face pairs?

[cyorg]

if i have pocket JJ-KK, what are the odds that someone has a higher pocket pair?

[MickeyHoldem]

Let's assume a full 10 player table for the following calculations:
For KK atleast one person will have a higher pair =
9 * 6/1225 - (9c2 / 50c4) = .043925

For QQ things get more difficult. Let's use a different method:
When you hold 2 cards, how many ways are there to distribute the other 9 hands?
D = 50c18 * 17!! = 622114224555779000000 (excel rounds off)

The problem becomes how many of these will contain a higher pair?
Sometimes there will be 4 higher pocket pairs...
p(iiii) = 3 * 3 * 42c10 * 9!! = 12514622485365

Sometimes there 3 pocket pairs...
p(iii)= 2 * 3 * 6 * 44c12 * 11! = 7892555247436860

Now 2 pocket pairs... more complications... the pairs could be of the same rank or different ranks...
different rank...
p2(ii) = 2 * 3 * 6 * 46c14 * 13!! = 1166970668728160000
same rank...
p1(ii) = 2 * 3 * 46c14 * 13!! = 194495111454694000


Now 1 pocket pair...
p(i) = 2 * 6 * 48c16 * 15!! = 54847621430223700000


Now what to do with these numbers?
There will be exactly two rank against us
R2 = p2(ii) - p(iii) + p(iiii) = 1159140686593150000
and exactly one rank
R1 = p(i) - p1(ii) - 2*R2 = 52334844945582700000

so with QQ the odds are (R1 + R2) / D = .085987402



For JJ there are even more possibilities for combinations of hands that will contain a higher pocket pair, so because I don't like typing...

Just the answer = .127125134

edited for typos

[gaming_mouse]

[ QUOTE ]

When you hold 2 cards, how many ways are there to distribute the other 9 hands?
D = 50c18 * 17!! = 622114224555779000000 (excel rounds off)


[/ QUOTE ]

Not sure this is right. Don't you have to use multi-nomial here?

50!/((2!)^9 * 32!)

I might be misunderstanding your approach, but I think that's right.

[MickeyHoldem]

ok... 50c18 is how to choose 18 cards to go into the remaining 9 hands
17!! is the number of ways to arrange those 18 cards into 9 hands of 2

17!! = 17*15*13*11*9*7*5*3*1


arranging 6 cards into 3 hands... 5!! ways = 5*3*1 = 15
arranging 8 cards into 4 hands... 7!! = 7*5*3*1 = 105
etc

[gaming_mouse]

hmmmm....

this gives a different answer from multi-nomial, tho. why would that be? i think there is something not right about this method, but i havn't really thought about it. i'm fairly sure multi-nomial is correct, tho.

how sure are you about your method? why would multi-nomial not work here?

[MickeyHoldem]

Are you thinking of the general formula?

n! / ((n/2)! * 2^(n/2))

First we choose the 18 cards from 50... then apply the formula

18! / (9! * 2^9) = 34459425

this is the same as 17!! = 17*15*13*11*9*7*5*3*1 = 34459425

This mumber must be multiplied by 50c18

[gaming_mouse]

[ QUOTE ]
Are you thinking of the general formula?

n! / ((n/2)! * 2^(n/2))

First we choose the 18 cards from 50... then apply the formula

18! / (9! * 2^9) = 34459425

this is the same as 17!! = 17*15*13*11*9*7*5*3*1 = 34459425

This mumber must be multiplied by 50c18

[/ QUOTE ]

No. I am thinking of the multi-nomial formula, which I recall as being the way to choose groups if size n1,n2,...nr from a total of N things, where n1+n2+...nr=N. The formula is:

N!/n1!*n2!*...nr!

You can also think of it like this. First choose 2 from 52. Now choose 2 more from the remaining 50. Now choose 2 more from the remaining 48, etc...

(52*51/2) * (50*49/2) * ....

This will simplify in this case to the multi-nomial forumula above. So I'm still pretty sure it is correct. Do you still see some error??

[gaming_mouse]

[ QUOTE ]
This will simplify in this case to the multi-nomial forumula above. So I'm still pretty sure it is correct. Do you still see some error??

[/ QUOTE ]

Mickey,

pzhon pointed out that I am calculating ordered sets of hands, while you are calculating unordered sets.

So your method is correct.

Using multi-nomial could work too, but then we'd have to do all the other calcs on ordered sets as well.

gm