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poker deck is 13 rank (2-Ace) and 4 suits for a total of 52 cards

Draw 3 random cards what is the chance of exactly 1 pair (same rank)?

using combination

$$\frac { \binom{4}{2} \binom{13}{1} \binom{48}{1}} {\binom{52}{3}} = 0.169411765 $$

using fraction 1 - not
first card is what it is 1/1
second card there are 48/51 that don't match the first
third card there are 44/50 that don't match the first or second
1 - (48/51 x 44/50) = 0.171764706

Why are those two numbers different?
Which is correct?
Are neither correct?

I can get the second number with match fractions
first card is what is it 1/1
second card match 3/51
second card does not match but third matches one of the first two 48/51 x 6/50
3/31 + 48/51 x 6/50 = 0.171764706

To get combination to match the other numbers I need to add exactly 52 to the numerator

if look at it as (p is pair o is other)
p p o
p o p
o p p
3 x (1 x 3/51 x 48/50) = 0.169411765
that matches combination
I think I believe combination but I think all 4 should match

See answer from MikeP
The trip is the 52 difference in the numerator
$$ \binom{4}{3} \binom{13}{1} = 52 $$

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The difference is the probability of a 3-of-a-kind event $\frac{3}{51}*\frac{2}{50}\approx0.0024$

0.1694+.0024=0.1718

The 1-"no pair" case includes this probability

The "pair" only case does not include this probabiliy

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  • $\begingroup$ And that is the 52 in the numerator of combination $\endgroup$ – paparazzo Mar 24 '16 at 13:19

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