# Poker Ace-King hole probabilities

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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?

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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".

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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,

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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]

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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!

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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

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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 :)

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the answer is 576/12900654000
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.

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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.

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agreed....well then that is the probablity that the first three players get that combo

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The prob of one person getting an ace & king is 2 *(4/52 * 4/51). The reason for the 2 is that they can get a Ace-King or a King-Ace. Second player is 2(3/50 * 3/49). Third player is 2(2/48 * 2/47).

With n players, there's C(n, 3) ways that this can happen. So I *think* that if you multiply all those numbers with C(n,3), you'll get the answer.

Although it worries me that this answer could probably go above 1 as n gets really big. Anytime a prob can go over 1, means you probably did something wrong :/

Also you'll probably have to compute the probability for 4 people getting ace-king and subtract it, since this value would be included.

Also a side note about suppossedly "astronomical" probabilities. If you really want to know the prob of something amazing happening, you have to consider all the different things that you would consider "amazing". Maybe if all 3 players got Ace-Queen, it would be equally as amazing. Maybe if 4 players had ace-ace, king-king, queen-queen, jack-jack, that would be amazing too. So you have to add all these together, if you really want to find out how rare one event is.

It's like the case of one person winning the lottery twice. If you look at just one person, the odds that they'll win the lottery twice is ridiculously small. But if you look at *all* the people playing the lottery, for every lottery, the odds that *someone* will win twice is actually not that small. I think someone calculated it at 1/13 per year.

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I'm not an expert by any means, but... all solutions presented here presume that a player should be dealt an ace first, then a king. But IMO it could be the other way around as well, increasing the chance of the first player's first card to 8/52 instead of 4/52.

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If you cube the percentage i gave earlier that should be right.

ace

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I'm going to say that it's about as likely as bidding on a run in pinochle only to have your opponent lay down the same run against you with the two nines of the suit split between the two partners. [wink] I had this happen to me the other day (although I was one of the partners).

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Quote:
Original post by kSquared
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.

Multiply by total number of players choose three then.

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If I undestood question correctly, all players have equal probability to get AK (or KA) regardless to in which order cards are given.

Probability is number of ways you can get 3 players with AKs, divided by number of all possible situations that must be equiprobable.

There is 13*4=52 cards.
Let's enumerate all equiprobable situations.
First card taken from deck can have 52 values. Second card can have 51 values, etc.
For N players, we have
M=product(52-2*N>n>=52, n) = 52!/((52-2N)!)
possible equiprobable cases.

Now, we need to count all cases when we get 3 (AK or KA) pairs.
It is number of ways 3 players can be picked out of N players, N*(N-1)*(N-2)/6 *, multiplied by number of ways 3 players can get AK (or KA) And latter is number of ways they can get AK pairs multiplied by 2^3=8 (each can get either AK or KA). edit: oops, forgot one important thing! there it must be multiplied by number of ways remaining players can pick cards.

First A can have 4 suits, second 3, etc. Same for K . For AKs we have this value squared. I.e. we have (4*3*2)^2=24^2

Finally we get 8*(24^2)*N*(N-1)*(N-2)/6 = 8*24*4*N*(N-1)*(N-2) = 768*N*(N-1)*(N-2) ways you can get 3 AK
So, we get final probability
8*24*4*N*(N-1)*(N-2)/M

so for 5 players it is
edit: wrong there, see below!
8*24*4*(5*4*3)/(52*51 *50*49 *48*47 *46*45 *44*43) ~= 8*10-13
(numbers grouped for clarity)
hm. hmm... hmmmm..... Weird. Looks like typo somewhere in the middle, or overflow in calculator I used. I have some work to do now and then I will go walk outside maybe will find what's wrong. I'd run a simulation but my computer is busy rendering (obligatory ad)my 3D clouds at this moment.

edit: Terribly OOOPS!!!!! Forgot to multiply by one thing(see above)
Probability should be
8*24*4*(5*4*3)*(46*45 *44*43)/(52*51 *50*49 *48*47 *46*45 *44*43) = 8*24*4*(5*4*3)/(52*51 *50*49 *48*47) ~= 3*10-6
i'm is not quite sure it's right now, tho.
edit: oops again.
edit2: and it's probability that 3 or more players get AK cards. (if 4 would, event would be even more amazing so this is also "probability of amazing event")

Probability that exactly 3 players get AK is
8*24*4*(5*4*3)*(46*45 *44*43 - 2*2*44*43)/(52*51 *50*49 *48*47 *46*45 *44*43)
(if i'm not mistaken again)
that is also close to 3*10-6.

[Edited by - Dmytry on August 31, 2005 1:44:27 PM]

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Just happened to look at these forums but..

I do not know the terms used in english propabilities so i'll use Casio's terms.

Any combination of the 3 players out of 5 is:

5C3 (

then gets a bit complicated for me to explain but..

now let's imagine the three hands with 2 cards each. The order of aces and kings does matter.

player1: [ace] [king]
player2: [ace] [king]
player3: [ace] [king]

N(players_all_get_ace) = 4P3 ( = 3! * 4C3 )
N(players_all_get_king) = 4P3 ( = 3! * 4C3 )

all the possible combinations of cards are.. oh god i'm so not able to tell for sure..

N(E) = 52P10 (10 = all the cards dealt to the 5 players)
= 52! / (52!- (52-10)!)
= 52*51*50*49*48*47*46*45*44*43

N(A) = 5C3 * 4P3 * 4P3
P(A) = N(A) / N(E)
= ~ 2.787 * 10^-15

So the actual probability for n-number of players when n > 3

P(A) = (nC3 * 4P3 * 4P3) / ( (52! / (52! - (52-n*2))
= (nC3 * 12) / ((52! / (52! - (52-n*2))

And no. I do not have enough stamina to make that any simpler.

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oh yes. sorry.

2^3 * to the N(A) also becouse the order of the ace and king doesnt matter so it is

= (nC3 * 96) / ((52! / (52! - (52-n*2))

and therefor it's 1.67 * 10^-14
which is in percentages 1.67 * 10^-12 %

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My results matched Dymetry's to eight digits which was all I checked. Given the nature of the problem it isn't all that unlikely for two people to make the same mistake, but they are fairly drastically differant approachs.

I used a recursive routine to run through every scenerio for which three, and only three, players have ace-king hands. I used four categories of hands. These were: ace no king; king no ace; ace-king; and neither ace nor king. The recursive routine took the number of aces and kings remaining, number of ace-king hands remaining, hands played and current chance for the scenerio.

There are ten ways to fill 3, and only three, of the 5 positions with ace-king hands. The first of the remaining slots could be ace no king, king no ace or neither a king nor ace. If the first one is ace no king then the second one can only be king no ace or neither king nor ace and similarly if the first is ace no king. If the first is neither a king nor ace then it can be any of the three for the other. So seven ways to fill the other two slots making 70 scenerios.

I used the following for my chances for each category. Ace no king used 2 * (CardsLeft - AcesLeft - KingsLeft) * AcesLeft / (CardsLeft * (CardsLeft-1)). King no ace used 2 * (CardsLeft - AcesLeft - KingsLeft) * KingsLeft / (CardsLeft * (CardsLeft-1)). Ace-king used 2 * AcesLeft * KingsLeft / (CardsLeft * (CardsLeft-1)). Neither ace nor king used (CardsLeft - AcesLeft - KingsLeft) * (CardsLeft - AcesLeft - KingsLeft - 1) / (CardsLeft * (CardsLeft-1)). No two times because don't care what card it is as long as it is neither an ace nor king.

The routine just tries each hand in each slot. An ace no king hand is only valid if there will be both enough aces and hands left to deal all the ace-king hands remaining and similarly for king no ace hands. Neither ace nor king is valid as long as enough hands remain to deal all the ace-king hands remaining. Ace-king hands are valid as long as all the ace-king hands have not already been dealt.

Finally if hands remain a recursive call is made otherwise the probability for the scenerio is added to the total probability. When a recursive call is made the chance passed is the product of the chance for the scenerio up to the current hand and the chance for the current hand. The initial call is 100%.

It is a brute force approach, but not much of a computation since the recursive function was only called 104 times.

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the odds of getting any of the cards as your first card is 1/52. the odds of getting any of another is 1/51

so the odds of getting 2 pairs is (i think)

4 of same rank in deck so 52/4 = 13

1/13 * 1/13 = 1/169 for one pair

for two pairs = *2 = 1/338 but i could be way off

[Edited by - scheisskopf on September 28, 2005 2:10:26 PM]

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The chance of drawing a pair of a given rank from a full deck is 4/52*3/51=1/221. The chance of drawing a second pair of the same rank once a pair of that rank has already been removed is 2/50*1/49=1/1225. The chance of dealing two two-card hands from a full deck each a pair of a given rank is (4*3*2*1)/(52*51*50*49)=1/221*1/1225=1/270275.

You multiplied the chance of getting a card by the chance of getting a card. You should have multiplied the chance of getting a pair times the chance of getting a pair. You used the wrong chance on the cards so you wouldn't have gotten the right number, but it would have been a whole lot closer. I mainly point that out because you had it right and then you switched. It's hard when you don't have a feel for what the right number should be, but often if you try numbers like 0%, 50% and 100% you can catch error like that. If the chance of one pair is 100% then is would the chance of two pair be 200%?

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