# Poker Ace-King hole probabilities

This topic is 4858 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Yesterday I was channel surfing and came across a World Poker Tour event that caught my eye. The televised contests typically show each player's pocket hand next to their name. This was a five-player game; two players had already folded. The remaining players each had hole cards of an Ace and a King, so there were three Ace-King hole card hands. My friend watching with me made a remark along the lines of, "wow, what are the odds of that happening?" I tried to work it out in my head, but it got too complicated. Right now, I'm stuck at this point. I'm not sure how to abstract the idea of "three players out of n players getting AK hands" into math. Any pointers?

##### Share on other sites
It would be better to post exact rules there for people who don't play poker. (my only poker-related experience yet: I'm constantly finding referer-spam pointing to online poker sites in server logs. have no idea why.[grin])
My idea about poker rules is something like
if there is N players,
1: N*5 random cards is given to players. 5 cards to each.
2: players put some money (irrelevant) or leave game. Several rounds of this stuff.
3: players can choose to discard some of their cards. These cards are mixed with remaining cards.
4: players who have m<5 cards receive 5-m random cards.
5: again money-putting buzz, then cards is opened and winner is announced.
(correct me if i'm is wrong, i may be talking about entirely different game at all.)
I'm is not sure in precise details and have no idea what is "Ace-King hole cards".

##### Share on other sites
There are several poker variations. I think what kSquared is describing is where each player has two cards of their own that only they know about, and there are several cards out on the table that belong to everybody, and nobody gets to change what their cards are once they're dealt them.

27

no seriously,

##### Share on other sites
Sorry; I'll be more explicit about the poker rules for those who don't know how it works. In this case it's actually pretty simple.

In Texas Hold 'Em, each player is initially dealt two cards face down from a single deck (52 cards). A deck of cards contains thirteen different ranks (from lowest to highest: 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A) and four suits (hearts, spades, diamonds, clubs). Each (rank, suit) pair is a card in the deck, so cards are identified in this way; e.g., "the four of hearts" is the card with the (4, hearts) tuple.

The initial two cards are called the "hole cards". It's this probability that I'm interested in: what are the chances that out of the n players at the table, there are three players with Ace-King hole cards? (It doesn't matter which three players.)

It actually has nothing to do with poker per se. I could have rephrased the question as "I give n players envelopes containing 2 cards each from a single deck (without replacement). What is the probability that three of the n players have an Ace and a King in their envelopes?"

[Edited by - kSquared on August 31, 2005 9:57:24 AM]

##### Share on other sites
You don't need to know the rules of Poker to figure out the probability. All he's asking is... If 5 players were dealt 2 cards each (from a 52 card deck), what is the probability of 3 players getting an Ace and a King each?

As for the math, I can't help.

Edit: On second thought, I guess knowing the order in which they were dealt probably has an effect as you'd need to know how many cards were left when each one was dealt.You'd also need to know the position of the two that folded that didn't get Ace King. Gah, dammit!

##### Share on other sites
According to this page, the bottom table says that the odds of getting dealt AK suited or unsuited is 1.2% or 82 to 1.

ace

##### Share on other sites
I'm quite rusty on my probability but I think this is right:

First player:
P(A) = 1/13
P(K) = 1/13
P(A&&K) = 1/13 * 1/13 = 1/169

Second player:
P(A) = 3/50
P(K) = 3/50
P(A&&K) = 3/50 * 3/50 = 9/2500

Third player:
P(A) = 1/24
P(K) = 1/24
P(A&&K) = 1/24 * 1/24 = 1/576

Fourth player:
P(A) = 1/46
P(K) = 1/46
P(A&&K) = 1/46 * 1/46 = 1/2116

So for the nth player you would have to say something like:

x = (P(A&&K) || P(A&&!K) || P(!A&&K) || P(!A&&!K) )
f(n) = x && x-1

Hmm. I'm not so sure now. Curse my mind for not co-operating once again. I would love to see the correct solution though :)

##### Share on other sites
you can reduce that at your leisure
here is how i got it

initially there are 4 aces and 4 kings in a deck of 52 cards

we are going on the fact that the first two cards dealt to a player are the "hole cards".
the probability that the first person is dealt an ace first is 4/52. prob 2nd player is dealt an ace is 3/51. prob 3rd player is dealt an ace is 2/50. dont care what cards player 4 and 5 get so prob they get a card is 49/49 and 48/48 which is 1 so we dont even bother to multiply them in with the rest of the probablity. now back to player 1, prob dealt a king is 4/47, player 2 is 3/46, player 3 is 2/45. now the rest of the cards are dealt out but we dont care about there value because we already got the "hole cards" as an ace-king match for 3 players so those probabilities are always 1.

so you multiply out all those probs i mentioned above and you come out with the answer i mentioned. I dont feel like doing the math to reduce it though so go ahead.

##### Share on other sites
Quote:
 Original post by ncsu121978we are going on the fact that the first two cards dealt to a player are the "hole cards". the probability that the first person is dealt an ace first is 4/52. prob 2nd player is dealt an ace is 3/51. prob 3rd player is dealt an ace is 2/50. dont care what cards player 4 and 5 get so prob they get a card is 49/49 and 48/48 which is 1 so we dont even bother to multiply them in with the rest of the probablity.

I don't think that's right. What if the first, fourth, and fifth players were the ones getting the aces? You're computing the probability that the first three players (which singles out specific players) get the AK hand, but what we really want is the probability that any three (but exactly three) players get AK hands.

• ### What is your GameDev Story?

In 2019 we are celebrating 20 years of GameDev.net! Share your GameDev Story with us.

• 11
• 15
• 14
• 46
• 22
• ### Forum Statistics

• Total Topics
634054
• Total Posts
3015274
×