This is a problem from BrainStellar
I assumed $E[X_j]$ to be the expected number of letters in the box when $j^{th}$ trial happens.
So the problem reduces to finding $E[X_{j+1}]$(in this case)
$E[X_{j+1}] = E[E[X_{j+1}|X_{j}]]$.
$E[X_{j+1}|X_{j}] = (X_{j} + 1)(X_j/j) + X_j (1- (X_j/j))$.
Which gives,
$E[X_{j+1}|X_{j}] = (X_j/j) + X_j$;
So now we have the recurrence relation,
$E[X_{j+1}] = E[X_j/j + X_j]$ , given that $E[X_1] = 1$
Using this I am getting the answer as $n-1$ which is clearly wrong.
I would appreciate if you could find out the fault in the process.
PS: I know I did not use the property that the lower box has to have less number of letters. I did not know where to put it
Let $\ X_j $ and $\ Y_j\ $ be the number of letters in the two boxes after the $\ j^\text{th}\ $ distribution of letters, and $\ \displaystyle D_j=\left|X_j-Y_j\right|\ $. Then the expected number of letters in the box with no more letters than the other after the $\ j^\text{th}\ $ distribution is $\ E\left[\frac{j+1-D_j}{2}\right]\ $, and \begin{align} \ \\ E\left[D_j\right|\left.D_{j-1}\right]&=\displaystyle\frac{D_{j-1}}{j}\left(D_{j-1}-1\right)+ \left(1-\frac{D_{j-1}}{j}\right)\left(D_{j-1}+1\right)\\ &=1+\frac{(j-2)}{j} D_{j-1} \end{align} for $\ j\ge 2\ $. Therefore \begin{align} \ \\ E\left[D_j\right] &=E\left[E\left[D_j\right|\left.D_{j-1}\right]\right]\\ &=1+\frac{(j-2)}{j} E\left[D_{j-1}\right]\ . \end{align} Starting from $\ j=2\ $, the first few terms of this recursion are $\ 1,\frac{4}{3}, \frac{5}{3}, 2, \dots\ $, suggesting $\ \frac{j+1}{3}\ $ for the general term, which is easily confirmed by induction. The quantity you're looking for is therefore \begin{align} \ \\ E\left[\frac{N-D_{N-1}}{2}\right]&=\frac{N}{3}\ .\\ \ \end{align}