I thought of the following probability problem, which seems to have an answer of 1/e, and wonder if someone has an idea as to how to prove this.


Suppose a man has a bottle of vitamin pills and wishes to take a half pill per day. He selects a pill from the bottle at random. If it is a whole pill he cuts it in half, takes a half pill, and puts the other half back in the bottle. If it is a half pill, he takes that. He continues this process until the bottle is empty. What is the expected maximum number of half pills in the bottle? If the bottle starts with n pills, and M is the expected maximum number of half pills, then M/n appears to tend to 1/e as n tends to infinity. 



Probability Problem Involving e
One way of solving the problem (probably not the easiest) is via the "differential equation method" (see eg the paper by Nick Wormald, entitled "The differential equation method for random graph processes and greedy algorithms", in Lectures on Approximation and Randomized Algorithms, pp. 73-155).


Suppose there are $n$ pills in the beginning. Let
$$
P(\lceil tn\rceil) = \text{number of whole pills after $\lceil tn\rceil$ steps}
$$
and 
$$
H(\lceil tn\rceil) = \text{number of half pills after $\lceil tn\rceil$ steps}
$$


Clearly, $2P(\lceil tn\rceil) + H(tn) = 2n - \lceil tn\rceil$.


Moreover, let $p(t) = \frac1{n}P(\lceil tn\rceil)$, and define $h(t)$ similarly.


Now, given the state $S\_{\lceil tn\rceil}$ of the system after $\lceil tn\rceil$ steps, observe that
$$
\mathbb{E}[P(\lceil tn\rceil+ 1) ~|~ S\_{\lceil tn\rceil}] = P(\lceil tn\rceil) - \frac{P(\lceil tn\rceil)}{P(\lceil tn\rceil) + H(\lceil tn\rceil)}.
$$
This suggests that, up to negligible error terms, $p(t)$ satisfies the ODE
$$
p'(t) = -\frac{p(t)}{p(t) + h(t)}
$$
with initial condition $p(0) = 1$. Similalry, $h(t)$ satisfies
$$
h'(t) = \frac{p(t)}{p(t) + h(t)}-\frac{h(t)}{p(t) + h(t)}
$$
and $h(0) = 0$. With this it shouldn't be too difficult to get the maximum of $h(t)$.


