$X_i$ iid with $P(X_i=j)=p_j$, $j=1, \dots, m$. $\sum_{j=1}^m p_j = 1$. Define $N = \min\{n>0:X_n = X_0\}$, compute $E(N)$.
I have two solutions, but different answers:
Solution 1
$E(N) = E(N\mid X_1=X_0)P(X_1=X_0) + E(N\mid X_1\neq X_0)P(X_1\neq X_0)$
So $x = 1\cdot y + (1+x)(1-y)$, where $x = E(N), y = P(X_1=X_0) = \sum_{j=1}^m p_j^2$, and thus
$E(N) = 1/y = 1/\sum_{j=1}^m p_j^2$.
Solution 2
$E(N) = \sum_{j=1}^m E(N\mid X_0=j)P(X_0=j) = \sum_{j=1}^m 1/p_j \times p_j = m$
I cannot see where is wrong. Any help?