Solution
The Chance of a Voting Cycle
Show the problem again
Three voters each independently submit a strict ranking of candidates A, B and C, chosen uniformly at random from the 6 possible rankings. Majority preference between two candidates is decided by which one more voters rank higher (with three voters there are no ties). The majority relation is *cyclic* if it has the form X beats Y, Y beats Z, Z beats X for some ordering of the candidates (a Condorcet cycle). What is the probability that the majority relation is cyclic? Give a reduced fraction.
Worked solution
There are 6³ = 216 equally likely profiles. Consider the cycle A > B > C > A. Each of its three pairwise majorities needs at least two of the three voters, so 6 voter-agreements are needed in total. A transitive ranking agrees with at most 2 of the 3 cyclic preferences A > B, B > C, C > A (all three would be a cycle), and agrees with exactly 2 only if it is one of the rotations ABC, BCA, CAB. So every voter must hold a rotation, and each pair must get exactly 2 agreements. Each rotation misses a different pair (ABC misses C > A, BCA misses A > B, CAB misses B > C), so each must be missed exactly once: the three voters hold the three different rotations, in some order, giving 3! = 6 profiles. The reverse cycle A > C > B > A likewise arises only from ACB, CBA, BAC in some order, another 6 profiles. So 12 of the 216 profiles are cyclic (a full enumeration of all 216 confirms the count), and the probability is 12/216 = **1/18**.
Source: Condorcet's paradox (1785); the 1/18 figure for three voters and three candidates is a standard impartial-culture computation. Statement written for AxiomIQ.