Known at the time of drawing lots

You are not doing it right. Ask the questions to be drawn in order. Then the situation is divided into: first, A doesn't win, the probability is 1/3, and B and C must have tickets; Second, A has won, and the probability is 2/3, so there is one ticket left. The probability of B winning is 1/2, and at this time, 2/3U 1/3= 1/3. These two cases add up to two thirds. Note that because we don't know what A draws, the probability of B drawing has nothing to do with the order. If it is known that A draws one, then the probability of B is 1/2.