# 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!

## 2 Answers

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

or

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

• 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. Nov 11, 2018 at 8:57
• 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. Nov 11, 2018 at 14:02
• Thanks for the help! Nov 12, 2018 at 0:59