Well... I think this is right:
3 cards = 5/52 x 4/51 x 3/50 = 1 in 2,210
This calculation would be right if we drew only 3 card, but in fact we draw 5 cards and need 3 winning cards from this five.
If we take this into consideration and also put back the drawn cards into the deck as lnspin suggested, here is the formula for the probability of getting exactly n winning cards from the 5:
Probability for n winning cards = combin(5,n) x (5/52)[SUP]n[/SUP] x (47/52)[SUP]5-n[/SUP]
where combin(5,n) = 5! / (n! x (5-n)!)
So:
combin (5,0) = 1
combin (5,1) = 5
combin (5,2) = 10
combin (5,3) = 10
combin (5,4) = 5
combin (5,5) = 1
For example, combin (5,3) = 10 means that if we have 3 winning cards of 5, there are 10 possible combinations, here they are ('x' means winning and '.' means losing cards):
xxx..
xx.x.
xx..x
x.xx.
x.x.x
x..xx
.xxx.
.xx.x
.x.xx
..xxx
So, here are the formulas for the given number of winning cards:
Probability for 0 winning cards = 1 x (47/52)[SUP]5[/SUP]
Probability for 1 winning cards = 5 x (5/52) x (47/52)[SUP]4[/SUP]
Probability for 2 winning cards = 10 x (5/52)[SUP]2[/SUP] x (47/52)[SUP]3[/SUP]
Probability for 3 winning cards = 10 x (5/52)[SUP]3[/SUP] x (47/52)[SUP]2[/SUP]
Probability for 4 winning cards = 5 x (5/52)[SUP]4[/SUP] x (47/52)
Probability for 5 winning cards = 1 x (5/52)[SUP]5[/SUP]
And the results:
Probability for 0 winning cards = 60.32156% (or 1 in 1.65778)
Probability for 1 winning cards = 32.08594% (or 1 in 3.11663)
Probability for 2 winning cards = 6.826795% (or 1 in 14.6482)
Probability for 3 winning cards = 0.726255% (or 1 in 137.693)
Probability for 4 winning cards = 0.038631% (or 1 in 2588.62)
Probability for 5 winning cards = 0.000822% (or 1 in 121665)
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Total: 100%