A man decides to play the lottery every week for 30 years. That would be a total of $52\times30$ weeks. The lottery is selecting 6 digits out of 48. The order of the numbers are not important. I'm looking for the probability of winning at least once during those 30 years. First I calculated the probability for winning one week.
$\frac{6!(48-6)!}{48!} = \frac{1}{12271512}$
From there I just assumed that one could multiply by the number of weeks. But that doesn't give the correct answer. The complement would be the probability of not winning a single time during those 30 years, in case it makes sense to go at it that way.