Voting with 3-way ties From Peter Winkler's 'Mathematical puzzles' > Ashford,Baxter and Campbell run for election and end up in a 3-way tie. To break it, they solicit voters' second preference and there is also a 3-way tie. Ashford steps forward and says: 'since the total number of voters is odd, I propose that voters first choose between Baxter and Campbell, and I face the victor in a run-off' . Baxter complains that this gives an unfair advantage to Ashford. Why? I was trying to reason as follows but I cannot quite conclude... Call Ashford winning event A , Baxter winning event B and Campbell winning event C. Then from the first statement: $$P(A) = P(B) = P(C) = 1/3 $$ and from the second preference vote: $$P(A|B) = P(B|C)=P(C|A)= 1/3$$ So to solve the puzzle I thought I would compute $$P(A|B \cup C) =P(A \cap (B \cup C) ) / P (B \cup C) = P ( (A \cap B) \cup (A \cap C)) / (P(B) +P(C)) = 0 ??$$ which looks wrong ...

The problem is actually already solved by Jyrki's comment, since the only way to satisfy the given conditions is that the $3$ voters have the cyclical preferences given in his example, or the opposite preferences. Thus you just need to count the number of times each of the candidates wins under the proposed "solution", given such a preference set.

As you don't explain how the probabilities you introduce are to be interpreted (the problem statement has no probabilistic elements), I can't say much about your approach. The second displayed equation seems to make no sense, as it contains conditional probabilities of one candidate winning given that another candidate wins, which should be zero.