There is nothing wrong with your calculation; each player wins with probability 1/3. The point is that these are not disjoint events, so( wins)+( wins)+( wins)≠( or or wins). P(A wins)+P(B wins)+P(C wins)≠P(A or B or C wins). In fact, the sum of the probabilities is the expected number of players who win. (When there are no multiple winners, so the events are disjoint, this number is always 0 or 1 and so its expectation is the probability someone will win, but this is not true in general.) Here (no-one wins)=3×127+3!27=13 P (no-one wins)=3×127+3!27=13, (2 people win)=3×3×127=13P(2 people win)=3×3×127=13, and (1 person wins)=13P(1 person wins)13, so the expectation is exactly 11, consistent with your answer.
Hope this helped :)))