[BruceZ]

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If we assume independence it should be :

(1/5001)^2*5000/5001 + 5000/5001*(1/5001)^2*5000/5001 +
(5000/5001)*(1/5001)^2 = 0.0000001199

ie ppq,qppq,qpp

[/ QUOTE ]

[ QUOTE ]
Mine is the exact answer btw .

[/ QUOTE ]

Exact for what? If you are excluding cases where more than 2 players make aces, then you would need to exclude ppqp and qppp which are now being counted. No other case of 3 or more is currently being counted.