January 31st, 2019 at 10:41:06 PM
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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?

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?

February 1st, 2019 at 7:30:06 AM
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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.

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.

February 1st, 2019 at 12:05:42 PM
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If you have A-9 (off), the board has come A-7-3 rainbow and you have one opponent that is A-X, then the odds at showdown are:

You Win: 38.9%

Villain Wins: 47.7%

You Tie: 13.4%

The reason you are an underdog, is because the opponent (villain) is ahead of you if his 2nd card is a K, Q, J, 10 (16 cards) but also if it is a 7 or 3 ( 6 cards) or another Ace (2 cards.) So, 24 of the remaining 46 cards will put villain ahead of you, with 2 cards to come.

You Win: 38.9%

Villain Wins: 47.7%

You Tie: 13.4%

The reason you are an underdog, is because the opponent (villain) is ahead of you if his 2nd card is a K, Q, J, 10 (16 cards) but also if it is a 7 or 3 ( 6 cards) or another Ace (2 cards.) So, 24 of the remaining 46 cards will put villain ahead of you, with 2 cards to come.

Last edited by: gordonm888 on Feb 1, 2019

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

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