Poker Forum

hi guys, especially maths wizzs.

I'm trying to develop my own strategy, something I can call my own and stick with. so, a few questions.

How many different starting hands are there? like all poker hands, not just playable ones.

And how do I calculate how many different hands I would play if...lets say I only played pairs, two picture cards, or suited connectors.

If someone could give an example of how to calculate that kind of thing it would be great, as I'd love to be able to work that stuff out myself and work out an optimal % of hands played for a particular style of play.

Thanks everyone

Hi Ethan,

found this on Wikipedia. I love this site.

There are (52 × 51)/2 = 1,326 distinct possible combinations of two hole cards from a standard 52-card deck in hold 'em, but since suits have no relative value in poker, many of these hands are identical in value before the flop. For example, A♣J♣ and A♥J♥ are identical, because each is a hand consisting of an ace and a jack of the same suit. There are 169 nonequivalent starting hands in hold 'em (13 pocket pairs, 13 × 12 / 2 = 78 suited hands and 78 unsuited hands; 13 + 78 + 78 = 13 × 13 = 169). These 169 hands are not equally likely (see Poker probability (Texas hold 'em)). Hold 'em hands are sometimes classified as having one of three "shapes":

Pairs, (or "pocket pairs", which consist of two cards of the same rank (e.g. 9♠9♣). One hand in 17 will be a pair, each occurring with individual probability 1/221 (P(pair) = 3/51 = 1/17).
Suited hands, which contain two cards of the same suit (e.g. A♠6♠). Four hands out of 17 will be suited, and each suited configuration occurs with probability 2/663 (P(suited) = 12/51 = 4/17).
Offsuit hands, which contain two cards of a different suit and rank (e.g. K♠J♥). Twelve out of 17 hands will be nonpair, offsuit hands, each of which occurs with probability 2/221 (P(offsuit non-pair) = 3*(13-1)/51 = 12/17).
It is typical to abbreviate suited hands in hold 'em by affixing an "s" to the hand, as well as to abbreviate non-suited hands with an "o" (for offsuit). That is,

Ok, so I got my brain into gear, and I worked out that there are 169 different starting hands. (I think)

so if you just played Pairs and Broadway combo's, that's 25 hands

so... 25/169 = 14.79% of hands played

add some suited connectors....(4 to J) 7 hands

32/169 = 18.93% of hands

maybe AXs and KXs that's 18 more hands

50/169 = 29.59%

what else can I throw in....

suited one gapers? 4-Q that's 6 more

56 hands/169 = 33.14%

I'm just thinking out loud, and I'm not any kind of super poker player or anything. thoughts anyone? You can always take the help of a good

Poker Calculator like Poker Pro 2006 to work it out.

I hope you will just consider a possibly more realistic approach. From your questions concerning how to even calculate the simplest equations a statistical problem of this magnitude will require, I can only assume you are a very longgggg way from the Super genius's with super computers like the brain trust's from M.I.T and It could take a life time to earn enough hands on experience to really have any chance of aquiring the understanding you are talking about....But even more importantly "Why should you even try to out think the Brain trust's and super computers, or spend your life to gather information when it's already there for you to use?" Every one from Doyle to Ivey have books full of more information then anyone could want or need, and the Raw stats are compiled and even easier to access. have you considered studying whats already been done and then trying to move it even farther along?...I wish you luck either way and would love to see what ever you do come up with.

Good luck

Ken

Ironman_kg2002@yahoo.com