# Probability of getting a specific straight in poker (texas hold'em)

For those unfamiliar with texas hold'em, a straight is simply 5 cards in a numbered sequence - for example, {2,3,4,5,6}. We get those cards by choosing the best 5 of 7 cards, two being in your starting hand, and 5 in the table, drawn in succession.

My goal is to calculate the probability of getting a straight, given a specific starting hand (pre-flop).

For example, let's say we get the starting hand {2,3}. How can we calculate the probability of getting {2,3,4,5,6} using the community cards? We know we need the cards {4,5,6} to come up at least once on the community cards.

In order to (hopefully) simplify the calculations, I've decided to consider each card draw independent - let's pretend to be playing with a shuffled stack of a infinite number of standard 52-card decks. We can get the exact same card twice, including those we already have in our hand.

On such a game, having 5 random draws, and considering only the card number, we have a total of 13^5 possible results.

My question is two-fold:

• How to efficiently compute the number of times specific cards ({4,5,6}) come up in the 13^5 possible outcomes. I'm wondering if - since the draws are independent - the probability of getting {4,5,6} is simply (1-(12/13)^5)^3? I'm struggling to come up with something more plausible.

• Is the "infinite number of decks" a reasonable approximation, and is there a way to (still efficiently) compute the probability without it?

--EDIT--

Can this, by chance, be calculated using a binomial(5,1/13) distribution, where the value of the "reverse CDF" is 1 (that is, where the sum of the PDF from 1 to infinity == 1)?

A simulation seems to come up with the number 0.021

• Infinite deck does not make it easier and you can make a straight with A2345 Jan 19, 2017 at 10:09

Let's just do this by straightforward combinations. You need a 4, a 5, a 6, and any 2 other cards. $${{{4 \choose 1}{4 \choose 1}{4 \choose 1}}{38 \choose 2}} \over {50 \choose 5}$$
• Thanks for your clear explanation. For a smaller example of getting 2 specific cards out of 3 possible, in 3 draws, would this mean you'd get (1/3)^2 * 2 * 3 == 2/3? I did a exhaustive listing of the combinations for a smaller example, and the result is 4/9 May 4, 2014 at 15:39