Counting Condorcet

By Rebecca Embar and Doron Zeilberger

.pdf    .tex   
Appeared in- Enumerative Combinatorics and Applications 2:3 (2022) Article #S2R22

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.

Added April 29, 2022: As we said in the paper, the sequence enumerating Condorcet profiles with an odd number of voters (and three candidates) is not yet in the OEIS (we hope to put it there soon), but a search for the complementary numbers, those that do NOT lead to the Condorcet paradox namely


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!

Maple package

Sample Input and Output for Condorcet3.txt

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