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 the13^5
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