$\star$ Describe a procedure that takes as input two integers $a$ and $b$ such that $0 < a < b$ and, using fair coin flips, produces as output heads with probability $a/b$ and tails with probability $(b-a)/b$. Give a bound on the expected number of coin flips, which should be $\mathcal{O}(1)$. (Hint: represent $a/b$ in binary.)

This is actually tricky. Fortunatelly, this answer helper me.

You represent $a/b$ as binary. Then you start flipping coins and expect each flip to match a digit (0 for heads, 1 for tails). If you expect a 0, but get 1, you return tail. If you expect 1 and get a 0, you return heads. Otherwise, you keep doing it.

To put it otherwise, just the flips produce a random number in binary. If you expect 0, but get 1, then you have produced a number greated than $a/b$. Otherwise, you have produced a number less than it.

The probability of terminating is $1/2$ at each flip. The expected number of throws is:

$$ \sum_{r=1}^{\infty}\frac{r}{2^r} = \frac{1/2}{1/4} = 2 $$

Don't know when this came from? Well, for $k < 1$:

$$ \sum_{k=0}^{\infty}kx^k = \frac{x}{(1 - x)^2} $$

How do you get that, you ask? You just integrate this sum:

$$ \sum_{k=0}^{\infty}x^k = \frac{1}{1-x} $$

Mindblowing.