odds of opponent holding pocket Aces given you holdpocket Kings

I saw this question on a WPT broadcast.

(In hold em)

In a 6-handed game what are the odds of an opponent holding pocket Aces given you have pocket kings? I Think they said the answer was about 44 to 1.

Could someone explain how to figure the odds. I've tried it several ways I thought were correct but have come up with answers between 33 to 1 and 40 to 1 and have concluded I don't know how to solve this problem and/or I don't remember the answer given on the broadcast.

[LetYouDown]

This problem cannot be solved. Many have tried, all have failed.

[ QUOTE ]
This problem cannot be solved. Many have tried, all have failed.

[/ QUOTE ]
Doesn't some math society or other have a $100,000 prize for solving this problem? OP could make some money if he does manage to crack it.

for 100,000 I'll work on this a little longer.

[tonypaladino]

The probability of someone being dealt AA is 1/220. It is 1/220 regardless of what you have (maybe slightly less given that 2 cards are known that are not A's)

[BruceZ]

[ QUOTE ]
I saw this question on a WPT broadcast.

(In hold em)

In a 6-handed game what are the odds of an opponent holding pocket Aces given you have pocket kings? I Think they said the answer was about 44 to 1.

Could someone explain how to figure the odds. I've tried it several ways I thought were correct but have come up with answers between 33 to 1 and 40 to 1 and have concluded I don't know how to solve this problem and/or I don't remember the answer given on the broadcast.

[/ QUOTE ]


I was going to thank LetYouDown for fielding some of these so I don't always have to, but he let me down this time.

See this post (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=3446464&page=&view=&s b=5&o=&vc=1), among many others.

Of course, there's no answer to this problem. But if there were, it would look like this:

It's 1 in 40.833.

We get this answer from realizing there are 50 cards left in the deck. There are 4 aces in the deck. (4/50)*(3/49)*5 players remaining = 2.449% of the time, or one out of 40.833 times.

This is all theoretical of course. In real life questions like this don't have answers. Take your question, write it out in cursive and put it under your pillow and hope the statistics fairy leaves you a gleaming silver dollar in the morning.

[BruceZ]

[ QUOTE ]
The probability of someone being dealt AA is 1/220. It is 1/220 regardless of what you have (maybe slightly less given that 2 cards are known that are not A's)

[/ QUOTE ]

When you hold KK, the probability of a particular opponent having AA increases to 6/C(50,2) = 6/1225 =~ 1/204.

[tonypaladino]

[ QUOTE ]
[ QUOTE ]
The probability of someone being dealt AA is 1/220. It is 1/220 regardless of what you have (maybe slightly less given that 2 cards are known that are not A's)

[/ QUOTE ]

When you hold KK, the probability of a particular opponent having AA increases to 6/C(50,2) = 6/1225 =~ 1/204.

[/ QUOTE ]

i meant slightly more. whoops

[BruceZ]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
The probability of someone being dealt AA is 1/220. It is 1/220 regardless of what you have (maybe slightly less given that 2 cards are known that are not A's)

[/ QUOTE ]

When you hold KK, the probability of a particular opponent having AA increases to 6/C(50,2) = 6/1225 =~ 1/204.

[/ QUOTE ]

i meant slightly more. whoops

[/ QUOTE ]

BTW, the probability of being dealt AA is 1/221, not 1/220. The odds are 220-1.

[Toddy]

[ QUOTE ]
Of course, there's no answer to this problem. But if there were, it would look like this:

It's 1 in 40.833.

We get this answer from realizing there are 50 cards left in the deck. There are 4 aces in the deck. (4/50)*(3/49)*5 players remaining = 2.449% of the time, or one out of 40.833 times.

This is all theoretical of course. In real life questions like this don't have answers. Take your question, write it out in cursive and put it under your pillow and hope the statistics fairy leaves you a gleaming silver dollar in the morning.

[/ QUOTE ]

On the Broadcast they said on a 9 handed table its 1 in 24.

The question was for a six handed table. Multiple the original .0025 number or whatever it was by eight instead of five and you'll get the right answer.

[wrestler_118]

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

[BruceZ]

[ 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.

[evanski]

This is quite wrong.

[Sciolist]

So say why? I hate this kind of post.

[KenProspero]

[ QUOTE ]
It's 1 in 40.833.

[/ QUOTE ]

Sounds about right. If I were lazy, and didn't want to do the math, I'd run about 5 million random hands and find check the number of times any of my opponents got dealt AA.

I expect that I'd see AA in an opponent about 122752 times in a six handed game, with individual frequency ranging from 22,443 times to 24,778.

This would yield AA showing up about 1 in 40.7325 times. Now, this 1 in 40.7325 would be a touch low because I wouldn't have taken into account those times where 2 players have AA. But overall, I would expect this to be consistent with your calculated answer.

Of course, I would NEVER be so lazy as to run the simulation, but if I WERE so lazy, I think I'd see something like this.

[ QUOTE ]
This is quite wrong.

[/ QUOTE ]

Prove it or shut up.

[LetYouDown]

[ QUOTE ]
I was going to thank LetYouDown for fielding some of these so I don't always have to, but he let me down this time.

[/ QUOTE ]
Hey! I hit the wall with the 300th question asking this =). I was also in the process of moving across the country for the past few weeks...so cut me some slack /images/graemlins/grin.gif.

[AaronS]

Probability of one particular opponent having AA is 6/50c2 = .00489796
Probability of him not having AA is .9951
Probability of no one having AA is .9951^5 = .9757
Probability of one (or two) opponents having AA = 1-.9757

= .024525 = 2.45%

Look good?

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. /images/graemlins/laugh.gif

[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 (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=3446464&page=&view=&s b=5&o=&vc=1) 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 (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=0&amp;Number=1913645&amp;page=) and were given this answer.

For more details, see my other posts in this thread which link to the explanation of the inclusion-exclusion (http://forumserver.twoplustwo.com/showthreaded.php?Cat=0&amp;Number=3446464&amp;page=&amp;vc=1) principle.

Vince thanks for this post. I have never heard of the inclusion-exclusion principle but came up with 39.8x to 1 as the odds I was most confident in.

Dave