Probalility of a set when holding a pair.

[VivaLaViking]

Greetings,

I am verifying the integrity of the two major online gaming sites and I will share the software I have written and the results of my findings.

I want to verify my calculations first. My question is if you are dealt a pair of down cards what percentage of flops will result in a set without quads or a full house?

My calculation uses combinatorial math written as (x C y) stated as; choose y cards from x choices.

To continue:
To choose the set only tho cards remain, (2 C 1).
Any card must not result in quads, (48 C 1).
Any card must not result in quads or a full house, (44 C 1).
Total number of flops while holding a pair is (50 C 3).

(2 C 1)(48 C 1)(44 C 1) = 4,224
(50 C 3) = 19,600

4,224 / 19,600 = 21.55%

Can anyone verify this result?

TY

[LetYouDown]

I can verify that they are not correct. The probability of hitting at least one of your pair on the flop is significantly less than this...let alone eliminating full houses/quads. I believe you're double counting the 2nd/3rd cards...

Very very quickly I come up with:

2 * (C(48,2) - C(4,2) * 12)
--------------------------- = ~10.78%
C(50,3)

[VivaLaViking]

Thanks for you help. I have over 30,000 trials from these (questionable) sites and before I release the software and findings to this site the importance of being able to support my results can not be overstated.

You replied:

2 * (C(48,2) - C(4,2) * 12)
--------------------------- = ~10.78%
C(50,3)

The validity of subtracting the held pair, C(4,2), will be considered but the terms, C(48,2) * 12, seem to allow for a second flopped pair. In the software the other flop resuls from a held pair are calculated separatelty (quads, two pair, full houses); and I believe the terms C(48,1) * C(44,1) are correct.

Can you explain your calculation? As a note, using your result, ~10.78 in the sample trials will still astound you and most unsuspecting users of these sites.

TY

[LetYouDown]

I'm substracting the C(4,2) * 12. There's 72 ways for the last two cards to pair.

C(48,2) = The ways the last two cards can come, without the 6's.

C(4,2) = The # of ways the remaining cards can pair (combinations of 7,7 for example.)

12 = The remaining ranks that can pair.

C(50,3) should be obvious.

[mindflayer]

the probability of any one of the flop matching your pocket pair is 11.77%

48/50 the first does NOT match 96%
47/49 the second does not match 95.92%
46/48 the third does not match 95.83%

the probability NOne of them match 96% x 95.92% x 95.83% =
88.23%

The odds that at least ONE of them match your pair = 11.77%
This is 8.5:1
The odds that you have EXACTLY trips and not quads or a full house is some small fraction less than this

let you down says 10.78%
which makes the odds of you flopping a full house or quads when you hold a pocket pair at 1%. (seems about right)
let me know if im wrong.

[LetYouDown]

Or you could just summarize it as 1 - C(48,3)/C(50,3) and save yourself a ton of typing. That had nothing to do with the question, however. We skipped that trivial calculation entirely.

In response to your edit...that's essentially what I'm saying (the 1% bit). Quads are also included.

[VivaLaViking]

I can't verify any of your results without some supporting math. The question I am seeking is with a pocket pair what is the expectation that a set will result on the flop and nothing else quads, full, two pair. Thanks for your consideration of my query. Any help is greatly appreciated.

TY

[LetYouDown]

What "math" is missing?

[VivaLaViking]

I'm really not Anal Retentive, I just don't want to be sued. Thank you for all your consideration of my query but the results differ by a full percent ~10.78% vs ~11.76%, this would surely find me in a courtroom. As the flop is being dealt, there must me some way of using combinatorial math directly without intermediate steps (subtraction) as in the presumably incorrect [C(2,1) * C(48,1) * C(44,1)] / C(50,3).

I believe that this site will find the software useful enough to the users that they will make it availible for free download so the ability to support my findings is essential.

I will insert a credit on the appropriate page in the software for any viewer who can help solve this problem. The software presently can work with IGM (Party Poker and Empire Poker) and Poker Stars; all that remais is verifying my math.

TY

[LetYouDown]

The difference of 1% in the math is because we were answering two entirely different questions. I was answering what the odds that you will hit a set on the flop AND NOT have quads or a full house. If you include those two additional possibilities, his number is correct. If you hold a pair, the odds that you will flop AT LEAST one of the two remaining cards of that rank, including full houses and quads are:

1 - C(48,3)/C(50,3)...which translates to 1 - the probability that you won't hit one. It comes out to:

11.755102%

The reason his number is off very slightly is because he rounded and then multiplied.

The "full" number for the question you originally asked is:

2112/19600 ~= 0.10775510204081632653061224489796