Really simple question

[zigzag]

I have no idea how to figure out this really simple question.

Deal 5 cards out. What percent of the time will there be an ace?

What if you deal 10 cards out?

[bigpooch]

The simplest way to think about this is to determine when
you WILL NOT GET AN ACE.

For five cards, to NOT get an ace, the first card MUST NOT
be an ace and that probability is 48/52. Then, the second
card is not an ace with a probability of 47/51, etc. You
then simply multiply; if you punch it in a calculator, the
odds of NOT getting an ace if you deal five cards is about
0.658842 and for ten cards, it's about 0.413445. The odds
of GETTING AN ACE is just 1-Prob(not getting an ace).

With combinatorics, it's a bit easier: if C(n,r) = the
number of combinations of n things chosen r at a time, the
odds of not getting an ace in k cards is simply
C(48,k)/C(52,k) where k is any integer between 0 and 52
inclusive.