odds of opponent holding pocket Aces given you holdpocket Kings

Hi
the odds of getting one king is 4 in 50
getting the other is 3 in 49
so getting both is 4/50 * 3/49 0,0048979
times the number of players 6
0,02938
1 in 34

[checkmate36]

I always thought that for every 20 times I had KK someone would have AA one time. Good thing I don't raise KK pre-flop. [img]/images/graemlins/laugh.gif[/img]

[KenProspero]

FWIW -- This has happened to me the last two times I got KK in a live game.

First time at Foxwoods, about a month ago -- playing a 4-8 game and my brother kicked the crap out of me with his AA (at least no one at the table thought we were slow-playing). And last night, in the first hand of a home game.

I love it when you start the night in the hole like that .... but hey stuff happens!!!

[ QUOTE ]
[ QUOTE ]
with 50 unknown cards you can have a posiblity of 1225 hands 6 are AA. Against 1 opponent then (6/1225).

Against X number of opponents then aprox:
1 - (1219/1225)^x

If x approx.
2 then .00977
3 then .01462
4 then .01945
6 then .02903
9 then .04323

Matt

[/ QUOTE ]

x*6/1225 is more accurate for any number of players. The error is C(x,2)/C(50,4).

Your independence approximation works better for some problems where several players can have a hand, such as the probability that any pair is out.

[/ QUOTE ]

I would have guessed that the independence assumption would work better than the x*6/1225 that you suggest, but of course I haven't calculated it.

The precise calculation would get quite messy, I think. It would require calculating the probability of having a total of zero, one, or two Aces in four of other five hands (assuming none of these four hands contain AA), and then calculating the probability of AA in the 6th hand, and multiplying by five. Right?

Could you please explain how you calculated the error here, C(x,2)/C(50,4)?

I had AA tonight and the guy with KK hit a set on the flop. We both made heads up play and he sucked out on the river to win the SNG too. Oh well, that's life.

[BruceZ]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
with 50 unknown cards you can have a posiblity of 1225 hands 6 are AA. Against 1 opponent then (6/1225).

Against X number of opponents then aprox:
1 - (1219/1225)^x

If x approx.
2 then .00977
3 then .01462
4 then .01945
6 then .02903
9 then .04323

Matt

[/ QUOTE ]

x*6/1225 is more accurate for any number of players. The error is C(x,2)/C(50,4).

Your independence approximation works better for some problems where several players can have a hand, such as the probability that any pair is out.

[/ QUOTE ]

I would have guessed that the independence assumption would work better than the x*6/1225 that you suggest, but of course I haven't calculated it.

The precise calculation would get quite messy, I think. It would require calculating the probability of having a total of zero, one, or two Aces in four of other five hands (assuming none of these four hands contain AA), and then calculating the probability of AA in the 6th hand, and multiplying by five. Right?

[/ QUOTE ]

Nope, there are only 2 terms in the exact inclusion-exclusion calculation since only at most 2 players can have AA.


[ QUOTE ]
Could you please explain how you calculated the error here, C(x,2)/C(50,4)?

[/ QUOTE ]

The exact calculation is x*6/1225 - C(x,2)/C(50,4). See the post about the inclusion-exclusion principle. The first term double counts all cases where 2 players have AA, and the second term subtracts this off.

Here is a comparison of the methods.

<font class="small">Code:</font><hr /><pre>
#opponents N P(AA) exact N*6/1225 1-(1219/1225)^N


1 0.49% 0.49% 0.49%
2 0.98% 0.98% 0.98%
3 1.47% 1.47% 1.46%
4 1.96% 1.96% 1.94%
5 2.44% 2.45% 2.43%
6 2.93% 2.94% 2.90%
7 3.42% 3.43% 3.38%
8 3.91% 3.92% 3.85%
9 4.39% 4.41% 4.32%
10 4.88% 4.90% 4.79%
</pre><hr />

[Softrock]

This is not correct because you have to subtract the times when two people will get KK. For example the probability of hitting a flush draw with two of your suit on the flop is not 9/47 plus 9/46 but 9/47 plus 9/46 minus 9/47(9/46). Another example, if 6 people each roll a die once is the probablity of someone rolling a 6 100%? (Of course it isn't). By your reasoning we'd multiply 1/6 x 6 and get 100%. I'm not up for doing the calculations but I suspect this would get close to the 44-1 that was quoted.

Thanks for the explanation!

[Vincent Lepore]

The odds of being dealt A,A are 220 to 1. So the ~ odds of one of 5 opponents having A,A is 44 to 1. (220/5). In reality the odds off an opponent having Aces when you have kings might be a lot less than that and as usual it depends on the situation. Suppose for example you are playing in a game with David Sklansky. Suppose you, Mr tight butt, are utg and raise with your K,K and Mr Math type, tighter butt, Sklansky is next to act. The odds of Sklansky having A,A are 99 ro 1. The reason there is a 1 in a hundred chance that he has something else is because there might be a twentythree year old "working" girl sitting next to him that he is trying to impress. So in tha situation he might reraise with A,K. The point is that if you use "static' probability the odds are ~ 44 to 1. But if you consider all factors (conditional) when making this determination the odds might be a lot less.

Vince

[BruceZ]

[ QUOTE ]
The odds of being dealt A,A are 220 to 1. So the ~ odds of one of 5 opponents having A,A is 44 to 1. (220/5).

[/ QUOTE ]

That's incorrect. First of all, you neglected to take into account your own cards. The odds of being dealt AA is indeed 220-to-1, but the probability that one particular opponent gets dealt AA when you don't hold an A is 6/C(50,2) = 6/1225 or about 203-to-1. Then as an approximation, you can multiply this by 5 to compute the odds that one of your 5 opponents holds AA as 5*6/1225 =~ 39.8-to-1. This is a slight approximation because it double counts the cases where 2 opponents hold AA, which has probability C(5,2)/C(50,4), and this must be subtracted off. So the exact answer is 5*6/1225 - C(5,2)/C(50,4) =~39.9-to-1.

You asked about this here and were given this answer.

For more details, see my other posts in this thread which link to the explanation of the inclusion-exclusion principle.