Can probability be greater than one?

Disclaimer: don’t take this article too seriously. Obviously, a probability greater than one violates many properties of probability (including Kolmogorov’s axioms), and the reasoning here is far from rigorous. I wrote this partly as a joke, partly as food for thought. I do think, however, that this line of reasoning has something to it and could probably be formalized by someone who has more motivation.

Have you ever wondered whether a probability could equal, say, 2?

The (frequentist) definition of probability is the limit of the ratio between the number of successes, \(N_s\), and the number of trials, \(N\):

\[ p = \lim_{N\to\infty}\frac{N_s}N \]

If the number of successes is allowed to be higher than the number of trials, then nothing prevents \(p\) from exceeding 1.

Consider, for instance, the sleeping beauty problem:

Sleeping Beauty volunteers to undergo the following experiment and is told all of the following details: On Sunday she will be put to sleep. Once or twice, during the experiment, Beauty will be awakened, interviewed, and put back to sleep with an amnesia-inducing drug that makes her forget that awakening. A fair coin will be tossed to determine which experimental procedure to undertake: if the coin comes up heads, Beauty will be awakened and interviewed on Monday only. If the coin comes up tails, she will be awakened and interviewed on Monday and Tuesday. In either case, she will be awakened on Wednesday without interview and the experiment ends.

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. During the interview Beauty is asked: “What is your credence now for the proposition that the coin landed heads?”.

See also this video by Julia Galef, from which I originally found out about this problem.

Denote the events of the coin landing heads as \(H\), the coin landing tails as \(T\), and Beauty wakening up and being interviewed as \(I\). Then the problem lies in finding the conditional probability \(p(H|I)\).

Conditional probability problems of this kind are solved by the Bayes theorem:

\[ p(H|I)=\frac{p(I|H)p(H)}{p(I|H)p(H)+p(I|T)p(T)} \]

The prior probabilities \(p(H)\) and \(p(T)\) equal \(1/2\) for a fair coin, so

\[ p(H|I)=\frac{p(I|H)}{p(I|H)+p(I|T)}. \]

The probability \(p(I|H)\) of Beauty being interviewed under heads is 1: it happens with certainty.

But if the coin comes up tails, the interviewing happens twice as many times. Every time the experiment is conducted and the coin lands tails, the event \(I\) occurs two times. The only logical possibility is to assign \(p(I|T)\) the value of 2.

Therefore,

\[ p(H|I)=\frac{1}{1+2}=1/3. \]

I guess I’m a thirder then.