Probability of Straight of 4 in 5 Card Poker Hand Having a bit of an experimental/theoretical clash in combinatorics assignment right around now. Attempting to find the probability of getting a Straight of $4$ in a normal $5$-Card Poker Hand. 
At the moment, I've calculated that a Straight of $5$ would be $9C1 \times (4C1)^5$ ($9$ possible Straights of $5$, going from $2-3-4-5-6$ to $10-J-Q-K-A$, as $A-2-3-4-5$ is not possible since Ace is only a high card, $\left(4C1\right)^5$ is to choose a suit for each card). 
However, I've come into some problems with my theoretical for Straight of $4$ being $10C1 \times (4C1)^4 \times 48C1 $($10$ possible Straight of $4$, not counting $A-2-3-4$, this goes from $2-3-4-5$ to $J-Q-K-A$, $\left(4C1\right)^4$ is again the possible suits, and $48C1$ is for the remaining card in the hand). Then I'd subtract the Straight of $5$ value, which comes out to $9216$, from the total of Straight of $4$, $122880$, and I get value of $113664$, then put over the total hands possible $(^{52}C_5)$ is about $4.37\%$. 
However, in my Excel for the experimental probability, over $10,000$ attempts I get a probability of about $2.6\%$. What am I doing wrong? Thanks!
 A: The problem is that there is some double counting happening in your answer.
Consider the hand $2\diamondsuit, 3\diamondsuit, 4 \diamondsuit, 5 \diamondsuit, 2\clubsuit$. There is a straight of $2\diamondsuit, 3\diamondsuit, 4 \diamondsuit, 5 \diamondsuit$ with an extra card of $2\clubsuit$ but one could also see this hand as $2\clubsuit, 3\diamondsuit, 4 \diamondsuit, 5 \diamondsuit$ with an extra card of $2\diamondsuit$. You count both hand seperately, but it is only one possible hand.
A: From $2,3,4,5$ to $J.Q,K,A$ is $10$ possible straights. Each card in the straight has a possibility of $1$ out of $4$ cards so the total number of straights including straight flushes in a four card hand is:
$10\cdot 4^4 = 2560$ 
Excluding straight flushes this would be 
$2560 - 40 = 2520$
In a five card hand, assuming we are looking at just a single straight, there are $32$ other card possibilities for $2,3,4,5$ and $J.Q,K,A$ and $28$ for the other $8$ straights. Here the total is
$2\cdot 32\cdot 4^4 + 8\cdot 28\cdot 4^4 = 73728$
Excluding straight flushes is
$(2\cdot 4^4 - 8)32 + (8\cdot 4^4-32)28 = 72576$
So the probability is $$P = \frac{73728}{\binom{52}{5}} = .02837$$
or
$$P = \frac{72576}{\binom{52}{5}} = .02793$$ excluding straight flushes.
