Conditional probability -Basketball How does the number of shots made by a player in a game get distribute if we are given that the number of shots missed equal to the number of shots scored?
data:
P(score)=$0.83$, P(miss)=$0.17$
no. of shots in a game  $Z\sim Poi(6)$
game duration: 48 min
$\underline{\textrm{My work so far:}}$
from what is given Y=no.of misses=no. of hits=X   
so $X\sim Poi(6\cdot 0.83=4.98)$ and $Y\sim Poi(1.02)$
$$P(X|X=Y)=P(X=k,Y=k)/P(X=Y)=   \frac{P(X=k)P(Y=k)}{\sum\limits_{n=0}^{\infty}  P(X=n)\cdot P(Y=n)}$$
The problem is that i dont sure if $P(X=Y)=\sum\limits_{n=0}^{\infty}  P(X=n)\cdot P(Y=n)$.
 A: I assume that "the number of shots made by a player "  refers to the total number of shots, scored and missed:  $Z = X + Y$
Then we want 
$$
\begin{align}
P(Z =z| X = Y)&=\frac{P(X = Y|Z=z) P(Z=z)}{P(X=Y)}\\
\tag{1}\end{align}
$$
Let's compute $P(X = Y|Z=z)$ . First, we note that this is zero if $z$ is odd. 
Now, assuming $z$ is even, $P(X = Y|Z=z)$ corresponds to the event of having, in $z=2t$ tries of a Bernoulli experiment with hit probability $p$, $t$ misses and $t$ scores. The probability of this event is given by a Binomial: $\binom{z}{t} p^t (1-p)^t $
Then the numerator of $(1)$ is
$$ \binom{z}{z/2} (p q)^{z/2} e^{-\lambda} \frac{\lambda^z}{z!} \tag{2}$$
where $p=0.83$, $q=1-p$,  $\lambda = 6$ 
(actually, it's not clear what you mean by $Z∼Poi(6)$ and "game duration=48 minutes" - it would seem that the latter is not necessary - unless the poisson parameter is actually some rate  (scores per time unit) , but then the statement is confusing)
The denominator is just a normalization constant:
$$
\begin{align}
P(X=Y)&= 
\sum_{z \,{\rm even}} \binom{z}{z/2} (p q)^{z/2} e^{-\lambda} \frac{\lambda^z}{z!} \\
&=e^{-\lambda}  \sum_{t=0}^\infty \frac{a^t }{(t!)^2} 
\end{align}
\tag3$$
with $a=p q \sqrt{\lambda}$. I'm not sure if that has a simple closed form expression.
