If player 1 holds AA, the probability that player B has AA is?

[meow_meow]

It's something you see sometimes - two players both holding AA on the same hand. A little unusual, but nothing to write home about.
Anyway, just now I saw it happen on a 400NL 6max table - 3 players all-in preflop: AA vs AA vs KK.
Ok, no biggie - nobody at the table even makes a comment.
Anyway, 15 hands later it happens again. Same two players who had AA before go all-in preflop. Again, both have AA. Just a strange coincidence I'm sure.
In the 110k hands in my PT database, I've never had my aces run into another pair. Weird.


Sorry for the time waster

[kingstalker]

should be 1 in 1225 I think.

[cwes]

kingstalker is of course right.

The two aces remaining in the other 50 cards do not care about you holding the other two aces. So there are still '50 choose 2 = 1225' possible combinations of which one combination (AA) creates the (more or less) desired situation.

[elitegimp]

[ QUOTE ]
kingstalker is of course right.

The two aces remaining in the other 50 cards do not care about you holding the other two aces. So there are still '50 choose 2 = 1225' possible combinations of which one combination (AA) creates the (more or less) desired situation.

[/ QUOTE ]

This is only correct if you are only dealing 2 hands, one of which is already AA. I'm not 100% sure on how to include additional hands, and I've already met my quota for incorrect posts today, but if you search the probability forum for 'AA vs AA' or '2 AAs' or something, you should get approximately a gazillion hits.

edit: for example, for 3 players -- there is a 1/1225 chance that player 2 gets AA if player 1 doesn't have it. There is a 48*47/2450 chance that player 2 is not deal an A at all, in which case there is a 1/1128 chance that player 3 has AA.

So prob that either player 2 or player 3 has the last AA = 1/1225 + (48*47/2450)*(1/1128) = 2/1225.

Since these events can not both happen (players 2 and 3 cannot both have AA if player 1 does), there is no correction to be made.

Anyhoo, it looks like prob(2 players in n-person table both have AA given one player has AA) = (n-1)/1225. So in a 6-person game, once every 245 hands (and about once every 136 hands in a 10-person game).

I'd like to go on record saying I don't see AA vs AA every 136 times online, so party must be rigged (or I made another mistake).

edit 2: D'oh! every 136 given that one person already has AA! I'm a moron.

[cwes]

Hi elitegimp

Your computation also is correct. It is just not the answer to the OP. His question is: "If player 1 holds AA, the probability that player B has AA is?".

You answered to: "If player 1 holds AA, the probability that any other player has AA is?"

[elitegimp]

[ QUOTE ]
Hi elitegimp

Your computation also is correct. It is just not the answer to the OP. His question is: "If player 1 holds AA, the probability that player B has AA is?".

You answered to: "If player 1 holds AA, the probability that any other player has AA is?"

[/ QUOTE ]

I suppose it's a matter of interpretation -- you're probably right in assuming "Player B = specific other player", but I read it as "Player B = any other player." No biggie [img]/images/graemlins/smile.gif[/img]

[buckshot_009]

Well, for 1 player to not have AA is 1224/1225 so for two players to not have AA is (1224/1225 * 1223/1224) and so on. so for a full table against 9 other players, one of them having AA is = ~.007. or .7% (or ~ 1/136). Seems a bit high to me but if you figure that you get AA 1/221 hands this situation occurs 1 time every 30056 hands delt.

[MickeyHoldem]

So to summarize...
Given a 10 player table...

If player A(you) has AA, the odds of specific player B(me) having AA = 1/1225

If player A(you) has AA, the odds of any other player B having AA = 9/1225

The odds of any player A and any other player B both having AA = 10c2 * 6/52c2 * 1/50c2 = ~1 in 6016

Given a PT database of 110,000 hands, I make the probability of you not having your AA run into another AA... at worst(if you play all 10 handed games) to be = .0258 = ~1 in 39

You have been lucky!! ...on average you should have had it happened about 3 times!!!

[motorholdem]

[ QUOTE ]
So to summarize...
Given a 10 player table...

If player A(you) has AA, the odds of specific player B(me) having AA = 1/1225

If player A(you) has AA, the odds of any other player B having AA = 9/1225

The odds of any player A and any other player B both having AA = 10c2 * 6/52c2 * 1/50c2 = ~1 in 6016

Given a PT database of 110,000 hands, I make the probability of you not having your AA run into another AA... at worst(if you play all 10 handed games) to be = .0258 = ~1 in 39

You have been lucky!! ...on average you should have had it happened about 3 times!!!

[/ QUOTE ]

Got this from another site. It's very much related to this question, only 6handed with a KK to go with the AA and AA. What is the probability of this..

Keep in mind that this was a 6 hand table. I am Hero, and folded pre-flop
(thank God). Anyone of the math guys out there have the probablity of this occuring at a 6H table? Button said he had JJ - no reason not to believe him.

Has Anyone seen this before?

Results are in White

Party Poker 10/20 Hold'em (6 max, 6 handed) converter

Preflop: Hero is SB with 8, 7.
UTG calls, MP raises, 1 fold, Button calls, 1 fold, BB 3-bets, UTG caps, MP calls, Button calls, BB calls.

Flop: (16.50 SB) 4, 7, T (4 players)
BB bets, UTG raises, MP 3-bets, Button calls, BB caps, UTG calls, MP calls, Button calls.

Turn: (16.25 BB) T (4 players)
BB bets, UTG calls, MP raises, Button folds, BB calls, UTG calls.

River: (22.25 BB) 3 (3 players)
BB checks, UTG checks, MP bets, BB calls, UTG calls.

Final Pot: 25.25 BB

Results in white below:
BB has Ad Ah (two pair, aces and tens).
UTG has Kc Ks (two pair, kings and tens).
MP has Ac As (two pair, aces and tens).
Outcome: BB wins 12.62 BB. MP wins 12.62 BB.

[MickeyHoldem]

AA vs. AA vs. KK in a 6-handed game....

I make the odds of seeing this 1 in 848272

odds someone gets KK 6-handed = 6/221
odds KK faces 2 AA's 6-handed = 1/23030