OFFSET
0,3
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.
LINKS
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.
EXAMPLE
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].
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);
CROSSREFS
KEYWORD
nonn
AUTHOR
Dennis P. Walsh, Apr 29 2013
STATUS
approved