A Poisson Count Divisible by Three

$X \sim \mathrm{Poisson}(\lambda)$. For $n = 2$ the probability that $X$ is even comes from averaging the generating function at $z = 1$ and $z = -1$: $P(2 \mid X) = \frac{E[1^X] + E[(-1)^X]}{2} = \frac{1 + e^{-2\lambda}}{2}.$

Use the same idea with the cube roots of unity to compute $P(3 \mid X)$ when $\lambda = 1$.

Answer format: round to 4 decimal places.