Find the Conditional Expectation $\mathbb{E}[X=k|Y=n]$ Suppose $X$ and $Y$ are stochastic variables with a simultaneous distribution
$$\mathbb{P}(X=k, Y=n) = \frac{e^{-1}2^{n-k}}{3^nk!(n-k)!},\ \ \ \ \text{for 0 $\leq$ k $\leq$ n and n $\geq$ 0 }$$
Determine the conditional expectation $\mathbb{E}[X=k|Y=n]$.
Now my reasoning is the following:
$$\mathbb{E}[X=k|Y=n]=\sum_{k=0}^nk\cdot \mathbb{P}(X=k|Y=n),$$
Where
$$\mathbb{P}(X=k|Y=n)=\frac{\mathbb{P}(X=k,Y=n)}{\mathbb{P}(Y=n)}$$
Since I can rewrite 
$$\mathbb{P}(X=k, Y=n) = \frac{e^{-1}2^{n-k}}{3^nk!(n-k)!}=\frac{1}{e(n!)}\sum^{n}_{k=0}\binom{n}{k}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k,$$
Using the Binomium of Newton, I've found
$$\mathbb{P}(Y=n)=\frac{1}{e(n!)}$$
Which brings me back to the conditional probability:
$$\mathbb{E}[X=k|Y=n]=\sum_{k=0}^nk\cdot \mathbb{P}(X=k|Y=n) = \sum_{k=0}^nk\ \binom{n}{k}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k$$
Now I'm stuck. Is my reasoning correct and how can I solve this last equation?

Technically speaking, you should write
$$\mathbb{E}[X | Y = n] = \sum_{k=0}^{n} k \mathbb{P}[X=k|Y=n].$$
Now, you can do a similar 'trick' as for the calculation of $\mathbb{P}(Y = n)$:
$$k \binom{n}{k} = k \frac{n!}{k!(n-k)!} = k \frac{n(n-1)!}{k(k-1)!(n-k)!} =n \frac{(n-1)!}{(k-1)!(n-k)!} = n \binom{n-1}{k-1}$$
