Conditional Probability of Geometrically Distributed Random Variables

Question

Let $X_1, X_2, \ldots ,X_m \sim \text{Geom}(p)$ and they are IID. We want to find the probability $$P(X_1=x_1, X_2=x_2, \ldots ,X_m=x_m \; | \; \sum_{i=1}^mX_i=t).$$ I found that $$P(X_1=x_1, X_2=x_2, \ldots ,X_m=x_m \; | \; \sum_{i=1}^mX_i=t)=$$ $$=\frac{P(X_1=x_1, X_2=x_2, \ldots ,X_m=x_m, \sum_{i=1}^mX_i=t)}{P(\sum_{i=1}^mX_i=t)}=$$ $$=\begin{cases} \frac{P(X_1=x_1, X_2=x_2, \ldots ,X_m=x_m)}{P(\sum_{i=1}^mX_i=t)} \ \ \ ,\text{if} \ \sum_{i=1}^mX_i=t \\ 0 \hspace{5.54cm}, \text{otherwise}\end{cases}$$ and $$P(X_1=x_1, X_2=x_2, \ldots ,X_m=x_m)=(1-p)^{-m+\sum_{i=1}^m x_i}p^m=(1-p)^{-m+t}p^m,$$ $$P(\sum_{i=1}^mX_i=t)=p^m(1-p)^t\binom{t+m-1}{m-1}.$$ This implies that the resulting probability does not depend on $x_i$. I tested it out with simulations and it's not true, the distribution is not uniform. Any ideas where the mistake is?

Answer 1

Based on my calculations$$P\left(\sum_{i=1}^mX_i=t\right)=p^m(1-p)^{t+1}\cdot (t+m)\cdot \binom{t+m-1}{m-1}$$ and $$P(X_1=x_1, X_2=x_2, \ldots ,X_m=x_m)=(1-p)^{t}p^m$$