Imagine you have n balls in a bag that are colored from 1 to n. At each turn you take one ball uniformly at random from the bag and then a second uniformly at random from the remaining balls which have a *different* color to the first. You then color one ball the color of the other and put them both back in the bag.

What is the expected number of turns before all the balls have the same color if:

1. You always paint the first ball the color of the second? Or…
2. You always paint the second ball the color of the first?

There is a simple martingale argument which shows that if there is a scaling law as n^c, then c=3/4. I don’t think this is in the paper noah links, although that has some other nice ideas and precise results.

Suppose that when there are r red balls and b blue balls, you bet r to win b that you will remove a blue ball next. The total value accumulated from (n,n) to (r,b) does not depend on the path. You must have lost n, n-1, …, n-r+1 and won n, n-1, …, n-b+1. If you end up with k balls, you have accumulated +-k(k+1)/2.

When you make a fair bet of r to win b, this contributes rb to the variance. The total variance is between roughly (1/2)n^3 (winning n times straight) and (2/3)n^3 (alternating wins and losses). If you end up at roughly +-n^c(n^c+1)/2, and the variance is proportional to n^3, then 4c=3, and c=3/4.

Note that there is an easy recurrence so the problem can be solved numerically for any fixed n. Call the case of r red out of t total balls (r,t) and let a(r,t) be the expected number of balls left when we arrive at (0,a) (by convention swap the colors when needed to maintain 2r<=t) then we move from (r,t) to either (r,t-1) or (r-1,t-1) with probabilities r/t and (1-r/t) respectively.

The ODE dr/dt=1-r/t seems relevant. An initial condition of r(2n)=n is no help but we do know that the first step (up to color switch) takes us from (n,2n) to (n-1,2n-1) and with that as an initial condition I get r(t)=(t^2-2n+1)(2t). Then solving r(t)=0 for t yields $\sqrt{2n-1}.$ The numerical data does seem higher than this though.

As I final note, this reminds me of the amazing article The Toilet Paper problem by Donald Knuth October 1984 in the American Mathematical monthly http://www.jstor.org/stable/2322567.
Two kinds of people, one who always removes the majority color and one who always removes the minority color are represented in ratio p:1-p If we start out with n of each color , how many will be left when one color is depleted? I was mildly scandalized that the problem story involves big choosers and little choosers who take a square of toilet paper from one of two available rolls in a certain stall. So that's what they mean by applied math! The math is breathtaking as well and maybe there are ideas to exploit.

Peter, I think that here you go back to the implicit assumption of having a general purpose quantum device. There, modeling the noise (or risks of failure if you wish) can be fairly restricted. But if you want to express, in the language of quantum probability, the way your noise depends on the quantum device that carries the computation then the laws for the noise can be more involved.