken on Tuesday morning, interviewed and then put to sleep. On Wednesday she is woken and the experiment ends. In the three awakenings that may occur during the experiment the evidence available to Beauty is identical.
2. A ‘Halfer’ or a “Thirder’?
In the original version of the Sleeping Beauty story, on awakening she is asked; “What is your credence that the coin lands heads?” (See Note 1.) Elga (2000) has argued her answer should be 1/3 and Door (2003) has also defended this view. Lewis (2001) has concluded that her answer should be 1/2. All these authors have used a Principle of Indifference in their arguments; the different conclusions arise from differences in other premises. However Elga’s Principle of Indifference is not universally accepted. The aim of this essay is to see what conclusions might be drawn without using Elga’s Principle of Indifference.
3. Beauty’s Premises.
The awakenings and the probabilities of their occurrence during the experiment are:
H1 = The Sleeping Beauty wakes on Monday morning
and the coin lands heads.
T1 =
The Sleeping Beauty wakes on Monday morning and the coin lands tails.
T2 =
The Sleeping Beauty wakes on Tuesday morning and the coin lands tails.
(1) P(H1 OR (T1 AND T2)) = 1
P(H1) = P(T1 AND T2) = P(T1) = P(T2)
= 1/2.
In each awakening Beauty is asked, “What awakening is this and what is the probability that your answer is correct?”
In each awakening Beauty’s reasoning is identical, and proceeds as below:
Let the day be D and let the result of the coin toss
be C.
Let
A be this awakening, where A =
The Sleeping Beauty wakes on D morning
and the coin lands C.
IF D = Monday AND C = Heads THEN A = H1.
IF D = Monday AND C = Tails THEN A = T1.
IF D = Tuesday AND C = Tails THEN A = T2.
(2) P(A = H1 êH1) = 1.
P(A = T1êT1 AND T2) = 1 – P(A = T2êT1 AND T2) = x, where x Î {1, 0}.
Therefore P(A Î {T1, T2} êT1 AND T2) = 1.
In addition P(A = T1êT1 AND T2) ¹ P(A = T2êT1 AND T2), so Beauty cannot have equal credence that her current awakening is T1 or T2.
The first set of premises is Beauty’s model of the outcomes of the experiment and the second set of premises is Beauty’s model of her location within the experiment.
From (1) and (2):
P(A = H1) = P(A Î {T1, T2}) = 1/2.
Therefore Beauty can believe her awakening is H1 with credence equal to a half, but
the values of P(A = T1êT1 AND T2) and P(A = T2êT1 AND T2) cannot be derived from (1) and (2).
4. Beauty’s Answer.
In each interview Beauty selects her answer randomly by tossing a dice.
If the result is 1 or 2 her answer is “This
awakening is H1.”
If the result is 3 or 4 her answer is “This
awakening is T1.”
If the result is 5 or 6 her answer is “This
awakening is T2.”
Her selection can be represented by the random variable D(A):
(3) P(D(A) = (A = H1)) = 1/3.
P(D(A) = (A = T1)) = 1/3.
P(D(A) = (A = T2)) = 1/3.
Beauty has chosen this random variable such that:
(i) P(D(A) êT1 AND T2) has a unique value.
(ii) The probability of D(A) being true for any awakening is proportional to the probability that the awakening occurs in the experiment.
5. What is the probability that D(A)
is true?
The probability that D(A) is true can be inferred from (1), (2) and (3).
P(D(A)) = P(D(A) êH1) ´ P(H1) + P(D(A) êT1 AND T2) ´ P(T1 AND T2)
Evaluate each term on the right hand side separately:
P(D(H1))
= 1/3.
P(A = H1 êH1) = 1.
P(D(A)½H1) = 1/3.
Therefore P(D(A) êH1) ´ P(H1) = 1/3 ´ 1/2 = 1/6.
P(D(T1))
= P(D(T2)) = 1/3.
P(D(T1)½T1 AND T2) = P(D(T2)½T1 AND T2) = 1/3.
P(D(A) êT1 AND T2) Î { P(D(T1)½T1 AND T2), P(D(T2)½T1 AND T2)}.
Therefore P(D(A) êT1 AND T2) = 1/3.
Therefore P(D(A) êT1 AND T2) ´ P(T1 AND T2) = 1/3 ´ 1/2 = 1/6.
Therefore P(D(A)) = 1/6 + 1/6 = 1/3.
6. The Complete Distribution.
Table 1 summarises the probabilities with which different awakenings occur in the experiment and the probabilities with which the answers chosen by the dice give a true identification of the awakenings. Each row in the table, except the first and last, represents one of a series of mutually exclusive events and its associated probability.
Table 1.
|
First awakening |
Is the answer chosen by the dice true or false? |
Second awakening |
Is the answer chosen by the dice true or false? |
Probability |
|
H1 |
True |
|
|
3/18 |
|
H1 |
False |
|
|
6/18 |
|
T1 |
True |
T2 |
False |
2/18 |
|
T1 |
False |
T2 |
True |
2/18 |
|
T1 |
True |
T2 |
True |
1/18 |
|
T1 |
False |
T2 |
False |
4/18 |
|
Total |
1 |
|||
The table shows that:
(i) P(D(A)½H1) ´ P(H1) = 3/18 = 1/6.
P(D(A) êT1 AND T2) ´ P(T1 AND T2) = 2/18 + 1/18 = 1/6.
P(D(A)) = 3/18 + 2/18 + 1/18 = 1/3.
(ii) Call H1 a heads-awakening and T1 and T2 tails-awakenings.
Let NH be the number of times a heads-awakening is correctly identified in an experiment.
Let NT be the number of times a tails-awakening is correctly identified in an experiment.
Let E(X) be the function that gives the expected or mean value of the random variable X.
E(NH) = 3/18.
E(NT) = 2/18 + 2/18 + 2 ´ 1/18 = 6/18.
E(NH) / (E(NH) + E(NT)) = (3/18) / (3/18 + 6/18) = 1/3.
I would call E(NH) / (E(NH) + E(NT)) a proportion rather than a probability, since its calculation involves adding the probabilities of two events which are not mutually exclusive.
7. Another Complete Distribution.
Suppose that in each awakening Beauty is asked, “Is this a heads-awakening or a tails-awakening and what is the probability that your answer is correct?” In each interview Beauty selects her answer randomly by tossing a dice, such that the probability that her answer is correct will be the same in each awakening:
If the result is 1, 2 or 3 her answer is “This is a heads-awakening.”
If the result is 4, 5 or 6 her answer is “This
is a tails-awakening.”
Her selection can be represented by the random variable D*(A):
P(D*(A) = (A = H1)) = 1/2.
P(D*(A) = (A Î {T1, T2})) = 1/2.
Table 2 summarises the probabilities with which different awakenings occur in the experiment and the probabilities with which the dice gives a true identification of the awakenings.
Table 2 shows that:
(i) P(D*(A)½H1) ´ P(H1) = 2/8 = 1/4.
P(D*(A) êT1 AND T2) ´ P(T1 AND T2) = 1/8 + 1/8 = 1/4.
P(D*(A)) = 2/8 + 1/8 + 1/8 = 1/2.
(ii) E(NH) = 2/8 = 1/4.
E(NT) = 1/8 + 1/8 + 2 ´ 1/8 = 4/8 = 1/2.
E(NH) / (E(NH) + E(NT)) = (1/4) / (1/4 + 1/2) = 1/3.
Table 2.
|
First awakening |
Is the answer chosen by the dice true or false? |
Second awakening |
Is the answer chosen by the dice true or false? |
Probability |
|
H1 |
True |
|
|
2/8 |
|
H1 |
False |
|
|
2/8 |
|
T1 |
True |
T2 |
False |
1/8 |
|
T1 |
False |
T2 |
True |
1/8 |
|
T1 |
True |
T2 |
True |
1/8 |
|
T1 |
False |
T2 |
False |
1/8 |
|
Total |
1 |
|||
8. What Should Beauty Believe?
Let Cr(X) be Beauty’s credence in the proposition X. This is what Beauty can believe about her current awakening and about the answers chosen by the dice:
i) She can believe that A is a head-awakening or that A is a tails-awakening, each with credence of a half:
Cr(A = H1) = Cr(A Î {T1, T2}) = 1/2.
ii) She cannot have equal credence that her current awakening is T1 or T2.
Cr(A = T1êT1 AND T2) ¹ Cr(A = T2êT1 AND T2).
To be able to have a unique degree of belief that e.g. A = T1êT1 AND T2; Beauty must introduce a random element into the experiment, because a unique value for P(A = T1êT1 AND T2) is not defined by the experimental protocol. (See Note 2.)
iii) She can believe that her awakening in the experiment is what the dice tells her, with credence equal to the probability that the answer of the dice is correct:
Cr(D(A)) = 1/3.
Cr(D(A)½H1) ´ Cr(H1) = Cr(D(A) êT1 AND T2) ´ Cr(T1 AND T2) = 1/6.
Cr(D*(A)) = 1/2.
Cr(D*(A)½H1) ´ Cr(H1) = Cr(D*(A) êT1 AND T2) ´ Cr(T1 AND T2) = 1/4.
iv) She can believe that the value of the proportion E(NH) / (E(NH) + E(NT)) is equal to1/3.
v) Suppose the coin is biased and the probability that the coin lands heads is h:
Cr(D(A)) = 1/3.
Cr(D(A)½H1) ´ Cr(H1) = 1/3 ´ h.
Cr(D(A) êT1 AND T2) ´ Cr(T1 AND T2) = 1/3 ´ (1 – h).
E(NH) / (E(NH) + E(NT)) = h/(2 – h).
9. References.
Elga A. (2000) Self-locating belief and the Sleeping Beauty problem. Analysis 60(2) 143-147. http://www.princeton.edu/~adame/papers/sleeping/sleeping.html
Lewis D. (2001) Sleeping Beauty: reply to Elga.
Analysis 61 171-76. http://www.nottingham.ac.uk/journals/analysis/preprints/LEWIS.html
Dorr C.
(2003) Sleeping Beauty: in
defence of Elga. http://homepages.nyu.edu/~cd50/papers/SleepingBeauty.pdf
Note 1: A definitive collection of
Sleeping Beauty problems and links to related sites are presented on Nick
Wedd’s excellent website. http://www.maproom.co.uk/sb.html
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Note 2: The introduction of a random element increases the
number of possible outcomes of the experiment and decreases Beauty’s
uncertainty about the location of her current awakening in the experiment. For a discussion by Brian Weatherson, on the concept of
uncertainty and an example where gaining more knowledge about a situation
decreases the number of possible outcomes and increases uncertainty (i.e.
probabilities are calculated with the new fact being taken as given, thereby
excluding outcomes in which the fact is not true), click on: http://philosophyweblog.blogspot.com/2002_12_01_philosophyweblog_archive.html
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