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

[ihaveapigyo]

lol, it's 7-1 that u'll improve to trips