I have seen this question.
However, mine is slightly different.
Let $p,q\in (0,1), p+q=1$. Assume that random variables $X$ and $Y$ are independent and both have the geometric distribution with parameters $p,q$, i.e.,
$$P(X=k)=P(Y=k)=q^kp\,,$$
where $k\in\mathbb{N}_0=\{0,1,2,...\}$.
I would like to show that
$$P(X=k\mid X+Y=n)=\frac{1}{n+1}$$
The questions I have seen here seem to arrive to the conclusion that it's equal to $\frac{1}{n-1}$. How do I arrive to $\frac{1}{n+1}$?
We have: $$P(X=k\mid X+Y=n)=\frac{P((X=k) \cap (X+Y=n))}{P(X+Y=n)}$$ Now: $$\begin{align} P((X=k) \cap (X+Y=n))~&=~P((X=k) \cap (Y=n-k))\\&=P(X=k)\, P(Y=n-k)\\&=q^k p\;q^{n-k}p\\&=q^n p^2\end{align} $$ whereas $$\begin{align} P(X+Y=n)&=\sum_{i=0}^n P((X=i) \cap (Y=n-i))\\&=\sum_{i=0}^n P(X=i) \, P(Y=n-i)\\&=\sum_{i=0}^n q^i p\, q^{n-i} p\\&=\sum_{i=0}^n q^n p^2\\&= (n+1)\,q^n p^2 \end{align}$$ and you only have to divide to get the result.
By convolution, \begin{align*} P(X+Y=n)=\sum_{k=0}^n P(X=k)P(Y=n-k) \\ =\sum_{k=0}^n (q^kp)(q^{n-k}p) \\ =\sum_{k=0}^n q^np^2 \\ =q^np^2(n+1) \end{align*}
Thus, \begin{align*} P(X=k|X+Y=n)=\frac{P(X=k)P(Y=n-k)}{P(X+Y=n)} \\ =\frac{q^np^2}{q^np^2(n+1)}=\frac{1}{n+1} \end{align*}
Your confusion may be stemming from there being two widely used definitions for the Geometric PMF (see the Wikipedia for more info).
The question you linked is doing the same as you, but the OP is considering an alternative definition for the Geometric distribution (i.e., $k$ is starting at $1$ and not at $0$ as you defined). Hence, following the same procedure you will get the result you want.
This is not unique to this calculus: the expected value also depends on the definition you are considering (although the variance no). Check the properties at https://en.wikipedia.org/wiki/Geometric_distribution.
