Odds of AA running into KK at full table?

[arod4276]

could anyone please tell me what the odds of this hapenneing, along with how you figured it out. thank you very much, arod4276

[tylerdurden]

((4/52*3/51) * 10) * ((4/50*3/49) * 9)

Chances of any one player getting AA = 4/52 * 3/51 = 0.0045
Chances of any of ten players getting AA = 0.0045 * 10 = 0.0045

Chances of any of nine remaining players getting KK after AA is already out of the deck = (4/50 * 3/49) * 9 = 0.0041

Chances of both AA and KK in same round = 0.0045 * 0.0041 = 0.00198 = 0.19%

[tylerdurden]

BTW, in that 0.19% of hands, sometimes two different players will have AA in addition to the player with KK, sometimes two different players will have KK in addition to the player with AA, and sometimes two players will have AA and two players will have KK.

[DMBFan23]

I see this as two separate problems.

Problem 1: I have KK (or AA) and I want to know how often someone else has AA (or KK).

Problem 2: What are the odds in general that this will happen to two players?

The math for problem 1 sets up problem 2, so I'll try that one first.

Answer to Problem 1: I'll solve assuming I have KK and I'm curious if someone else has AA (the converse has the same solution)

if I have KK, the odds of one opponent having AA are roughly 1/204 (I'll justify this upon request). To know how often one of multiple opponents has AA, it is easier to find the odds that NO ONE has it. the rest of the time, someone will.

if p(Bullets) = 1/204, then p(Anything Else) = 203/204. so for 9 opponents to all have (Not Bullets), just multiply the odds of each having (Not Bullets). so we get (203/204)^9 = 95.6% chance that no one has it, so that means that 4.3% of the time we have KK, one of our nine opponents will have AA.

Answer to Problem 2: I thought about a ton of ways to solve this, and I'd love to hear Mason's or others' input on this solution.

Conditional Probabilty says: P (Player 1 has KK and Player 2 has AA) = P (KK given AA) * P (AA).

We know that P (AA) = 1/221.
We also know that P(KK given AA) = 1/204 (from problem 1).

so P (player 1 KK, player 2 AA) is 1/45804.

now we need to account for the fact that players 1 and 2 could be any two players at the table. to choose 2 players from 10 (notice that order matters since we specify that player 1 has KK and player 2 has AA) can be done in 10*9 = 90 ways (10 candidates for player 1, 9 remaining candidates for player 2). so it seems that at a 10 player table, the probability of KK running into AA are 90/45804, or about 1/500.

Fire away, let me know if I made a mistake.

[DMBFan23]

dammit pvn beat me to it while I was making my response look pretty. our numbers match though. [img]/images/graemlins/cool.gif[/img]

[tylerdurden]

[img]/images/graemlins/smile.gif[/img] Yeah, this morning I realized he might be asking if he looks at his hole cards and sees AA, what the chances are of someone else having KK, but you covered that possibility as well.

[tylerdurden]

I had some typos in my numbers:

((4/52*3/51) * 10) * ((4/50*3/49) * 9)

Chances of any one player getting AA = 4/52 * 3/51 = 0.0045
Chances of any of ten players getting AA = 0.0045 * 10 = 0.045

Chances of any of nine remaining players getting KK after AA is already out of the deck = (4/50 * 3/49) * 9 = 0.044 (BTW, this is also the chance that someone else has KK once you look at your cards and see that you have anything other than a king.)

Chances of both AA and KK in same round = 0.045 * 0.044 = 0.00198 = 0.198%

[newcool]

Chance that one person has A,A = [ (4/52) * (3/51)] *10 = 4.5%
Chance that another person has K,K = [ (4/52) * (3/51) ] * 9 =4%

Chance that someone has A,A and K,K = .045 * .04 = .0018 or 18/10,000 = roughly 1/555

[tylerdurden]

[ QUOTE ]

Chance that another person has K,K = [ (4/52) * (3/51) ] * 9 =4%

[/ QUOTE ]

Once you've already established that one person has AA, there's only 50 unseen cards left, four of which are Kings, so it's 4/50 * 3/49.

[David Sklansky]

Chances of any one player getting AA = 4/52 * 3/51 = 0.0045
Chances of any of ten players getting AA = 0.0045 * 10 = 0.0045

Just for the record this isn't perfectly accurate. See why?