Conditional probability of sum of $n$ Poisson variables given sum of subset of $n$ variables $P(X_1+X_2+...+X_n=z|X_1+X_3=y)$ I have $n$ Poisson random variables $X_i$ each sampled from an independent Poisson distribution parametrized by $\lambda_i$. I want to compute $P(X_1+X_2+...+X_n=z|X_1+X_3=y)$. The distribution of the unconditional sum would be a Poisson, but I am not sure how to condition it.

The sum of independent Poisson random variables is a Poisson random variable as shown here. So for $Z = X+Y$ where all are Poisson r.v., $P(Z=z)=\frac{\exp^{-\lambda}\lambda^z}{z!}$. Here, $\lambda=\lambda_x+\lambda_y$. If $P(Z)$ was conditioned with $X$, $P(Z|X)$ the solution would be deterministic as the sum of two variables is known and one of the variables is known. As per @whuber and @Xi'an suggested, $P(X_1+X_2+X_3+X_4+...+X_n=z|X_1+X_3=y)=P(X_2+X_4+X_5...+X_n=z-y)$.
So, the conditional can be reduced to an unconditional that is just a sum of r.v. which in turn is just an extension of the attached derivation. Thanks @Xi'an and @whuber, appreciate the help.
