A224711 Number of ballot results from n voters that prompt a run-off election when three candidates vie for two spots on a board.

1, 0, 6, 6, 18, 90, 150, 420, 1890, 3570, 10206, 42966, 87318, 252252, 1019304, 2172456, 6319170, 24810786, 54712086, 159906318, 614406078, 1390381278, 4077926034, 15403838346, 35579546262, 104633453340, 389788932240, 915500037120, 2698033909680, 9934966920960

Offset: 0

Author

Dennis P. Walsh, Apr 29 2013

Keywords

• nonn

Comments

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.

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.

Examples

			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].
		

Links

• Dennis P. Walsh, The probability of a run-off election when three equally-favored candidates vie for two slots

Programs

Maple

ind:= n-> piecewise(n mod 3=0, 1, 0):
u:= n-> floor(n/2+1)-floor(n/3+2/3)-1:
a:= n-> 3*add(binomial(n, 2*ceil((n-1)/2)-2*k)*
        binomial(2*ceil((n-1)/2)-2*k, ceil((n-1)/2)-k), k=0..u(n))
        -ind(n)*2*binomial(n, 2*n/3)*binomial(2*n/3, n/3):
seq(a(n), n=0..30);

Formula

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.