I'm trying to find the UMVUE of $e^{-2\lambda}$ with  

$X\_1, X\_2,\ldots,X\_n \sim \operatorname{Poisson}(\lambda)$ being independent. 

So since $T(X) := \sum\_{i=1}^n X\_i$ is a complete sufficient statistic for the Poisson distribution, and also $W(X):= \mathbf{I}(X\_1 + X\_2 = 0)$ is an unbiased estimator of $e^{-2\lambda}$, then by Lehmann\_Scheffe,  

$$\tau(T)= \mathbb{E}(W(T)\mid T(X))
$$
is the UMVUE of $e^{-2\lambda}$. 

I'm having trouble computing $\tau$ so I would like to see if this is correct:  

First, to find the expectation, I need the PDF.  

$$\mathbb{P}(W(T)=s\mid T(X)=t) = \frac{\mathbb{P}(W(T)=s \ \cap \ T(X) = t)}{\mathbb{P}(T(X) = t)} \\ = \frac{\mathbb{P}(\mathbf{I}(X\_1 + X\_2 = 0) = s \ \cap \ X\_1 + X\_2 + \cdots + X\_2 = t)}{\mathbb{P}(X\_1 + X\_2 + \cdots + X\_n = t)}\\ = \frac{\mathbb{P}(\mathbf{I}(X\_3 + X\_4 + \cdots + X\_n = t)=s)}{\mathbb{P}(X\_1 + X\_2 + \cdots + X\_n = t)} 
$$
The denominator is just $\frac{e^{-n\lambda}(n\lambda)^t}{t!}$.  

When $s=1$, we have 
$$\mathbb{P}(\mathrm{I}(X\_3+X\_4+\cdots+X\_n=t) =1) = \frac{e^{-(n-2)\lambda}((n-2)\lambda)^t}{t!} 
$$
So 
$$\mathbb{P}(\mathrm{I}(X\_3+X\_4+\cdots+X\_n=t) =0) = 1 - \frac{e^{-(n-2)\lambda}((n-2)\lambda)^t}{t!} 
$$
After this, I'm unsure of how to find the expectation. Is my approach correct?