prob. of pocket aces against pocket kings heads up

[BruceZ]

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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)

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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

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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 .