By Rebecca Embar and Doron Zeilberger
Written: April 28, 2022
We give an elegant bijective proof that the number of vote-count profiles that lead to the famous Condorcet paradox with three candidates and 2n-1 voters equals 2*binomial(n+3,5). We then use this bijection to efficiently enumerate the total number of Condorcet voting profiles with a given number of (odd) voters, and related probabilities.
seq(6^(2*n1-1)-NuCo(n1),n1=2..8);
that is:
6, 204, 7236, 258936, 9291876, 333840744, 12001884264, 431639416944,
lead to the following interesting
article (in Japanese)
by Toshio Urata, entitled "The probability that no Condorcet winner emerges in an election"
[We thank Shigeru Mochida for translating the title and deciphering the author's name]
However we are sure that they can't compute the 10000-th term, while we can do it in second!
Note that is agrees with Guilbaud's exact value if 1/4-3/(2*Pi)*arcsin(1/3)