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!


2 Answers 2


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.


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 $$P = \frac{73728}{\binom{52}{5}} = .02837$$


$$P = \frac{72576}{\binom{52}{5}} = .02793$$ excluding straight flushes.

  • $\begingroup$ How would this work for Straight of 3 in a five card hand, with the same rules? Would it be 2 x 36 x 4^3 + 9 x 32 x 4^3 and then the last card being anything? Not sure how the last card would work. $\endgroup$
    – OKNEM
    Nov 11, 2018 at 8:57
  • $\begingroup$ No, the last card can't be anything if you are looking at just a "single" 3 card straight. It has to be a card that doesn't make it another 3 card straight or a 4 or 5 card straight. That is, if 4 cards are say, 5,6,7 and 9, then the last card can't be a 4,5,6,7, or 8. $\endgroup$
    – Phil H
    Nov 11, 2018 at 14:02
  • $\begingroup$ Thanks for the help! $\endgroup$
    – OKNEM
    Nov 12, 2018 at 0:59

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