[BruceZ]

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Bruce - pococurante got it right, there are two A's unaccounted for. By using combinations the probability of the villain pairing his 5 is 50.9%.

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Look at the post you mentioned on the Stud forum. Player 1 has the A [img]/images/graemlins/spade.gif[/img], player 6 has the A [img]/images/graemlins/heart.gif[/img], and hero has the A [img]/images/graemlins/diamond.gif[/img]. Only the A [img]/images/graemlins/club.gif[/img] is unaccounted for.

Your statement that there are 12 bicycle cards unaccounted for is incorrect. There are 6 exposed, and since we are told that opponent's first 3 cards are different, we know that he doesn't hold another 2, which leaves 20 - 6 - 3 = 11 bicycle cards.


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But now comes the second part of my question, is there a shortcut for calculating this probability on the fly rather than typing out the combinations by hand?

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Not if you want to take into account the exact cards that are unaccounted for.