In the Ehrenfest model, let $X_n$ denotes the number of balls in the left urn. And there are $N$ balls total. When we calculate $P(X_{n+1}=i+1|X_n=i, X_{n-1}=i_{n-1},...,X_0=i_0)$, why don't we take account of the probability of choosing the other urn?
I mean, why $P(X_{n+1}=i+1|X_n=i, X_{n-1}=i_{n-1},...,X_0=i_0)=(N-i)/N$ instead of $1/2(N-i)$ because the probability of choosing the right urn is 1/2 and the probability of selecting one ball from the right urn is $1/(N-i)$ so $P(X_{n+1}=i+1|X_n=i, X_{n-1}=i_{n-1},...,X_0=i_0)=(1/2)\cdot1/(N-i)$. I know this question may be a kind of silly but I really want to know the answer since this is the only model that I find confusing in stochastic process class.
At time $n$, one ball will change to the other urn. Assume you choose a ball randomly, i.e., every ball is chosen with probability $1/N$. There are $N-i$ balls in the right urn, so the probability of choosing a ball in the right urn is $(N-i)/N$. This gives the result.
You propose something different: You first choose the urn with equal probability, and then move an arbitrary ball in this urn to the other side. In any case, if you choose the right urn with probability 1/2, then the left urn will always have one ball more in the next time step (unless the right urn is empty). So you would get a probability of exactly 1/2.