How do you calculate this?

[paperchamp]

Sorry if this kind of question gets old but I'd like some help from any of you that are statistic/math geeks. What are the odds that in a 6 handed game two players would be dealt KK and one player would have AA? I just saw this and was trying to crunch the numbers. 220-1 is the odds of getting dealt any pocket pair. However, if someone already has KK wouldn't the odds of someone else getting dealt KK be decreased since there are only 2 kings left? I figure you have to multiply 220 X 220 X ? (? being the chance to be dealt KK when the other 2 kings are out) and then somehow factor in that there are 6 hands dealt. Am I way off?

[Jordan Olsommer]

OK here's what I got:

How I looked at this problem was we've got the six-handed table, and we're going to make three decks of cards: one deck comprised of solely the four kings, one of solely the four aces, and the third deck of the 44 remaining cards plus whichever two aces happen to not get dealt.

Now we pick two people at random of the six at the table to be the lucky (or unlucky, depending on your point of view) people to receive KK. there are 6c2 ways of doing this.

Now we deal out the four kings to the two players - there are 6 ways of doing this (4c2)

Now we pick a lucky player to receive AA. There are four players left, so there are four ways of doing this.

Next, we have to see which two aces he gets out of the four available - again, 4c2 = 6 ways.

now we put the remaining two aces back in our mini-deck of 46 cards and deal out the rest of them to the three remaining poor sods at the table. This can be done (46c2)*(44c2)*(42c2) ways.


OK, so now after all that we have the number of ways we can deal out KK to two players and AA to one.* So now to get the probability that this will happen, we have to divide this number by the total number of ways we can deal 6 two-card holdem hands out of a deck of 52: (52c2)(50c2)(48c2)(46c2)(44c2)(42c2)

And after all that, the answer that I get is approximately .000001, or approximately one in a million.

* note that this probability still includes the slim probability that someone of the remaining three players will be dealt AA as well (making for two KKs and two AAs at the table). Perhaps someone who is more mathematically inclined can correct for this.

btw, thanks for the puzzle /images/graemlins/smile.gif

-EDIT: Actually, now that I think about it, all you'd have to do to correct for the possibility of one of the remaining three getting dealt the other two aces is subtract three from the numerator, since there are only three ways of giving one of the three remaining players the only remaining AA, so it would have a negligible effect on the odds.

[BruceZ]

That's very close. You calculated the KK probability right. Your AA probability actually double counts the cases where 2 players get AA. Here is an exact solution:

The probability that 2 particular players get KK is 1/C(52,4), so the probability that 2 of the 6 players get KK is exactly C(6,2)/C(52,4). The probability that at least 1 player has AA given that 2 have KK is 4*6/C(48,2) - C(4,2)/C(48,4). So all together, the probability of 2 KK and at least 1 AA is:

C(6,2)/C(52,4)*[4*6/C(48,2) - C(4,2)/C(48,4)]

= 849,502-to-1.

Note, it you want 2 KK and exactly 1 AA, this would be:

C(6,2)/C(52,4)*[4*6/C(48,2) - 2*C(4,2)/C(48,4)]

= 850,737-to-1.

See inclusion-exclusion principle (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&Board=&Number=417383&page=&v iew=&sb=5&o=&fpart=) for more details.

[paperchamp]

Thanks! I actually figured it would be even more statistically unlikely then that but that's still a pretty rare occurence. I wonder how much that differs from the odds of having a player with AA, a player with KK, and a player with QQ in a six handed game. I'm guessing that occurs with much more frequency.

[Jordan Olsommer]

[ QUOTE ]
That's very close. You calculated the KK probability right. Your AA probability actually double counts the cases where 2 players get AA.

[/ QUOTE ]

Hmm, I'm not following where the double-count comes in.

We deal the two KK out to two people, then we deal 2 of the four aces out to one other person. Now we have three people left, and 46 cards remaining in the deck. If we didn't care what hands the three got, it would be (46c2)(44c2)(42c2). But we want to exclude the possibility that one of the remaining three get AA as well, so we can calculate that probability and then subtract it.

There are three ways to choose one of the remaining three to receive AA, and (44c2)(42c2) ways to deal hands to the two others after one of them gets the two remaining aces.

so (46c2)(44c2)(42c2) - (3)(44c2)(42c2) is our number of ways of dealing hands to the three remaining players while making sure that none of those three get AA.

But I still get .000001. Is there a way you can explain to me where I did the double-count in terms that I can understand?

[AaronBrown]

Your intuition is correct. The chance of a single player getting a pair is 1/17 (the first card can be anything, the second must be one of 3 cards out of 51). Once a player does that, the chance of a second player getting the same pair is 2/50 * 1/49 = 1/1,225. The chance of a second player getting a different pair is 48/50 * 3/49, which is 72 times as big.

Therefore, it's approximately true that the same scenario (3 pairs out of 6 players) is 72 times as likely if all the pairs are different than if two of the pairs are the same. If three of the pairs are the same, shoot the dealer.

[shday]

[ QUOTE ]
What are the odds that in a 6 handed game two players would be dealt KK and one player would have AA?

[/ QUOTE ]

This actually happened in a big tournament not long ago! (There was also a JJ!).

http://cardplayer.com/poker_magazine/archives/?a_id=14467&m_id=65553

it's hand #3

[blaze666]

[ QUOTE ]


C(6,2)/C(52,4)*[4*6/C(48,2) - 2*C(4,2)/C(48,4)]
= 850,737-to-1.


[/ QUOTE ]


what does 'C' represent?

[Jordan Olsommer]

'C' stands for "combination". C(4,2) means "the number of ways you can choose 2 things from a group of 4 things", discounting the order in which you receive them. For example, if you look down at the ace of spades and the ace of clubs in holdem, you really don't care which one the dealer gave you first /images/graemlins/smile.gif. So C(4,2) can be used to answer the question "how many ways can I be dealt AA, given that I know I'm getting aces?"

[BruceZ]

[ QUOTE ]
Hmm, I'm not following where the double-count comes in.

We deal the two KK out to two people, then we deal 2 of the four aces out to one other person. Now we have three people left, and 46 cards remaining in the deck.

[/ QUOTE ]

That's where the double counting comes in since there are aces in the 46 cards. Suppose player A gets AsAc. Then out out of the 46 cards, player B gets AdAh. Notice that player B could have gotten AdAh first, and then player A gets AsAc from the 46 cards, so this case is counted twice. This is a very common mistake.


[ QUOTE ]
If we didn't care what hands the three got, it would be (46c2)(44c2)(42c2). But we want to exclude the possibility that one of the remaining three get AA as well, so we can calculate that probability and then subtract it.


[/ QUOTE ]

Subtracting off the probability of 2 people having AA is what I'm doing with inclusion-exclusion. Note that your expression for AA (before subtraction) is the same as my term 4*6/C(48,2). This term purposely double counts the cases of 2 AA, and my next term subtracts this off to get the answer for at least 1 AA. Subtracting off 2 times the second term gives the answer for exactly 1 AA.

Note that with inclusion-exclusion, it is not necessary to explicitly consider any hands other than KK or AA.


[ QUOTE ]
There are three ways to choose one of the remaining three to receive AA, and (44c2)(42c2) ways to deal hands to the two others after one of them gets the two remaining aces.

so (46c2)(44c2)(42c2) - (3)(44c2)(42c2) is our number of ways of dealing hands to the three remaining players while making sure that none of those three get AA.

[/ QUOTE ]

Including the 4 players who could have AA times 6 ways to have AA, this should be:

4*6*(46c2)(44c2)(42c2) - (4c2)*6*(44c2)(42c2)

With the correction, this agrees with my answer for at least 1 AA. The first term double counts the cases of 2 AA, and the second term subtracts this off so that these are only counted once. If you subtract off 2 times the second term, then you get my answer for exactly 1 AA.

[Jordan Olsommer]

OK, I don't know what you're saying here because I think you're changing the formula I had at the end of my last post for four players when it's only three that remain, so let me start again.

here is what I'm looking at as what I came up with to find the probability:

http://img296.echo.cx/img296/6605/formula5ng.jpg

here you have:

(# of ways to choose 2 people to get KK) * (# of ways we can give them KK) * (# of ways to choose 1 person to get AA) * (# of ways to give them AA) * [ (# of ways we can give the remaining three people hands) - (# of ways we can give the remaining three people hands that involve one of them getting the other AA)]

all divided by the number of ways we can deal out six hands.

So what I'm asking is where in that is the double-count of the AA. I'm subtracting all the possible ways that someone else can get AA too in the numerator.

[BruceZ]

[ QUOTE ]
So what I'm asking is where in that is the double-count of the AA.

[/ QUOTE ]

I explained that in my first paragraph, so you need to reread that one. Think about it carefully until you understand it. I'll try again here:

[ QUOTE ]
(# of ways to give them AA) * [ (# of ways we can give the remaining three people hands)

[/ QUOTE ]

The same 2 pairs of aces can be dealt by these two terms, just in reverse order. The same aces are dealt to the same players. That's where the double counting is coming from. Again, it's OK to double count as long as you subtract off what you are double counting.


[ QUOTE ]
OK, I don't know what you're saying here because I think you're changing the formula I had at the end of my last post for four players when it's only three

[/ QUOTE ]

I'm including the 4th player who has the ace. That's why I've multiplied 4*6 by the first term, same as you did. You need to recognize that this term double counts the cases where 2 players have AA as I explained in the first paragraph. Then in the second term, you have to consider C(4,2) players who could both have AA, times 6 ways they can have it. You tried to consider only 3 players in the second term, but this was incorrect. You actually have to consider both players who have AA in this term. I was showing how to fix it to get the correct answer.

You didn't need to show your whole formula as I already understood it completely and corrected it to agree with mine. Here is the corrected formula, by your method. Note that only the second term is different from yours:

2 KK + at least 1 AA:

C(6,2)*C(4,2)*[
C(4,1)*C(4,2)*C(46,2)*C(44,2)*C(42,2) -
C(4,2)*6*C(44,2)*C(42,2) ] /
C(52,2)/C(50,2)/C(48,2)/C(46,2)/C(44,2)/C(42,2)

= 849,502-to-1.


2 KK + exactly 1 AA:

C(6,2)*C(4,2)*[
C(4,1)*C(4,2)*C(46,2)*C(44,2)*C(42,2) -
2*C(4,2)*6*C(44,2)*C(42,2) ] /
C(52,2)/C(50,2)/C(48,2)/C(46,2)/C(44,2)/C(42,2)

= 850,737-to-1.

[paperchamp]

[ QUOTE ]

This actually happened in a big tournament not long ago! (There was also a JJ!).

http://cardplayer.com/poker_magazine/archives/?a_id=14467&m_id=65553

it's hand #3

[/ QUOTE ]

Thanks for the link! Kind of ironic that they were exactly six-handed as well.