NoRiverNoFish Joined: Jan 31, 2019
• Posts: 1
January 31st, 2019 at 10:41:06 PM permalink
This question assumes 9-max Texas Hold'em as the game.

Let's say that I have an ace in my hand, and there's an ace on the flop -- what are the odds of any other player in a 9-max game also having an ace?

If I have an 9 as my kicker, what are the odds of my kicker being beaten?
ThatDonGuy Joined: Jun 22, 2011
• Posts: 4463
February 1st, 2019 at 7:30:06 AM permalink
There are 47 cards besides the flop and your two hole cards. Of these, two are aces, which means 45 are not aces.
There are (47)C(16) possible "sets" of the opponents' 16 hole cards.
Of these, there are (45)C(16) sets of 16 cards that have no aces.
The probability that there are no aces = (45)C(16) / (47)C(16) = (31 x 30) / (47 x 46) = about 0.43
The probability that there is at least one ace in the other hole cards = 1 - the probability that there are none = about 0.57.

Another way to look at it:
Let the other 8 players be A, B, C, D, E, F, G, H
Again, there are 47 cards remaining, of which 45 are not aces
The probability that A's first hole card is not an ace is 45/47
The probability that A's second hole card is not an ace, given his first is not, is 44/46
The probability that B's first hole card is not an ace, given A's two are not, is 43/45
And so on, with 42/44, 41/43, ...
The probability that H's second hole card is not an ace, given that none of A-G's are and H's first is not, is 30/32
The total probability of none of A-H's hole cards being aces is 45/47 x 44/46 x ... x 30/32 = about 0.43, so the probability that at least one is an ace is about 0.57.
gordonm888 Joined: Feb 18, 2015