# of two card starting hands

[SeanSkill]

I was under the impression that there are 1326 possible 2 card starting hands, 13 pairs x 6 ways to make ea = 78, 78 unsuited combinations x 12 ways to make ea = 936, 78 suited combinations x 4 ways to make ea = 312, 78+936+312=1326.

I was reading the Sklansky Chubakov rankings in NLHE T+P and there is a chart that has the number of hands out of 1225 that should fold. I understand that obviously if you have AA there can only be one other combination of AA not 5 others but I am confused as to how you get down to 1225. Obviously you are going to eliminate some of the AX combos but I am having trouble getting the math right. If any of you big brain types that could explain this better to me I would appreciate it.

[New001]

52C2 is 1326, isn't it? I'm really really drunk right now though so i could be sliightly off.

[Absolon]

There are 1326 two card hands when all 52 cards are available to choose 2 from.

But the SC numbers computing how many hands can profitable call a given hands all in is a situation where we need to know how many two card hands there are once two cards have been removed from the deck (the two in the fixed hand). Once one player's hand is fixed, there are 50 cards remaining in the deck to choose two more from. Therefore the number of two card starting hands remaining is 50 choose 2 = 50 * 49 / 2 = 1225.

[SeanSkill]

Thanks Absolon! One more thing if 52 * 51 / 2 = 1326 how do you figure # of starting hands for stud 52 * 51 * 50 / ?

[New001]

52C3 should be 52*51*50/(2*3) right?

[FortunaMaximus]

That good a textbook, eh?

[Absolon]

It depends on whether or not you want to distinguish the upcard. Do you consider (A7)A the same as (AA)7? You shouldn't, as wired aces are more disguised than split aces so, in one sense, a better hand. While they might have the same hot cold equity they definitely don't have the same value. But I digress.

If you distinguish the upcard then the number of choices is (52 choose 2) * 50 = 52 * 51 * 50 / 2 = 66300. Note that this is the same as first choosing the upcard and then choosing the downcards.

If you do not distinguish the upcard then the number of choices is 52 choose 3 = 52 * 51 * 50 / (3 * 2) = 44200.

This is the first glimpse of a phenomenon known as combinatorial explosion. By adding just one extra choice we see an explosion in the number of possible choices.

[SeanSkill]

How would the number of starting hands be effected if you were playing razz where straights and flushes do not count. IE As 2s 3s is no different than As 2d 3c?

[New001]

[ QUOTE ]
It depends on whether or not you want to distinguish the upcard. Do you consider (A7)A the same as (AA)7? You shouldn't, as wired aces are more disguised than split aces so, in one sense, a better hand. While they might have the same hot cold equity they definitely don't have the same value. But I digress.

If you distinguish the upcard then the number of choices is (52 choose 2) * 50 = 52 * 51 * 50 / 2 = 66300. Note that this is the same as first choosing the upcard and then choosing the downcards.

If you do not distinguish the upcard then the number of choices is 52 choose 3 = 52 * 51 * 50 / (3 * 2) = 44200.

This is the first glimpse of a phenomenon known as combinatorial explosion. By adding just one extra choice we see an explosion in the number of possible choices.

[/ QUOTE ]
The things i forget while drinking! good correction.

[Absolon]

That straights don't matter is irrelevant to the number of starting hand combinations.

To count the number of hand combinations, we do the following. (I will assume that we distinguish the upcard; again, what is relevant is that hands like (28)A and (A2)8 are dramatically different). There are three possibilities for a razz starting hand. All three cards are of different rank, two of the cards are of the same rank, all three cards of the same rank.

The number of combinations in which all three cards are different is (13 choose 3) * 3 = 858. That is, we are choosing three different ranks and then one of the ranks is chosen as the upcard.

The number of combinations in which two of the cards are of the same rank is (13 choose 2) * 2 = 156. That is, we are choosing two different ranks and then one of the ranks is chosen as the upcard.

The number of combinations in which all three cards of the same rank is 13.

Therefore the number of starting hands is 858 + 156 + 13 = 1027.

Razz is a "small" game lending it very easily to a thorough mathematical analysis.

[SeanSkill]

I really wish I had payed attention during math now. One more question to figure the # of possible flops in holdem would be 50*49*48 / (3*2) correct? but 117600 / 6 = 19600 is this right?

[JayTee]

[ QUOTE ]
I really wish I had payed attention during math now. One more question to figure the # of possible flops in holdem would be 50*49*48 / (3*2) correct? but 117600 / 6 = 19600 is this right?

[/ QUOTE ]

a cool little google trick (http://www.google.com/search?hl=en&q=50+choose+3&btnG=Google+Search)

[Absolon]

[ QUOTE ]
I really wish I had payed attention during math now. One more question to figure the # of possible flops in holdem would be 50*49*48 / (3*2) correct? but 117600 / 6 = 19600 is this right?

[/ QUOTE ]

Yes, that is the number of possible flops given that you know two of the cards.

[SeanSkill]

Math Tard here, how do you go about eliminating equivalent combinatorial cases in the stud and razz calculations. I Know that with with 2 cards for practical purposes there are only 169 unique hands out of 1326 possibilities. How would this relate to razz and stud hands if we were to distinguish the upcard.

[SeanSkill]

Bump absolon help me here

[Absolon]

[ QUOTE ]
Math Tard here, how do you go about eliminating equivalent combinatorial cases in the stud and razz calculations. I Know that with with 2 cards for practical purposes there are only 169 unique hands out of 1326 possibilities. How would this relate to razz and stud hands if we were to distinguish the upcard.

[/ QUOTE ]

Let's start with Hold'em to make sure that we understand the process.

We will split the hands into three types: paired, unpaired non suited and unpaired suited.

There are thirteen ranks and therefore thirteen different pairs.

For an unpaired non suited hand we must choose two ranks from the thirteen possible ranks and so there are 13 * 12 / 2 = 78 different hands.

Similarly there are 78 unpaired suited hands.

So the number of hands is 13 + 78 * 2 = 169.

For stud we split the hands into a few more types: unpaired rainbow, unpaired two suited with upcard of dominant suit, unpaired two suited with upcard of submissive suit, unpaired monotone, wired pair rainbow, wired pair two tone, split pair rainbow, split pair two tone with upcard of dominant suit, split pair two tone with upcard of submissive suit, rolled up.

For an unpaired rainbow we have to choose three ranks from thirteen so there are 13 * 12 * 11 / (3 * 2) = 286 hands.

There are the same number of unpaired two suited with upcard of dominant suit, unpaired two suited with upcard of submissive suit and unpaired monotone. In total there are 286 * 4 = 1144 unpaired hands.

For a wired pair rainbow we have to choose two ranks from thirteen so there are 13 * 12 / 2 = 78.

There are the same number of wired pair two tone so in total there are 169 wired pair hands.

There are the same number of split pair hands as there are wired pair hands but there are three different types of split pair hands so there are 234 split pair hands in total.

There are thirteen rolled up hands.

In total there are 1560 possible starting hands.

The analysis of razz is similar, just exclude the analysis of suitedness.