prob. of pocket aces against pocket kings heads up

[mmix85]

The subject says it all. In a heads up match what are the chances that one player holds pocket aces against the others pocket kings? (It happened to me 2 times today)

[jay_shark]

It isn't too difficult to figure out .

If there are n players , the probability is a little less than
6*n/1326 .

[Spete]

[ QUOTE ]
It isn't too difficult to figure out .

If there are n players , the probability is a little less than
6*n/1326 .

[/ QUOTE ]

It depends on how you choose to look at the question. If it's like: "Wow, kings! I wonder if my opponent has aces, though...", then you'll use the formula above (which, for n=1, unsurprisingly gives .0045: the probability a player gets pocket aces).

On the other hand, if it's like "I'd like to play HU, but seems like it'll always be KK vs. AA and I just can't take that anymore", i.e. if you are only interested to find the probability that YOU get dealt kings while HE gets the rockets, I'd like to think that the math goes:

P("I get kings and he wakes up with rockets") = .0045 x .0045 = 0.00002025 (approx. one in fifty thousand).

By the way, did you manage to fail to suck out both times?

[TrainHardDieHard]

[ QUOTE ]
[ QUOTE ]
It isn't too difficult to figure out .

If there are n players , the probability is a little less than
6*n/1326 .

[/ QUOTE ]

It depends on how you choose to look at the question. If it's like: "Wow, kings! I wonder if my opponent has aces, though...", then you'll use the formula above (which, for n=1, unsurprisingly gives .0045: the probability a player gets pocket aces).

On the other hand, if it's like "I'd like to play HU, but seems like it'll always be KK vs. AA and I just can't take that anymore", i.e. if you are only interested to find the probability that YOU get dealt kings while HE gets the rockets, I'd like to think that the math goes:

P("I get kings and he wakes up with rockets") = .0045 x .0045 = 0.00002025 (approx. one in fifty thousand).

By the way, did you manage to fail to suck out both times?

[/ QUOTE ]

-Is this method correct in solving it? this is exactly how i solved it.

(6/1326)*(6/1326) = .0000205 or .00205%.

[findingneema]

No. The probabilities are not independent, since you remove cards from the deck. In the exact heads up situation, one player gets AA (or KK, doesn't really matter) 6/1326. The other player gets the other hand 6/1225 (see how many fewer hands are possible once two cards are removed?). Because player A could get AA or KK, you have to double it. So,

(6/1326) * (6/1225) * 2 = 0.0000443 = 0.00443%
or ~ 22,559 to 1

i.e., for every 22,560 hands, you will expect one AA vs KK matchup. Same for any particular pp vs any other pp.

Now, of course if you already have KK, the odds of the other guy getting AA is exactly 6/1225 or ~ 203 to 1.

[BruceZ]

[ QUOTE ]
The subject says it all. In a heads up match what are the chances that one player holds pocket aces against the others pocket kings? (It happened to me 2 times today)

[/ QUOTE ]

Here is the exact solution.

[mmix85]

sorry not much of a math guy but what does the c mean and what do the commas signify in the equation you gave?
9*6/C(50,2) - C(9,2)/C(50,4) =~ 21.8-to-1

[BruceZ]

[ QUOTE ]
sorry not much of a math guy but what does the c mean and what do the commas signify in the equation you gave?
9*6/C(50,2) - C(9,2)/C(50,4) =~ 21.8-to-1

[/ QUOTE ]

I recommend jay_shark's answer for you. [img]/images/graemlins/smile.gif[/img] The exact solution is seldom necessary.

Otherwise see C(x,y) or xCy

[jay_shark]

If you hold kings , then there are 50c2 =1225 hand combos remaining .

The probability is 6n/1225.

Here is how I arrived at 1225 .

There are 50 unknown cards remaining since you hold kings . The total number of two-card combos is 50*49/2 .

Think of a tree diagram and under card 1 list all the 50 cards . Then under card 2 there are 49 branches for each card. Can you visualize this ?

But you must remember that for every hand (ie 4c5h) , there is a 5h4c which is the same hand so you must remember to divide 50*49 by 2 which brings your total to 1225 .

If this doesn't make sense , think of an easier example . Think of a deck of 4 cards{1,2,3,4 of hearts} . You're interested in 2 card combos.
12,13,14,21,23,24,31,32,34,41,42,43 . Thats 12 .

But if you look carefully , for every hand of the form xy , there is a hand of the form yx . This is the same hand so you divide 12 by two to give you 6 . Which is just 4*3/2 .

[jodine]

In the last two days (approximately 3k hands) in 6 handed games I have had KK vs. AA 4 times. I would expect this would happen

216 X 39.1 X 4 = in about 33,800 hands. Is my math correct?