Basic holdem combination question.

[t_rex]

OK. For your hole cards in holdem, there are 169 distinct combinations. I don't understand how this number is mathematically calculated though.

I get that there are 13 pairs, and 78 suited cards (2 card combinations out of 13 cards - suit is irrelevant, so no need to multipy by 4). What I don't really understand is how the 78 unsuited combinations are calculated. Could someone explain this to me?

Thanks

[BruceZ]

I get that there are 13 pairs, and 78 suited cards (2 card combinations out of 13 cards - suit is irrelevant, so no need to multipy by 4). What I don't really understand is how the 78 unsuited combinations are calculated.

Exactly the same way as the suited cards. For every suited hand there is a corresponding offsuit hand, and visa versa. Of course there are 3 times more offsuit hands (12 vs. 4) but we are counting all 12 of them as the same hand.

[t_rex]

i have seen the formula "13*(4 2)" used to calculate the number of offsuit cards (78) [the numbers in paren indicate a combination of 2 cards out of 4]. this doesn't make sense to me. Can you explain?

Thanks

[StevieG]

(4 2) or C(4 2) is a specific case of the function C(n k) (often read as "n choose k"). If you have n distinct objects, and choose k from that set, and the order does not matter, the function tells you how many different combinations of k items you get:

C(n k) = n!/(k! * (n-k)!)

here ! indicates factorial
n! = n * (n-1) * (n -2) * .. * 1

in other words, multily all thenumbers from 1 to n together

so C(4 2) = 4!/(2! * (4-2)!) = 4*3*2*1/(2*2) or 6.

[StevieG]

So now we try to calculate the possible pairs. There are 13 different ranks that can pair. A pair is 2 suits out of four, so for each rank we have (4 choose 2) 6 pairs avaailable. Hence, 78.

[StevieG]

[ QUOTE ]
So now we try to calculate the possible pairs. There are 13 different ranks that can pair. A pair is 2 suits out of four, so for each rank we have (4 choose 2) 6 pairs avaailable. Hence, 78.

[/ QUOTE ]

D'oh! Lousy brain

You want distinct combinations, not total numbers of pairs.

The number turns out to be 78 but thta's a quirk of teh number. the calcualtion is different.

The number of non-pair combos is
13 (start with any rank)
* 12 (next rank must be different(
/2! (the number of ways to arrange two cards)

which also equals 13*6 or 78.

So total number of starting hands =
13 pairs
+ 78 combos (suited)
+ 78 combos (offsuit)
----
=169

[t_rex]

i follow that, i think, but i've seen the calculation for unsuited combinations (not pairs) given as 13*(4 2) - choose 2 from 4. The result is 78. I see how you've arrived at the answer, but the logic seems slightly different than 13*(4 2). Could you explain?

[BruceZ]

It is just (13,2) = 13*12/2 = 78. That is, there are 13 ways to pick the first rank, and 12 ways to pick the second rank, and then you divide by 2 because otherwise you would count every hand twice since you can get the cards in either order (like J,T or T,J are the same hand). You ignore suits entirely and just deal with 13 numbers. This is the same as 13*(4,2) = 13*4*3/2 = 13*12/2 = 78. I can't think of why anyone would come up with that right off.

[Foo King]

Disregarding individual suits. 13 possible ranks for the first card X 13 possible ranks for the second card = 169 combinations

13 x 13 = 169 /images/graemlins/grin.gif

[Foo King]

To calculate the number of possible hands and taking the suit into account, the calculation would be 52 possible cards for the first card X 51 possible cards for the second card = 2652 individual two card hands are all that there is possible from a standard 52 card deck.

52 X 51 = 2652 /images/graemlins/grin.gif

Hope these two posts help.

[BruceZ]

Is this your attempt to imitate Cyrus' attempt to confuse? If his was this direct, it would be a welcome change, and I would enjoy responding to it even it was just him with a different handle.


Disregarding individual suits. 13 possible ranks for the first card X 13 possible ranks for the second card = 169 combinations

13 x 13 = 169

That is a valid but confusing (to beginners)way to compute the total number of offsuit and suited hands including pairs. 13 cards for the first card, and 13 cards for the second card which can be the same as the first card to make a pair, and since we are not dividing by 2, we are counting each non-pair hand twice, which counts both suited and offsuit cards. The more normal way is to count 13*12/2 = 78 offsuit hands and 78 suited hands, and then add 13 pairs and you get 169 total hands. I suspect, based on your smiley and the second calculation, that you thought you were wrong? Or trying to lure me into arguing incorrectly? Am I fighting monkeys, or only shadows? (Circle of Iron reference).


To calculate the number of possible hands and taking the suit into account, the calculation would be 52 possible cards for the first card X 51 possible cards for the second card = 2652 individual two card hands are all that there is possible from a standard 52 card deck.

52 X 51 = 2652

You have to divide that by 2, or else you will count every hand twice depending on the order you got the two cards. 52*51/2 = 1326.

[Foo King]

Are you the bouncer? /images/graemlins/grin.gif

No, I'm not Cyrus. Nor is this an "attempt to imitate an attempt to confuse". (This qualifies as one of the funniest things I've ever read /images/graemlins/grin.gif)

Just so we understand each other, the smiley faces mean lightheartedness on my part. There are no attempts to attempt any insults or ill will.

Once again, thank you so much for the warm welcome to the forums.

Foo = a metasyntactic variable /images/graemlins/wink.gif
King = it's good to be /images/graemlins/grin.gif

[Cyrus]

My humble advice about Bruce's posts is : take in the math advice but skip the commentary.

In a month's time, if you're still around, you will thank me.

Welcome. /images/graemlins/cool.gif

[ACPlayer]

My advice, listen to cyrus. He is usually right.

[Foo King]

I seem to have gotten off on the wrong foot.

Thanks guys. /images/graemlins/grin.gif

[BruceZ]

Hi Foo King,

If you are not Cyrus (and you know if you are or not) I extend to you the warmest welcome to this forum. You have not started on the wrong foot with me.

My beef is not with you, but with another poster who I thought you might be. I figured only he would use a word such as "metasyntactic", but then I realized that it and "foo" come from the world of UNIX and PERL and the like, and that would be way way out of his realm. Of course, these words may have pervaded other areas as well.

Since you've decided to take my math advice, my math advice to you is to skip anything Cyrus posts on this math forum, and anything on any forum that involves logic for that matter, which would include about everything he posts. He has been thoroughly discredited in these areas, and he is an insidious force on these forums. Not all members understand this, because they have not had the in- depth experience with him that I have had, and the evidence is old, scattered, and of a fairly subtle nature. If you PM me, I can send you the evidence, or you can find it on other topics, but I don't feel it's appropriate to link to it here, and I do not wish to engage in this matter on this forum any more than necessary. It's a funny thing, but if you act to discredit someone, even when you are right, you yourself can lose credibility.

Anyway, sorry you had to join in the middle of this. Enjoy this forum, and I hope you will find it useful and informative.

-Bruce

[southerndog]

List 'em like a multiplication table.. Suited cards across the top, unsuited down the side..

[Foo King]

I appreciate the welcome. Thanks.

I've come to these forums to learn. Many thanks to all of you posters who take the time to teach the rest of us.

And now, let the fur fly. /images/graemlins/grin.gif