Biased coin flip from an unbiased coin flip Von Neumman's method allows us to generate a fair coin flip from any unbiased coin flip using only two bits (two tosses) of information (http://en.wikipedia.org/wiki/Fair_coin). 
Is the reverse possible? Generating any arbritrary Bernoulli random variable with a fair one ($p = \frac {1} {2}$) with two bits of information. I know we can approximate a uniform distribution with infinite fair Bernoulli tosses. I'm not so sure if this can be done using two bits of information.
 A: I think it can be done using a method similar to von Neumann's.
Suppose you want to produce a Bernoulli trial - a simulated coin toss - with probability of heads being $p = \dfrac{a}{b}$, with $a \leq b$, both non-negative integers, $b \neq 0$ (so that $0 \leq p \leq 1$).
Begin a series of toss sequences with a fair coin, each sequence being $b$ tosses in length.
At the completion of each sequence, if it has not had exactly $a$ heads, discard that sequence and begin a new sequence. However, if it has had exactly $a$ heads, then the process stops and we take the result of the first toss in that sequence. The probability of the result being a head should be $\dfrac{a}{b}$, as required.
This makes it achievable for rational $p$. I suspect transcendental $p$ (e.g. $\frac{1}{\pi}$) would be impossible to generate but that algebraic irrationals (e.g. $\frac{1}{\sqrt{2}}$) might be possible.
A: It seems to me that with two flips of a fair coin (two bits of information) you can only simulate the probabilities $\frac14, \frac12, \frac34$ (and, trivially $0$ and $1$).  
This is because you have an equiprobable sample space of $4$ outcomes, so all events have probability $\frac{k}4$ for some $k\in\{0,1,2,3,4\}$.
