Ehrenfest urn model expectation question

Question

Consider the Ehrenfest urn model in which $M$ molecules are distributed between two urns, and at each time point one of the molecules is chosen at random and is then removed from its urn and placed in the other one. Let $X_n$ denote the number of molecules in urn $1$ after the $n$th. Let $\mu_n = E[X_n]$

show that : $\mu_{n+1} = 1 +(1- 2/M)\mu_n $

part of the solution(which is not mine):

use that: $E[X_{n+1}] = E[E[X_{n+1}\mid X_n]]$

given $X_n$

$$ X_{n+1} = \begin{cases} X_n +1 &\text{with proability} \frac{M-X_n}{M} \\[6pt] X_{n} - 1 & \text{with probability } \frac{X_n}{M} \end{cases} $$

(now comes the part i can't understand)

$$E[X_{n+1}\mid X_n] = X_n + \frac{M-X_n}{M} - \frac{X_n}{M}. $$

why do one need to subtract these probabilities? how do i interpret the subtraction? : $$\frac{M-X_n}{M} - \frac{X_n}{M} $$

Answer 1

You have probability $\dfrac{M-X_n}{M}$ that you will add $1$ and probability $\dfrac{X_n}{M}$ that you will add $-1$.

So the first probability times $1$ plus the second probability times $-1$ is the expected value of the amount you will add.