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?


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

  • $\begingroup$ Infinite deck does not make it easier and you can make a straight with A2345 $\endgroup$ – paparazzo Jan 19 '17 at 10:09

Let's just do this by straightforward combinations. You need a 4, a 5, a 6, and any 2 other cards. $${{{4 \choose 1}{4 \choose 1}{4 \choose 1}}{38 \choose 2}} \over {50 \choose 5}$$

Evaluating this yields 0.0212, agreeing with your simulated value.

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  • $\begingroup$ 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 $\endgroup$ – goncalopp May 4 '14 at 15:39
  • $\begingroup$ You are correct - my method does some double counting. I will need to rethink the answer as it will be too high. $\endgroup$ – soakley May 4 '14 at 19:00
  • $\begingroup$ The result is indeed similar, but I'm having some trouble understanding the expression. Where do the numbers 50 and 4 come from (I assume 38==50-4*3)? How would you expand this to the smaller example? $\endgroup$ – goncalopp May 7 '14 at 19:32
  • $\begingroup$ I went back to the finite deck. Your hole cards reduce the deck from 52 cards to 50. The three 4's represent the four 4s, 5s, and 6s. So the solution as given now really only counts the straights that have one 4, one 5, one 6, and 2 other cards. It doesn't count the straights with two 4's, for example. $\endgroup$ – soakley May 8 '14 at 16:52
  • $\begingroup$ I don't think it necessarily extends to the smaller example since you are not using cards. In it, for the cases starting with 0, there is a (1/3)(1/3)(1/3) chance for 001, (1/3)(1/3) chance for those starting with 01, and (1/3)(1/3)(1/3) chance for 021. That adds up to 5/27. Same logic applies for cases starting with 1 so that is also 5/27. Then for cases starting with 2 it's (1/3)(2/3)(1/3), or 2/27. Total is then 12/27, or 4/9 as you got from counting. $\endgroup$ – soakley May 9 '14 at 17:30

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