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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 $p^x$?

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    "Let us continue this discussion in chat." Mwahaha. Are you for real?2016-09-02

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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.

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    This is great! Though I think it could be formulated$a$bit more concisely, how about "$\sqrt{1/2}$ is the probability that of two infinite sequences $a$ and $b$ of unbiased random bits, the exact number $n$ of leading zeroes in $a$ appears in $2n$ leading bits from $b$".2013-01-22