Probability of being picked more than once in a simple random sample. A school has $500$ girls and $500$ boys. A simple random sample is obtained by selecting names from a box (with replacement) to a get a sample of $10$. 

Find the probability of someone being picked more than once.

My working is: 
\begin{align}
P(\text{being picked more than once}) &amp;= 1 - P(\text{not being picked}) - P(\text{being picked exactly once})\\
&amp;=1 - \left(\frac{999}{1000}\right)^{10} - 10\left(\frac{1}{1000}\right)\left(\frac{999}{1000}\right)^9 = 0.00004476
\end{align}
However, this is not correct. Any ideas?
This is a variation on the Birthday Problem, with names instead of birthdays and drawings from the hat instead of people in a room.

The answer is $1$ minus the probability that all ten names are different,
which is the product of the probabilities that the $n$th name is different from all previous names, given that all previous names were unique.
The probability for the $n$th name is $\frac{1000 + 1 - n}{1000},$ so the answer is

$$ \frac{1000}{1000} \cdot \frac{999}{1000}
\cdot \frac{998}{1000} \cdot \frac{997}{1000}
\cdot \frac{996}{1000} \cdot \frac{995}{1000}
\cdot \frac{994}{1000} \cdot \frac{993}{1000}
\cdot \frac{992}{1000} \cdot \frac{991}{1000}
= \frac{1000!}{1000^{10}\,990!}.$$
Probability of anybody being picked more than once:

$$
\frac{\binom{10^{3}}{10}\times 10!}{\left(10^{3}\right)^{10}}
$$

Probability of a particular person being picked more than once:

$$
1-\frac{\left(10^{3}-1\right)^{10}+10\times\left(10^{3}-1\right)^{9}}{\left(10^{3}-1\right)^{10}}
$$