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Old 02-28-2007, 03:20 PM
jay_shark jay_shark is offline
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Join Date: Sep 2006
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Default Re: prob. of pocket aces against pocket kings heads up

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 .
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