A pill bottle with large and small pills Alright here's the exact question:
A bottle initially contains $48$ large pills and $76$ small pills. Each day a patient randomly chooses one of the pills. If a small pill is chosen, it is eaten. If a large pill is chosen, the pill is broken in two; one part is eaten and the other part is returned to the bottle, and is now considered to be a small pill. Let $X$ be the number of small pills in the bottle after the last large pill has been chosen and its smaller half is returned. Find $\operatorname{E}(X)$. 
Now here is my thought process so far:
$X_i$ = the time at which the $i$th large pill is chosen and then broken. Then
$X = \sum_{i = 1}^{48}X_i$, so $\operatorname{E}(X) = \sum_{i = 1}^{48}\operatorname{E}(X_i)$. From this I gather that $X\sim \operatorname{Geo}(\frac{48-i+1}{76+i-1})$, but I have no idea how I would actually go about computing such a ridiculous amount of geometric random variables.
 A: The process is not well-defined since you didn't specify a distribution. It stands to reason that a patient randomly choosing a pill out of a bottle is more likely to pick a large one than a small one. For this answer, I'll assume that you meant to imply that the pills are chosen independently and with uniform distribution.
Let there be $k$ large and $n$ small pills. A small pill survives until the end if it is chosen after all large pills. For the $n$ original small pills, there are $k$ large pills they need to survive, and the probability for this is $\frac1{k+1}$ by symmetry. By linearity of expectation, this yields a contribution $\frac n{k+1}$ to the expected number of surviving small pills.
Then we also have to take into account the small pills that are produced during the process. The small pill that is produced when the $j$-th to the last large pill is broken needs to survive $j-1$ large pills, with probability $\frac1j$, for a contribution of
$$
\sum_{j=1}^k\frac1j=H_k\;,
$$
where $H_k$ is the $k$-th harmonic number. In total, the expected number of small pills is
$$
\frac n{k+1}+H_k\;,
$$
which in your case with $k=48$ and $n=76$ is
$$
\frac{76}{49}+H_{48}=\frac{18624692152821783046631}{3099044504245996706400}\approx6.01\;.
$$
Your approach is wrong in two respects; $X$ is the number of surviving small pills, not a time; and the sum of times at which the large pills are chosen has no significance. If you wanted to pursue this approach, you'd need the sum of the times it takes to choose the next large pill after the previous large pill. This is the approach taken in the coupon collector's problem, but I doubt that it would prove fruitful in this case.
