searched but did not find, possible hands

[ccwhoelse?]

how many possible starting hand combinations are there, thanks?

[NoChance]

I think there are 169 possible hands. This chart is an interesting read.

169 hands

[ccwhoelse?]

[FastCards]

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???

[BruceZ]

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.

[FastCards]

Thanks Bruce. You are a credit to this forum.