Answer
.34
Work Step by Step
We'll first compute the probability that the hand contains no ace. There are $\binom{52}{5} = \frac{52!}{47!5!}$ equally likely poker hands. For the number of hands that contain exactly no aces, there are $\binom{48}{5} = \frac{48!}{43!5!}$ choices. So the probability is
$$\frac{48!}{43!5!} \frac{47!5!}{52!} \approx .66$$
Hence the probability of at least one ace is 1 - .66 = .34.
