Probability of choosing two random cards of the same value I know this question was asked here: What is the probability of choosing two cards of the same value?
But I'm trying to understand it in terms of combinations.
Two cards are randomly chosen from a deck of 52.  What is the probability that they have the same value?
I'm thinking:
$\frac{\binom{13}{1} \binom{3}{1}}{\binom{52}{2}}$
Out of 13 possibilities you choose the first card (say it's a 7).  Then for the second card you need to pick another 7 out of the remaining three 7's.
Can anyone explain to me where I'm going wrong in my thinking?
Thanks
 A: Your thinking is very close to what is needed. The issue is, the choice of your second card is not independent from the first. When you pick the first card, you are forcing an ordering where that first card must be picked first, then the second card must be picked. Parcly Taxel's method adeptly shows how you can carefully include the consideration of the order in which the cards are selected. That approach works really well, and I highly recommend it.
However, if your goal is truly for an approach using only combinations, then it is important to avoid inducing any ordering.
The goal is to pick both cards at the same time instead of one then the next.
Instead of picking the first card then the second, pick the value of the two cards, but do not actually pick a card. There are $\dbinom{13}{1}$ ways of choosing the value. Regardless of which value is chosen, there are four cards of that value. Now, we select both cards at once in $\dbinom{4}{2}$ ways. This gives the probability of selecting two of the same card:
$$\dfrac{\dbinom{13}{1}\dbinom{4}{2}}{\dbinom{52}{2}} = \dfrac{1}{17}$$
This gives the exact same answer as Parcly Taxel's solution. In fact, it is counting the same thing, just in a different way.
A: When calculating probabilities for drawing without replacement it is better to treat the draws in sequence, so instead of $\binom{52}2$ in the denominator you have $52×51$.
The first factor in the numerator is $52$ since one card doesn't constrain the possibility of both cards having the same value. The second factor, for the second draw, is $3$ since that is the number of remaining cards sharing a value with the first draw. Thus we have $\frac{52×3}{52×51}=\frac1{17}$.
