This is a combination of your probability of winning and the chance that the item you want is still available.
Takeing them in reverse order, if you win on the $i$th draw ($i\le10$) the probability that the prize is still available is:
$$P_i=\frac{11-i}{10}$$
The probability that the $i$th draw is the first one you win is:
$$W_i=\frac{n}{501-i}\prod_{j=0}^{i-1}\frac{500-n}{500-j}$$
So the probability that you get the prize you are after is:
$$\begin{align} A&=\sum_{i=1}^{10} W_iP_i\\\ &=\sum_{i=1}^{10}\frac{n}{501-i}\prod_{j=0}^{i-1}\frac{500-n}{500-j}\frac{11-i}{10} \end{align}$$
Expand this, substitute the chance you want and solve for n.