What is the probability that a person is dealt 13 cards of the same suit out of the 26 cards he is dealt to, from a shuffled 52 cards deck?

Question

I know that to the probability of drawing 13 cards and having them in the same suit from a deck can be found by $\frac{4}{(52!/13!\cdot 39!)}$, but what about from drawing 26 cards? I originally thought $\frac{4}{(52!/26!\cdot 26!)}$ would be the answer but it seems too small/missing something. Then I got $\frac{4\frac{39!}{26!13!}}{(52!26!\cdot 26!)}$, where $4(\frac{39!}{26!13!})$ is the number of ways that at the deck would have 13 cards of least one suit, which seems better, but I am still not sure.

Answer 1

What you have is almost right.

There are $4$ ways to pick a suit, and then there are $\binom{39}{13}$ ways to pick $13$ cards from the rest of the deck, so at first sight there should be $4\binom{39}{13}$ hands of $26$ cards that contain a full suit. The problem is that there are $\binom42$ hands that contain two full suits, and each of these has been counted twice, once for each suit. Thus, the actual number of $26$-card hands containing at least one full suit is

$$4\binom{39}{13}-\binom42=32,489,710,776-6=32,489,710,770\;.$$

The denominator of $\binom{52}{26}$ is about $4.959\times 10^{14}$, so that correction of $-6$ makes a negligible difference in the probability, of course.