We give several approaches. The first one is closest in spirit to yours.
You have bought your $3$ tickets. Now the lottery corporation is choosing the $4$ winning tickets. These can be chosen in $\binom{n}{4}$ ways. If the lottery is well run, the process makes more or less sure that the choices are all equally likely.
How many ways can they choose the $4$ winning tickets so that none of them are yours? Clearly $\binom{n-3}{4}$. So the probability you win no prize is
$$\frac{\binom{n-3}{4}}{\binom{n}{4}}\tag{A}.$$
(The expresssion in (A) can be considerably simplified.) The probability you win at least one prize is therefore $1$ minus the answer of (A).
Another way: Imagine the lottery corporation picks the winning tickets one at a time. The probability the first ticket it picks is bad (not one of yours) is $\frac{n-3}{n}$.
Given that the first ticket picked was bad, there are $n-4$ bad left out of $n-1$, So given that the first ticket was bad, the probability the second is bad is $\frac{n-4}{n-1}$. Thus the probability the first two are bad is
$\frac{n-3}{n}\cdot\frac{n-4}{n-1}$.
Continue two more rounds. The same reasoning shows that the probability all our tickets are bad is
$$\frac{n-3}{n}\cdot\frac{n-4}{n-1}\cdot\frac{n-5}{n-2}\cdot\frac{n-6}{n-3}.$$
Still another way: Let's switch points of view.
Imagine that the winning tickets have already been determined. So there are $4$ "good" tickets and $n-4$ bad. We will find the probability that you pick all bad.
There are $\dbinom{n}{3}$ ways that we could choose our three tickets. There are $\binom{n-4}{3}$ ways to choose them so they are all bad. So the probability that we choose all bad is
$$\frac{\binom{n-4}{3}}{\binom{n}{3}}.\tag{B}$$
The probability of at least one good is $1$ minus the number in (B).
Remark: We could use a strategy like yours: Find the probability of winning exactly $1$ prize, exactly $2$ prizes, exactly $3$ prizes, and add up,
We calculate the probability of winning exactly one prize. The other calculations are roughly similar. So you have bought $3$ tickets. We find the probability exactly one is good,
There are $\binom{n}{4}$ ways for the corporation to choose $4$ tickets. How many ways are there to choose $1$ that you have and $3$ that you don't have?
The one you have can be chosen in $\binom{3}{1}$ ways. For each such choice, the $3$ you don't have can be chosen in $\binom{n-3}{3}$ ways. Thus the probability of exactly one winning ticket is
$$\frac{\binom{3}{1}\binom{n-3}{3}}{\binom{n}{4}}.$$
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.
Best Answer
Your chance of not winning on the first ticket is $\frac 5{10}$ because there are five other tickets than yours. Assuming you lose the first, your chance of losing the second is $\frac 49$ and assuming you lose that your chance of losing the third is $\frac 38$. Your chance of winning at least one prize is $1$ minus the product of these $$1-\frac 5{10}\cdot \frac 49 \cdot \frac 38=\frac {11}{12}$$