#1
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searched but did not find, possible hands
how many possible starting hand combinations are there, thanks?
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#3
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*N/M* thanks
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#4
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Possible hands in detail + the chances of getting \"no cards\"
169 different starting hands.
There are (52*51)/(2*1) ways of selecting 2 cards from 52, so there are 1326 distinct starting hands. Breakdown: Pairs: 13 different, 6 combinations each Suited non-pair: 78 different, 4 combos each Unsuited pair: 78 different, 12 combos each (13*6)+(78*4)+(78*12) = 1326. Only slightly related to this (but hopefully of some interest), a friend said last night that he had played 250 hands in the Pokerstars WCOOP £500 Hold'em and "never saw a pair bigger than TT [img]/images/graemlins/frown.gif[/img]". He wanted to know the probability of this happening. * 24 of the 1326 hands are AA,KK,QQ or JJ so the chances of getting one of these hands is (24/1326) = 1.8%. So, the chances of not getting one of these hands in 250 deals is (1-(24/1326) to the power 250 = 1.04% A step further; the chance of not getting AA,KK,QQ or JJ in n hands would seem to be (1-(24/1326) to the power n. This gives the chance of not getting one of these hands as: 1 hand: 98.2% 10 hands: 83.3% 25 hands: 63.3% 50 hands: 40.1% 100 hands: 16.1% 150 hands: 6.5% 200 hands: 2.6% 250 hands: 1.0% 300 hands: 0.4% If you include AKs and AKo, then you have a probability of 40/1326 for each hand 1 hand: 97.0% 10 hands: 73.6% 25 hands: 46.5% 50 hands: 21.6% 100 hands: 4.7% 150 hands: 1.0% 200 hands: 0.2% * As it happens, I had AA 3 times, KK at least twice and saw QQ and JJ each at least once but unfortunately 128th out of 548 runners paid £0. [img]/images/graemlins/wink.gif[/img] Perhaps someone with more time/skill than me could produce a formula to work out the chance of getting AA/KK/QQ or JJ x times in y deals??? |
#5
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Re: Possible hands in detail + the chances of getting \"no cards\"
Perhaps someone with more time/skill than me could produce a formula to work out the chance of getting AA/KK/QQ or JJ x times in y deals???
By the binomial distribution, the probability of exactly x in y deals is: C(y,x)*(24/1326)^x*(1 - 24/1326)^(y-x) For at least x in y deals, use binomdist in Excel to compute the probability of at most x-1, and subtract this from 1. |
#6
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Re: Possible hands in detail + the chances of getting \"no cards\"
Thanks Bruce. You are a credit to this forum.
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