The expected value of an Ehrenfest Chain at state $X_{n+1}$ can be shown to be $$E[X_{n+1}] = 1 + (1 - 2/M)E[X_n]$$ Using recursion, one can find the closed form solution of $E[X_n]$ which is done below.
\begin{align*} E[X_n] &= 1 + \left(1 - \frac{2}{M}\right)E[X_{n-1}]\\ &= 1 + \left(\frac{M-2}{M}\right)E[X_{n-1}]\\ &= 1 + \left(\frac{M-2}{M}\right)\left[1 + \left(\frac{M-2}{M}\right)E[X_{n-2}]\right]\\ &= 1 + \frac{M-2}{M} + \left(\frac{M-2}{M}\right)^2E[X_{n-2}]\\ &= 1 + \frac{M-2}{M} + \left(\frac{M-2}{M}\right)^2 + \left(\frac{M-2}{M}\right)^3E[X_{n-3}]\\ &= \sum_{i=0}^{n-1} \left(\frac{M-2}{M}\right)^i + \left(\frac{M-2}{M}\right)^n E[X_0]\\ &= \sum_{i=0}^{n} \left(\frac{M-2}{M}\right)^i - \left(\frac{M-2}{M}\right)^n + \left(\frac{M-2}{M}\right)^n E[X_0]\\ &= \frac{1}{1 - \frac{M-2}{M}} - \left(\frac{M-2}{M}\right)^n + \left(\frac{M-2}{M}\right)^n E[X_0]\\ &= \frac{M}{2} + \left(\frac{M-2}{M}\right)^n\left[E[X_0] - 1\right] \end{align*}
Here's the problem. I'm pretty sure that final term is supposed to be M/2 instead of 1, but that would require a geometric sum to be attached to the $-\left(\dfrac{M-2}{M}\right)^n$ term, but I'm not sure how that would happen. Was there an error made in the algebra?