Thread: number of pairs
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Old June 18th, 2008, 10:35 AM
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Quote:
Originally Posted by robkitch View Post
if u are dealt 5 cards what is the possible way of getting a pair which a pair can be a 2 and a 3

EDIT:
Soroban's post gave me greater insight just now

\frac{ {4 \choose 1} {4 \choose 1}{44 \choose 3} }{{52 \choose 5}} = 0.08153 = 8,15 \ percent \ approx
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Last edited by janvdl; June 18th, 2008 at 10:48 AM.
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