Probability of Winning a Lottery Prize with High Probability, Given $n$ Plays Edit: It has been pointed out that the answer to this problem is simply $1 - (1 - \Pr(\text{success}))^n$.
I've recently been toying around with probabilities related to the lottery, and I was wondering about how many plays it would take to win any prize in a drawing. Formally, suppose we have $n$ plays in a typical 6/49 lottery game in which a prize is awarded for matching at least $3$ balls. What is the probability of winning a prize?
If we let $A_k$ denote the event of at least one play matching exactly $k$ balls and let $B$ denote the event of at least one play winning any prize, then we have
$$\Pr(B) = 1 - (\Pr(A_0) + \Pr(A_1) + \Pr(A_2)) = 1 - \frac{\binom{6}{0} \binom{43}{6}}{\binom{49}{6}} - \frac{\binom{6}{1} \binom{43}{5}}{\binom{49}{6}} - \frac{\binom{6}{2} \binom{43}{4}}{\binom{49}{6}} =: p_1.$$
Edit: For general $n$, we simply have $\Pr(B) = 1 - (1 - p_1)^n$. This question was quite straightforward in retrospect.
Since the original question turned out to be so simple, are there any results related to the following?
A similar question which is discussed on the first linked Wikipedia page is the following: What is the minimum number of plays to guarantee one wins a lottery prize? Wikipedia states that this is an open problem. If we instead ask "What is the minimum number of plays to win a lottery prize with high probability (say, $0.95$)?" then what can we say?
 A: For each $k\in \{1,\dots,6\}$, let $q_k$ be the probabililty matching at least $k$ numbers on a single lottery draw. Explicitly,
$$
q_k={\binom{6}{k}\binom{43}{6-k}+\binom{6}{k+1}\binom{43}{6-k-1}+\dots+\binom{6}{6}\binom{43}0\over \binom{49}6}
$$
Now, suppose you choose $n$ tickets randomly, and independently of each other. Then the probability of matching at least $k$ on at least one ticket is
$$
1-(1-q_k)^n
$$
This completely answers your first two questions. However, for your last question about winning at least once out of $n$ tickets with probability at least $95\%$, you cannot use the same methods as above. Above, we assumed the tickets were chosen independently at random, but one can do much better than that by cleverly distributing numbers across the tickets.
For example, suppose that "with high probability" means $99\%$. The probability of matching at least $3$ on one of $n$ random tickets is $1-(1-q_3)^n$. If you set this equal to $0.99$ and solve for $n$, you find that you need $245$ randomly chosen tickets in order to get at least a $99\%$ chance of matching $3$ at least once. However, if you play the $163$ tickets listed at this website, then you would be guaranteed to match $3$ at least once. The minimum number of tickets it takes to match $3$ at least once with probability at least $99\%$ is likely an open problem, just like the $100\%$ variant mentioned in the Wikipedia article.
