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.

[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?

[BarronVangorToth]

[ QUOTE ]
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?

[/ QUOTE ]


Don't question the math, David -- ever.

Barron Vangor Toth
www.BarronVangorToth.com
"Hates It When 9 out of 10 have Pocket Aces but he's the 1 out of 10"

[Prevaricator]

If player 1 doesn't get dealt AA, then player 2 is slightly more likely to get dealt AA.

[BruceZ]

[ QUOTE ]
If player 1 doesn't get dealt AA, then player 2 is slightly more likely to get dealt AA.

[/ QUOTE ]

You are saying that the events of players being dealt AA are not independent. That is true, but it is NOT the reason why multiplying by 10 is not exact. The reason is that more than one player can have AA. In other words, the events of players being dealt AA are not mutually exclusive. When we multiply by 10, we double count all the hands where 2 players hold AA.

To be exact, we have to subtract off the small probability that 2 players hold AA. The probability that two specific players hold AA is 1/C(52,4). There are C(10,2) = 45 ways to pick 2 players out of 10. Now notice that these 45 pairs are mutually exclusive since only 1 pair can have the 4 aces. Therefore, the probability that 2 players have AA is exactly 45/C(52,4). The probability of at least 1 player having AA is exactly 10*6/C(52,2) - 45/C(52,4) =~ 4.51%.

Your answer would correctly explain why the approximation 1 - (220/221)^10 = 4.43% is not exact. Multiplying these probabilities requires that the hands be independent.

It is possible for events to be mutually exclusive without being independent. For example, if you hold AA, the probability that a specific opponent holds AA with you is 1/1225, and the probability that one of your 9 opponents holds AA is exactly 9/1225. This is exact because only 1 opponent can have AA. The events of different opponents holding AA are mutually exclusive, even though they are not independent.

[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.