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$Old$ 06-18-2008, 10:17 AM

robkitch $robkitch is offline$

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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

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$Old$ 06-18-2008, 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; 06-18-2008 at 10:48 AM.

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$Old$ 06-18-2008, 10:41 AM

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Hello, robkitch!

Your English is not quite readable . . .

Quote:

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

I'll take a guess at what you meant . . .

We want five cards: one Deuce, one Trey, and three Others.

To get one Deuce, there are: ${4\choose1} = 4$ ways.

To get one Trey, there are: ${4\choose1} = 4$ ways.

To get three Others, there are: ${44\choose3} = 13,244$ ways.

Therefore, there are: . $4 \times 4 \times 13,244 \:=\:\boxed{211,904\text{ ways}}$

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