Quick Question About AA vs. KK

[MickeyHoldem]

From a search of previous posts... The odds of finding yourself up against AA when holding KK vs. 7 opponents was given as...

(7 * (6 / (50 choose 2))) - ((7 choose 2) * (6 / (50 choose 2)) * (1 / (48 choose 2))) = 0.0341945289

which I believe simplifies for n opp. to...

n*6/1225 - (n choose 2)/(50 choose 4)

Question: Is this the odds for being up against atleast 1 AA or exactly 1 AA.

[gaming_mouse]

That calculation is for at least 1.

[MickeyHoldem]

Thank-you!!

So exactly 1 would be...

n*6/1225 - 2*(n choose 2)/(50 choose 4)

[gaming_mouse]

[ QUOTE ]
Thank-you!!

So exactly 1 would be...

n*6/1225 - 2*(n choose 2)/(50 choose 4)

[/ QUOTE ]

Assuming your simplification is correct, yes.

[BruceZ]

[ QUOTE ]
Thank-you!!

So exactly 1 would be...

n*6/1225 - 2*(n choose 2)/(50 choose 4)

[/ QUOTE ]

Exactly, and your simplification is correct.