Probability that hand contains Ace and King of at least one suit?

Compute the probability that a hand of 13 cards (drawn randomly from a standard deck of 52) contains both the ace and the king from at least one suit.

I think I would use the inclusion exclusion principle here, but I'm not sure how to start. Would it be something like: P(Ace+King of suit 1) + P(Ace+King of suit 2) + P(Ace+King of suit 3) + P(Ace+King of suit 4) - ... so on.?

• Do the ace and king have to be from the same suit, or can they be from different suits? – Brian Tung May 8 '15 at 0:52

If $p$ is the probability that you get at least one ace and one king, consider $$1-p = p_{\lnot A}+p_{\lnot K}-p_{\lnot AK}$$ where $p_{\lnot A}$ is that probability that you didn't get any ace, $p_{\lnot K}$ is the probability that you got no king, and $p_{\lnot AK}$ is the probability that you got no ace and no king.
So $$p_{\lnot A} = p_{\lnot K}=\frac{\binom{48}{13}}{\binom{52}{13}}$$ and $$p_{\lnot AK}=\frac{\binom{44}{13}}{\binom{52}{13}}$$
So $$p = \frac{\binom{52}{13} - 2\binom{48}{13} + \binom{44}{13}}{\binom{52}{13}}$$