Mea culpa. Instead of arguing with the shifting sands of ambiguous and imprecise English, it's better to use equations and evidence. P(A) = first ticket wins

Mea culpa. Instead of arguing with the shifting sands of ambiguous and imprecise English, it's better to use equations and evidence. 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.