Probability that a 5-card poker hand is a straight

Question

I'm trying to find the probability that a 5-card poker hand contains 5 numbers in a numerical sequence.

For the first card, there are 52 options. For the second, there are 4 on either side of the first, so you have $\frac{8}{51}$. For the third, there are 3 on either side of the second, so you have $\frac{6}{50}$.

The end result is:

$$P(Straight)= 52\cdot{8\choose 51}\cdot{6\choose50}\cdot{4\choose49}\cdot{2\choose48}=\frac{19968}{5997600}=0.0033$$

Have I done this correctly? What makes me doubtful is the exact answer I've seen evaluates to 0.0039. Any help is appreciated.

Answer 1

"Straight" in poker is generally taken to exclude "straight flush" and royal flush"

However, in the body of the question, you have written "5 numbers in a numerical sequence." Taking this to be your intention, i.e. 1-2-3-4-5 through 9-10-11-12-13, the computation, ignoring various rules of poker, would just be

$4^5\times 9/\binom{52}{13}$

Note

I have deliberately used numbers 1-13 for illustration to avoid detailed rules for poker, eg under high rules an ace could count as high or low (changing the possible runs of five numbers to $10$), and the question of whether royal flush and straight flush are to be included or not.
I trust you can add these niceties of poker rules, having grasped the basic concept.

Answer 2

There are ${52\choose 5}=2,598,960$ total possible hands.

How many hands contain a straight (including straight flushes)? Well, first off you'd have to have at least one card card A-10 (assuming Aces can be considered a 1), which can occur in ${10\choose 1}{4\choose 1}=40$ ways. Then we need to pick one of each of the successive ranks - there are ${4\choose 1}=4$ ways to do this with each rank, so that's $4^4$ total arrangements. All told then, there are ${10\choose 1}{4\choose1}*{4\choose 1}^4={10\choose 1}{4\choose 1}^5=10,240$ such arrangements.

Probability of a straight: $\frac{10,240}{2,598,960}\approx 0.0039400.$ This agrees with the probabilities the OP has seen.

If you wanted to exclude straight flushes, you'd just need to calculate how many of those are possible and factor that in. As we see above, there are ${10\choose 1}{4\choose 1}^5$ possible straights, so then there should just be ${10\choose 1}{4\choose1}^1=40$ possible straight flushes (ie - instead of each card choosing its suit, we just choose one suit and all of the cards must be that). So we'd say that there are only 10,240-40=10,200 possible straights excluding straight flushes (note that a royal flush is a special type of straight flush, and thus is factored in here).

Thus the probability of a straight that isn't a straight flush would be $\frac{10,200}{2,598,960}\approx 0.0039246$. This agrees with the value given in: https://en.wikipedia.org/wiki/Poker_probability