%I #30 May 01 2013 02:32:58
%S 1,0,6,6,18,90,150,420,1890,3570,10206,42966,87318,252252,1019304,
%T 2172456,6319170,24810786,54712086,159906318,614406078,1390381278,
%U 4077926034,15403838346,35579546262,104633453340,389788932240,915500037120,2698033909680,9934966920960
%N Number of ballot results from n voters that prompt a run-off election when three candidates vie for two spots on a board.
%C We assume each of n voters cast two votes, one each for two of three candidates. A run-off election is necessitated if all 3 candidates receive the same number of votes or if there is a tie for the second to most votes. The total number of ballot results is 3^n since each voter must choose two of three candidates. The number of ballot results that necessitate a run-off election is derived in the note "The probability of a run-off election.." cited in the link section below.
%C The sequence A103221 is used in the derivation. Note that we assign the value 1 to a(0) because if no voters cast ballots on election day another election is needed.
%H Dennis P. Walsh, <a href="http://capone.mtsu.edu/dwalsh/VOTENEW.pdf">The probability of a run-off election when three equally-favored candidates vie for two slots</a>
%F a(n) = 3*sum(C(n,2*b(k)) *C(2*b(k),b(k)), k=0..u(n)) -2*C(n,2n/3) * C(2n/3,n/3) I[3|n] where b(k) = ceiling((n-1)/2)-k, u(n) = floor((n+2)/2) - floor((n+2)/3)-1 = A103221(n)-1, and I[statement] equals 1 if the statement is true and equals 0 otherwise.
%e For n=3, a(3)=6 since a three voter election has 6 possible ballot results that necessitate a run-off. Let A, B, and C denote the three candidates, and, for example, let [AB|AC|BC] denote a ballot result in which voter 1 votes for candidates A and B, voter 2 votes for candidates A and C, and voter 3 votes for candidates B and C. The 6 ballot results that necessitate a run-off election are then given by [AB|AC|BC], [AB|BC|AC], [AC|AB|BC], [AC|BC|AB], [BC|AB|AC], and [BC|AC|AB].
%p ind:= n-> piecewise(n mod 3=0, 1, 0):
%p u:= n-> floor(n/2+1)-floor(n/3+2/3)-1:
%p a:= n-> 3*add(binomial(n, 2*ceil((n-1)/2)-2*k)*
%p binomial(2*ceil((n-1)/2)-2*k, ceil((n-1)/2)-k), k=0..u(n))
%p -ind(n)*2*binomial(n, 2*n/3)*binomial(2*n/3, n/3):
%p seq(a(n), n=0..30);
%K nonn
%O 0,3
%A _Dennis P. Walsh_, Apr 29 2013
|