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#1
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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 |
#2
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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) |
#3
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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 |
#4
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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. |
#5
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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. |
#6
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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. |
#7
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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 |
#8
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lol, it's 7-1 that u'll improve to trips
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