1

Building a poker program and would like to make sure I have my calculation correct. If a player is holding pocket pair, how does one calculate the probability of flopping a set (one card out of the three will match the player's pair). Ex: Let's say I'm holding pocket Aces.

The way I am attempting to calculate this is by using the binomial formula and finding all 3-card combinations that include exactly one of the two remaining aces and divide that by all possible 3-card combinations (50 Choose 3 = 19,600)

  • So given the Aces I'm already holding, I simply do 2 Choose 1 and calculate the number of combos I can have with the remaining Aces in the deck.
  • Next, for the remaining two cards on the flop that cannot repeat, I do 12 Choose 2
  • Then for each of these two cards there's 4 different possible suits so I do 4 Choose 1 = 4.
  • Multiplying all of these together I get: 2 * 66 * (4 ^ 2) = 2,112

Finally, I get 2,112 / 19,600 = 10.78%

A few places online state that the chances of flopping a set with pocket pairs is around 11.5 - 11.8%. However I cannot find an in-depth explanation of the calculation that would help me modify the calculation I will use in my program.

Can someone please explain to me where I'm going wrong in my calculation? And if there's a simpler way to calculate this probability please explain? Thanks!

Improve this question

2 Answers 2

Reset to default
3

total combinations = 19,600

assuming you have red 2s

flops with 2s but no 2c [2s, 48, 47] = 48c2 = 1128

flops with 2c but no 2s [2c, 48, 47] = 48c2 = 1128

flops with 22x [2c, 2s, 48] = 48

2256/19600 = 11.5102%

flops quads 48/19600 = .2%

combined = 11.755102%

I also saw it represented as the inverse of the probability that there will not be a 2 on the flop

48/50 = .96

47/49 = .95918

46/48 = .95833

= 103776/117600 = .88244897

inverse is .11755 or 11.75%

Improve this answer
2
  • In your first example, 48c2 should be 1128, which in the end would be 11.5% to flop a set. My problem was that I was calculating the probability of flopping ONLY a set, but your method makes more sense, is more simple, and explains where everyone got the 11.5%. Just what I was looking for, thanks!
    – jfviray
    Dec 12, 2017 at 20:50
  • good catch. yeah that makes sense. I was wondering why it was off as the two should give the same result
    – Kenny Hammerlund
    Dec 12, 2017 at 20:52
0

jfviray, Your calculations are correct for flopping set exactly. 11.8% is the probability of flopping set or better which includes quads and full house too.

The above calculation by Kenny shows that

Improve this answer

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.