equation help!!

[prowannabie]

can someone tell me how this works for holding a pocket pair? (disregard the dots, only way i could get them to indent

.2
C =6
.4

[basementproject]

It's a combination. There are 6 combinations of a specific pocket pair that you can be dealt.

I have no idea what the theory is behind this. Ask pzhon for the mathematical proof.

What i do know is how to use it. Say you want to determine the # of ways you can be dealt 6 6. There are four 6s in the deck. You want two of them in your hand. You want to figure out how many different ways you can be dealt 6 6.

On any decent calculator there's a button "nCr". This is the combination function. You're gonna do the following to figure out the ways in which you can be dealt 4 4.

4 nCr 2 = 6

where n = the number of elements to choose from and,
r = the number of elements to choose.

You can do this to find out the number of ways you can be dealt any specific holecards. To figure out AK, for instance:

8 nCr 2 = 28

If you want to know how many holdem hand combinations there are in holdem:

52 nCr 2 = 1326

Just as another example to drive the point home, if you want to see how many possible boards there are in holdem, you calculate as such:

52 nCr 5 = 2598960

To take your original question a step further, let's calculate the probability of being dealt a pocket pair rather than just the # of ways we can get one. The probability of getting 6 6 is going to be:

4 nCr 2 / 52 nCr 2 = 6 / 1326

[jay_shark]

The definition of C(n,r) is n!/(r!*(n-r)!) .

The number of ways of selecting r objects from a set of n objects is C(n,r) . Or equivalently , the number of ways of arranging r 0's and (n-r) 1's is C(n,r) . So we may label each 0 and 1 to distinguish them from each other .

ie , 01_02_03_..._0r and 11_12_13_..._1(n-r) .
So let x denote the number of unique possible arrangements .
We have that x*r!*(n-r)! = n! , solve for x and we get
x = n!/(r!*(n-r)!)

[pzhon]

[ QUOTE ]

You can do this to find out the number of ways you can be dealt any specific holecards. To figure out AK, for instance:

8 nCr 2 = 28

[/ QUOTE ]
No, that counts any hands where both cards are a king or ace, AA and KK in addition to AK. You can count the AK hands by 8C2-4C2-4C2 = 28-6-6=16, or just 4x4 since there are 4 choices for the ace and 4 choices for the king.

[BlueBear]

[ QUOTE ]
On any decent calculator there's a button "nCr". This is the combination function.

[/ QUOTE ]

Another way to calculate this is to use the COMBIN function in Excel. So this would be =COMBIN(4,2)

What does C(4,2) actually mean?

It means there are 6 different combinations when we pick 2 cards from 4 different cards.

Let's say the 4 different cards are

As, Ac, Ad, Ah

How many different ways can we make pocket pairs from these aces? If we were to select sets of 2 cards from the 4 cards above, the different combinations are now

(As,Ac) , (As,Ad) , (As,Ah) , (Ac,Ad) , (Ac,Ah) , (Ad,Ah)

which is a total of 6 combinations.