#11
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Re: Texas Holdem Hands Probability
OK here are my workings...
a) Suits = 13 different cards with 4 choose 2 suits = 13 x 4c2 = 78 hands b) Suited connector = 13 combinations with 4 suits = 13 x 4 = 52 hands c) All paint = 16 paint cards choose 2 = 16c2 = 120 hands d) A-x suited = A with 12 other cards by 4 suits = 12 x 4 = 48 hands Total = 298 hands Agree now with all except (c), am confident of these workings! Then the overlaps of A-x suited agree it is 12 overlaps Then the overlaps of Suited connectors with All paint is 12 overlaps (A-K, K-Q & Q-J by the 4 suits) Then the overlaps of Suits with All paint is 24 overlaps (A-A, K-K, Q-Q & J-J by 4 suits choose 2, i.e. 4 x 6 = 24) Therefore we have = 298 - 12 - 12 -24 = 250 hands! |
#12
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Re: Texas Holdem Hands Probability
[ QUOTE ]
c) All paint = 16 paint cards choose 2 = 16c2 = 120 hands Agree now with all except (c), am confident of these workings! [/ QUOTE ] Your calculation includes things like AA, KK, etc, which are already counted in the pairs. |
#13
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Re: Texas Holdem Hands Probability
[ QUOTE ]
(78 + 52 + 86 + 48) - 12 - 8 -12 = 232 [/ QUOTE ] In my orig post I made an arithmetic error. 86 above should be 96. The correct final answer is 242. |
#14
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Re: Texas Holdem Hands Probability
1. any pair -- 6*13 = 78 hands
2. any suited connector -- 13*4 = 52 hands 3. all paint hole cards -- 16C2 = 120 hands 4. A-x suited -- 12*4 = 48 hands 1&2 -- 0 1&3 -- 6*4 = 24 1&4 -- 0 2&3 -- 3*4 = 12 2&4 -- 2*4 = 8 3&4 -- 3*4 = 12 1&2&3 -- 0 1&2&4 -- 0 1&3&4 -- 0 2&3&4 -- 4 1&2&3&4 -- 0 Total = 78 + 52 + 120 + 48 - 24 - 12 - 8 - 12 + 4 = 246. Does this agree with any of your answers? |
#15
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Re: Texas Holdem Hands Probability
Did you subtract AKs one too many times?
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#16
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Re: Texas Holdem Hands Probability
[ QUOTE ]
Did you subtract AKs one too many times? [/ QUOTE ] Yep. I forgot the third term in inclusion-exclusion. Nice catch. |
#17
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Re: Texas Holdem Hands Probability
Does everyone agree?
ANSWER: P(any pair) = 13 different cards with 4 suits to choose from, with only 2 cards available = 13 x 4C2 = 13 x 6 = 78 hands; P(suited connector) = 13 combinations within one suit by 4 suits = 13 x 4 = 52 hands; P(all paint) = there are 6 ways to combine A, K, Q & J, 4 choose 2. For each, there are 16 possible hands = 16 x 4C2 = 16 x 6 = 96 hands; and P(A-x suited) = A goes with 12 other cards in that suit and there is 4 suits = 12 x 4 = 48 hands. These hands are not all mutually exclusive therefore remove: a) A-x suited overlaps with All Paint in 12 situations (A-K, A-Q & A-J by the 4 suits); b) All paint overlap with Suited Connectors in 12 situations (A-K suited, K-Q suited & Q-J suited by the 4 suits); c) A-x suited overlaps with Suited Connects in 8 situations (A-K and A-2 by the 4 suits);and d) All paint overlap with Pairs in 24 situations (A-A, K-K, Q-Q & J-J by 4 suits choosing 2, i.e. 4 x 4C2 = 4 x 6 = 24). Then since we have removed all A-K suited so need to add back A-K by the 4 suits = 1 x 4 = 4 Therefore the total hands are equal to: = 78 + 52 + 120 + 48 - 12 - 12 - 8 - 24 + 4 = 246 hands The total possible combination of hands is 52 choose 2 = 52C2 = 1,326 Therefore you will play 246 in 1,326 hands or 18.195 % of the time. Therefore, you come out betting approximately 2 hands per round. |
#18
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Re: Texas Holdem Hands Probability
Another way to look at it
Probability working is as follows: P(pair) = 1 x 3/51 P(suited connector) = 1 x 2/51 P(A-x suited) = 1/13 x 12/51 x 2 but we remove A-2 and A-K which was covered by suited connector, = 1/13 x 10/51 x 2 P(all paint) = (1/13 x 3/51 x 2) x 6 (for AK, AQ, AJ, KQ, KJ, QJ) = 18.25% |
#19
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Re: Texas Holdem Hands Probability
ANSWER:
P(Pair) = 13 different cards with 4 suits to choose from, with only 2 cards available = 13 x 4C2 = 13 x 6 = 78 hands; P(Suited Connector) = 13 combinations within one suit by 4 suits = 13 x 4 = 52 hands; P(All Paint) = 16 paint cards in the deck, with only 2 to choose from = 16C2 = 120 hands; and P(A-X Suited) = Ace goes with 12 other cards in that suit and there are 4 suits = 12 x 4 = 48 hands. Therefore, a total of: = 78 + 52 + 120 + 48 = 298 hands However, some of these hands are not mutually exclusive therefore we remove: a) A-x Suited overlaps with All Paint in 12 situations (A-K, A-Q & A-J by the 4 suits); b) A-x Suited overlaps with Suited Connectors in 4 situations (A-2 by the 4 suits) (n.b. it may seem you should remove A-K in this scenario too however, doing this will remove A-K entirely in which case you will need to add it back); c) All Paint overlap with Suited Connectors in 12 situations (A-K suited, K-Q suited & Q-J suited by the 4 suits); and d) All Paint overlap with Pairs in 24 situations (A-A, K-K, Q-Q & J-J by 4 suits choosing 2, i.e. 4 x 4C2 = 4 x 6 = 24). Therefore, we remove a total of: = 12 + 4 + 12 + 24 = 52 hands Therefore, the total hands that meet the criteria are: = 298 - 52 = 246 hands The total possible combination of hands is 52 choose 2 = 52C2 = 1,326 Therefore you will play 246 in 1,326 hands or 18.55% of the time. OR Using probabilities... P(Pair) = 52/52 x 3/51 P(Suited Connector) = 52/52 x 2/51 P(All Paint) = (16/52 x 15/51) - (16/52 x 3/51) {to remove Pairs} - (4 suits x 2! order x 3 cards (A-K, K-Q & Q-J) = (4 x 3 x 2) / (52 x 51) i.e. P(All Paint) = (16/52 x 15/51) - (16/52 x 3/51) - (4 x 3 x 2)/(52 x 51) P(A-x Suited) = 1/13 x 12/51 x 2 but we remove A-2 which was covered by suited connector and we remove A-K, A-Q & A-J suited which was covered by All Paint = 1/13 x 8/51 x 2 = 1/17 + 2/51 + 42/663 + 16/663 = 123/663 = 18.55% Therefore, you come out betting approximately 2 hands per round. |
#20
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Re: Texas Holdem Hands Probability
As a side note, many people consider 32 to be a connector, but it is no more connected than 52. In fact, 32 makes fewer straights than 52 as there are fewer one-card straights, just as many potential two-card straights, but more two-card straights by 32 are counterfeited.
http://twodimes.net/h/?z=872198 5[img]/images/graemlins/heart.gif[/img] 2[img]/images/graemlins/heart.gif[/img] beats A[img]/images/graemlins/spade.gif[/img] A[img]/images/graemlins/club.gif[/img] 6.94% with all other hearts, 2s and 5s removed from the deck. The only way to win is with a straight. http://twodimes.net/h/?z=872201 3[img]/images/graemlins/heart.gif[/img] 2[img]/images/graemlins/heart.gif[/img] beats A[img]/images/graemlins/spade.gif[/img] A[img]/images/graemlins/club.gif[/img] 5.19% with all other hearts, 2s and 3s removed from the deck. The only way to win is with a straight. I consider 32 (with 42, KQ, and KJ) to be a two-gapper and 43 (or QJ) to be a one-gap connector. |
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