The probability that a Condorcet Scenario will occur with an odd number of voters and 3 Candidates

Where each voter chooses one of the six rankings with the same probability (i.e. 1/6)



                              By Shalosh B. Ekhad



Theorem: Suppose three candidates, 1,2,3, are running for office, and there \
    2n-1 voters, each of them choosing one of the 3!=6

possible ranking with the same probability (1/6). The votes are counted and \
    it turns out that in the vote of

                                 1 vs. 2, 1 won

                                 2 vs. 3, 2 won

                                 1 vs. 3, 3 won



                  In other words, there is a cycle 1->2->3->1

Let a(n) be the probability of that happening times 6^(2n-1) (i.e. the numer\
    ator)

             The sequence a(n) satisfies the following recurrence



                                                      2
  1296 n (2 n + 3) (2 n + 1) a(n)   36 (2 n + 3) (22 n  + 33 n + 12) a(n + 1)
- ------------------------------- + -----------------------------------------
                        2                                      2
         (n + 1) (n + 2)                        (n + 1) (n + 2)

              2
       4 (19 n  + 57 n + 45) a(n + 2)
     - ------------------------------ + a(n + 3) = 0
                         2
                  (n + 2)



                              and in Maple format



-1296*n*(2*n+3)*(2*n+1)/(n+1)/(n+2)^2*a(n)+36*(2*n+3)*(22*n^2+33*n+12)/(n+1)/(n
+2)^2*a(n+1)-4*(19*n^2+57*n+45)/(n+2)^2*a(n+2)+a(n+3) = 0


Using this recurrence it is easy to compute many values, for example, the pr\
    obability for 1999 voters is

                            0.043848798672392226718




                  The asymptotic expression is estimated to be

                                            0.021101164
                        0.043869914022955 - -----------
                                                 n



                             and in Maple notation



.43869914022955e-1-.21101164e-1/n


But it is also possible to have 1->3->2->1, so the probability of a Condorce\
    t scenario is



                                            0.042202328
                        0.087739828045910 - -----------
                                                 n



                              And in Maple format

.87739828045910e-1-.42202328e-1/n


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