How many bit-flips are required to achieve random distribution? If I have a binary number W bits wide, initially all set to zero, and I repeatedly pick a random bit and toggle it from zero to one or vice versa, how many times would I need to do this to achieve maximum entropy?
I hope I am using the term "maximum entropy" correctly -- what I mean is  a point where the distribution of ones and zeros is as random as possible, and no amount of continued toggling will make the distribution any more random.
 A: After $N=2T$ toggles, the average value of any bit is 
$$\sum_{k=0}^{T-1}\frac{(W-1)^{2T-2k-1}}{W^{2T}}{2T\choose 2k+1}\\
=\frac12-\frac12\left(\frac{W-2}W\right)^{2T}$$
and the second term is roughly $\frac12e^{-2N/W}$.  Use enough toggles to make that difference small enough.
A: Starting with $N$ zero bits and performing $F$ flips.  Let's look at a single bit.
There are $(N-1)^{F-k}\binom{N}{k}$ ways that our bit can be flipped $k$ times.  There are $N^F$ ways the random choices can go, so define $p_k$:
$$p_k = \frac{(N-1)^{F-k}}{N^F}\binom{N}{k}$$
The generating function $G(x)$ for the number of flips is:
$$G(x) = \sum_k p_kx^k = \left(\frac{N-1}{N} + \frac{x}{N}\right)^F$$
We're interested in distinguishing the cases where the bit is flipped an even number of times and an odd number of times.  This is easy to do.  For any polynomial $p(x)$, $\frac{p(x) + p(-x)}{2}$ creates a new polynomial containing only the even terms of $p$, and $\frac{p(x) - p(-x)}{2}$ contains only the odd terms of p.
$$p_\text{even} = \frac{G(1) + G(-1)}{2}\\
p_\text{odd} = \frac{G(1) - G(-1)}{2}$$
Obviously, $G(1) = 1$.  But $G(-1) = \left(\frac{N-2}{N}\right)^F > 0$.  So you will always have:
$$p_\text{even} = \frac12 + \left(\frac{N-2}{N}\right)^F\\
p_\text{odd} = \frac12 - \left(\frac{N-2}{N}\right)^F\\
$$
$$ \lim_{N \rightarrow \infty}\left(\frac{N-2}{N}\right)^F = e^\frac{-2F}{N}$$
