Understanding what $\sqrt{p}$ means for an event of probability $p$

Question

Say I have a random event $E$ with probability $p$. There is a natural interpretation in terms of $E$ for the probability $p^2$: it's the probability that $E$ occurs twice if I perform two independent trials.

Is there a similar interpretation for the probability $\sqrt{p}$? More generally, given $x \in ]0, 1[$, is there an interpretation of $p^x$?

Answer 1

As already said in the comments, the answer depends very much on the simulation procedures one allows. Assume for example that $p=\frac12$ and that one wants to simulate a bit 0 or 1 with probability $\sqrt{p}=\frac1{\sqrt2}$ of being 1. Since $\frac1{\sqrt2}$ is not dyadic, this is impossible from a finite collection of unbiased independent bits, but, as soon as one allows a stream of independent unbiased bits with finite but unlimited length, basically everything becomes possible.

To see this in the example at hand, consider the series expansion $$ \frac1{\sqrt2}=\sum\limits_{n\geqslant0}{2n\choose n}\frac1{2^{3n+1}}. $$ This suggests the following procedure. First simulate some independent unbiased bits with values 0 or 1, and count the number N of 0s before the first 1. If N=0 (that is, if the first bit is 1), return B=1. Otherwise, simulate 2N other independent unbiased bits with values 0 or 1 and consider their sum S. If S=N, return B=1, otherwise, return B=0. Then, B has probability $\frac1{\sqrt2}$ to be 1.

The mean number of unbiased bits needed to generate N is 2, hence the mean number of unbiased bits needed to get each (biased) bit B is 2+4=6.