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?

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.

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]

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 :)

Quote:Original post by ncsu121978
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.

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.