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.
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.