Many of well-known probability paradoxes are related to selection effects, self-locating belief and/or anthropic reasoning, for example:
Statistically, the probability of any one of us being here is so small that you’d think the mere fact of existing would keep us all in a contented dazzlement of surprise. We are alive against the stupendous odds of genetics, infinitely outnumbered by all the alternates who might, except for luck, be in our places (Lewis Thomas 1978).
Sleeping Beauty Problem
A person, called Sleeping Beauty, takes part in an experiment and is given the following details.
On Sunday she will be put to sleep. Depending on the outcome of a toss of a fair coin, she will be awakened once or twice during the experiment:
- on Monday only, if the outcome of the toss is heads
- on Monday and Tuesday, if the outcome is tails
After each awakening, Sleeping Beauty is interviewed, and put back to sleep with a drug that makes her forget that awakening. So each time Sleeping Beauty is awakened and interviewed she does not know whether she has been awakened before. She also does not know which day of the week it is.
During the interview Sleeping Beauty is asked: “What is your belief now that the outcome of the coin toss was heads?”
This problem has produced a lot of debate. Two very different possible solutions are defended by the so-called halfer and thirder camps.
Thirders argue that the probability of heads is 1/3 (Elga 2000). They reason as follows.
Upon awakening Sleeping Beauty is in one of three situations:
H1 heads and it is Monday
T1 tails and it is Monday
T2 tails and it is Tuesday
Suppose Sleeping Beauty is informed that the coin landed tails. She then knows she is in situation T1 or T2. Given that the outcome of the toss is tails, her belief that it is Monday should equal her belief that it is Tuesday, since being in one situation would be subjectively indistinguishable from the other, so
P(Monday | tails) = P(Tuesday | tails)
or P(T1 | T1 or T2) = P(T2 | T1 or T2)
therefore P(T1) = P(T2) (1)
Suppose now that Sleeping Beauty is informed upon awakening that it is Monday. The objective chance of heads is equal to the chance of tails, so
P(heads | Monday) = P(tails | Monday) = 1/2
or P(H1 | H1 or T1) = P(T1 | H1 or T1) = ½
therefore P(H1) = P(T1) (2)
From (1) and (2) it follows that
P(H1) = P(T1) = P(T2)
Since these are probabilities of all possible situations, they sum to 1, so
P(H1) = P(T1) = P(T2) = 1/3
Halfers argue that the probability of heads is 1/2 (Lewis 2001). They reason as follows.
Sleeping Beauty is told the details of the experiment and receives no new information when she is wakened. Since her belief before the experiment was P(heads) = 1/2, she ought to continue to have a belief of P(heads) = 1/2 since she gains no new relevant evidence when she wakes up during the experiment. This directly contradicts one of the thirder’s premises, since it means P(tails | Monday) = 1/3 and P(heads | Monday) = 2/3.
Elga, A. (2000). Self-locating Belief and the Sleeping Beauty Problem. Analysis. 60 (2): 143–147. http://www.princeton.edu/~adame/papers/sleeping/sleeping.pdf
Lewis, D. (2001). Sleeping Beauty: reply to Elga.Analysis. 61 (3): 171–76. http://www.fitelson.org/probability/lewis_sb.pdf
Wikipedia. Sleeping Beauty problem.