If you bought all the tickets, do you think you'd have over 100% chance of winning?
Edit: looks like leephillips beat me to it
P(A) = first ticket wins
P(B) = second ticket wins
P(A|B) + P(A|^B) + P(^A|B) = 1 - P(^A|^B) = 1 - (1 - 1/N) * (1 - 1/(N-1)) = 2 / N
It is slightly > 2 / N if the ticket is revealed before the next one is chosen because then it does throw away that possibility.
And most lottery games allow numbers to be re-used, so you haven't "eliminated" anything. If you exhaustively bought all N number combinations, you have a 100% chance of winning, but you also have a decent chance to split the pot with someone else who also bought the winning numbers.
Next time your family screams and shouts at you that you're wrong, maybe you should actually listen to what they're saying?
Your piling on isn't helpful either, so can it.
(See above.)