Re: odds of opponent holding pocket Aces given you holdpocket Kings
 

[ 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.

Re: odds of opponent holding pocket Aces given you holdpocket Kings
 

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.

Re: odds of opponent holding pocket Aces given you holdpocket Kings
 

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

Re: odds of opponent holding pocket Aces given you holdpocket Kings
 

[ 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.

Re: odds of opponent holding pocket Aces given you holdpocket Kings
 

This is quite wrong.

Re: odds of opponent holding pocket Aces given you holdpocket Kings
 

So say why? I hate this kind of post.

Re: odds of opponent holding pocket Aces given you holdpocket Kings
 

[ 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.

Re: odds of opponent holding pocket Aces given you holdpocket Kings
 

[ QUOTE ]
This is quite wrong.

[/ QUOTE ]

Prove it or shut up.

Re: odds of opponent holding pocket Aces given you holdpocket Kings
 

[ 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 [img]/images/graemlins/grin.gif[/img].

Re: odds of opponent holding pocket Aces given you holdpocket Kings
 

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?