Odds of Ace King vs pair, 6 handed

Hi,

My question is:

In a 6 handed poker cash game, if you hold Ace King what is the % chance of one of your opponents holding any pocket pair between, pocket 5's and pocket Queens.
I know that when you hold an ace there is a 50% chance, one of your opponents holds an ace as well, in any 6 handed game but what about pairs?
The odds of any players being dealt a pocket pair are 1 in 17 but does this change in this scenario?
Any feedback would be appreciated. Thank you

Quote: PokerPro32

Hi,

My question is:

In a 6 handed poker cash game, if you hold Ace King what is the % chance of one of your opponents holding any pocket pair between, pocket 5's and pocket Queens. <snip>

PokerPro32,

Well, you know 2 of 52 cards: an A and a K. That means the remaining 50 cards contain 3 each of A's and K's, and 4 each of 2's through Q's. In particular, the total number of 5's through Q's is 8*4 = 32.

Therefore, the chance that any given opponent, say Fred, has pocket 5's through Q's is this:

P(5-Q pair, 1 opp) = (32/50)*(3/49) = 96/2450 = 0.03918...

That means that the probability that Fred does NOT have pocket 5's through Q's is

P(no 5-Q pair, 1 opp) = 1 - 96/2450 = 0.96082...

In a 6-hand game, you have 5 opponents, and each of them has the same probability of NOT having pocket 5's through Q's. Therefore, the probability that NONE of your opponents has pocket 5's through Q's is

P(no 5-Q pair, 5 opp) = [P(no 5-Q pair, 1 opp)]^5 = (1 - 96/2450)^5 = (0.96082...)^5 = 0.81886...

Finally, the probability that at least one opponent has pocket 5's through Q's is

P(at least one 5-Q pair, 5 opp) = 1 - 0.81886... = 0.18114..., or about 18.1%.

Hope this helps!

Dog Hand

The analysis (above) is necessarily simplified.

If the first opponent is defined as not have a pair, that affects (reduces) the odds of the next four opponents having a pair. The probabilities of getting a pair for multiple players are coupled.

But Dog Hand's answer is a reasonable approximation.