AAKKKK 3 handed

[BugsBunny]

In a 3 handed game what's the probability of getting AA KK KK?

I'm pretty sure the following is correct:

AA KK KK has 6 possible combinations

C(52, 6) = 20358520 total possible 3 handed deals

20358520/6 = 3393086.66666667

so odds against are 3393085.66666667 to 1 (3 handed game so only 1 combo of players is possible)

But when I try to confirm this using fractions I keep coming up with much larger numbers.

For example:
1st player will get AA 1/221

2nd player will get KK, when first player has AA
1/221 * 4/50 * 3/49 = 1/45120.8333

3rd player will get KK, when 1st has AA and 2nd has KK
1/45120.8333 * 2/48 * 1/47 = 1/50896300

What adjustment am I forgetting about here? (I can think of one, but it's not enough, so I'm leaving out a piece that I'm pretty sure I'm missing here)

OR

Chance of getting exactly 1 AA dealt in a 3 handed game:
(4/52 * 3/51 * 3) - (2/50 * 1/49 * 2) = 0.0119420076 = 1/73.665034

Once the AA is dealt KK KK will be dealt to the remaining 2 players:
4/50 * 3/49 * 2/48 * 1/47 = 0.00000000043421624 or 1/230300

Combining the 2 we get:
1/73.665034 * 1/230300 = 1/16965057.3302


I hate using fractions, but still want to know where I'm going wrong here.

Thanks

[Cobra]

I came up with a different answer but I am not positive about this. Hopefully someone else will check my math and correct me if it is wrong.

There are 52c6 flops and 4c4*4c2 number of ways to flop aakkk. So the probability of three players seeing a flop with 2 aces and 4 kings is 6/20,358,520 = 1 in 3,393,086

Now we have a flop of AAKKK and the probability that one person gets AA out of this flop is 3*2c2/6c2 = .2

So we just take answer one X answer two and we get the probability of seeing this flop on your next deal.

1 in 16,965,433.33

Cobra

[jay_shark]

AA|KK|KK; KK|AA|KK; KK|KK|AA

There are 6c3 ways to select 3 players from 6 . The probability players 1,2 and 3 have exactly one pair of aces and two pair of kings in some order is [6/52c2*6/50c2*1/48c2]*3 . Now we multiply this number by 6c3 .

[jay_shark]

As a reminder , you have to subtract the times when 4 players have two pairs of aces and kings from the inclusion/exclusion .


[6/52c2*1/50c2*6/48c2*1/46c2]*4c2*6c4

[BugsBunny]

[ QUOTE ]
AA|KK|KK; KK|AA|KK; KK|KK|AA

There are 6c3 ways to select 3 players from 6 . The probability players 1,2 and 3 have exactly one pair of aces and two pair of kings in some order is [6/52c2*6/50c2*1/48c2]*3 . Now we multiply this number by 6c3 .

[/ QUOTE ]

I said it's a 3 handed game. You don't have 6 players. You only have 3.

[jay_shark]

Fair enough . My solution is even better which allows you to solve this for n players . If you read it carefully , you should have no additional questions regarding this problem .

For my original post , the answer should omit the factor of 6c3 . The answer is 3c1*6/52c2*6/50c2*1/48c2 .

Is this more clear to you now ?

[BugsBunny]

OK - I got that and it makes sense. I should have read your original a bit closer. I saw the 6 players in there and jumped to a conclusion.

Why doesn't 6/C(52,6) work in this case?

C(52,6) is the total possible combinations in a 3 handed deal. Ooops - never mind. That's not true is it? Looking at it this way A2 34 56 is the same as A3 45 62, but that isn't the case because each hand is an individual item. So I'm under counting the number of possible deals - by a lot.
Sometimes I'm an idiot [img]/images/graemlins/frown.gif[/img]


Thanks.

[BugsBunny]

That answer is correct. Get the same thing using the formulas Jay talks about. And also doing it this way:

Chance of 3 players getting KKKK:
1/C(52,4) * C(3,2) = 3/270725 = 1/90241.6666666667

Once you have KKKK the remaining player will get AA:
4/48 * 3/47

So:
3/270725 * 4/48 * 3/47 = 1/16965433.3

So the answer checks now whichever way I do it.