# Probability, why my solution doesn't work out? (P of drawing a pair)

The task is simple, the probability of drawing a pair of cards. You draw two cards from a stack, what is the chance that you get two kings or two fours.

My idea was the following. There are 13 different valued cards. The probability of getting lets say a pair of two is following.

$\frac{1}{52} \cdot \frac{3}{51}$

The chance of getting first card 2 is $\frac{1}{52}$, the chance of getting one of the three other cards that would make this a pair is $\frac{3}{51}$. That sounds reasonable to me. And now to account for all 13 different types you just multiply this with 13. Or add it up 13 times.

$13(\frac{1}{52} \cdot \frac{3}{51})$

This is wrong and I don't understand why. The probability of getting a random pair should be the sum of getting every type of pair.

A correct solution would be the following.
$\frac{52 \cdot 3}{52 \cdot 51}$

I try to stick with the 'count the number of beneficial outcomes in every step and multiply method' but I got stuck on why my way of thinking didn't work out here.

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The probability that the first card is a 2, is $4/52$, not $1/52$. Your reasoning works out, with this correction. – David Mitra Nov 20 '11 at 13:14
Silly me. Lalalala. Thanks. – Algific Nov 20 '11 at 13:16
It doesn't matter what first card drawn is. Probability of matching is $3/51$. – André Nicolas Nov 20 '11 at 15:56
Shouldn't it be $\frac{2}{13}\cdot\frac{3}{51}$ where the $\frac{2}{13}$ is the probability that the first card is a four or a king? – Henning Makholm Nov 20 '11 at 16:00
@David, why not post that as an answer, so it may be accepted? – Mr.Wizard Nov 22 '11 at 9:54