What is the probability that a domino randomly selected from the set will be a double? A domino is a rectangular tile composed of two squares. An integer is represented on both squares, and each integer 0-9 is paired with every integer 0-9 exactly once to form a complete set. A $\textit{double}$ is a domino that has the same integer on both of its squares. What is the probability that a domino randomly selected from a set will be a $\textit{double}$? Express your answer as a common fraction. I thought the answer was 1/10, because the first square can be any number, and the second number has 1/10 chance of being the same as the first number. Why am I wrong?

There are $\binom{10}{2}$ dominoes with two different numbers and $\binom{10}{1}$ with the same number on each square. Hence, the probability that a domino is a double is $$\frac{\dbinom{10}{1}}{\dbinom{10}{2} + \dbinom{10}{1}}$$ As for why your approach does not work, while you have counted each double once, you have counted each domino with two different numbers twice, once for each way you could pick one of those numbers as the first square.