I did the following exercise, but I would like to extend the question to the variance of the variate.
Bob and Bub each has his own coin. Chance of coming up "heads" is $\rho$ for Bob's coin and $\tau$ for Bub's. They flip alternatively, first Bob, then Bub, then Bob again, etc. Let Bob's flip followed by Bub's flip constitute a round, and let $R$ denote the number of rounds until each gets "heads" at least once. For $\rho = 1/3$, $\tau = 2/5$, what is the expectation of $R$?
General answer for the expectation is:
$$\mathbb{E}[R]=\frac{1 + \frac{\rho}{\tau} + \frac{\tau}{\rho} - (\rho + \tau)}{\rho + \tau - \rho \, \tau}$$
This agrees with Monte Carlo simulation I did (with $10^5$ repeats), which approximates expectation and variance (with $\rho = 1/3$, and $\tau = 2/5$) to $3.84647$ and $6.48666$ respectively.
Is anybody able to calculate variance symbolically?