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(2010-03-06)   Condorcet's Paradox
A group of voters may prefer A to B, B to C and C to A !

When faced with a choice among three options A, B and C, each individual may express his or her own preferences in one of six distinct ways.  For the sake of  future  convenience, we may tally the number of voters in each of those 6 categories using the 6 quantities  a,b,c,x,y,z  introduced in the following table:

Notations

Order of preferences	Number of voters
C A B	a + x
B A C	a
A B C	b + y
C B A	b
B C A	c + z
A C B	c

With those notations, the outcomes of the three possible straight votes are:

Thus, the aforementioned paradoxical result occurs if and only if  x, y and z  verify the so-called  triangular inequalities  (which state that the sum of any pair of quantities is never less than the third).

In that case, those 3 triangular inequalities do imply that x,y,z are positive.  For example, we have  2x+y > x+z > y  which implies  x>0.
 
The other paradoxical case  (B>A, A>C, C>B)  occurs when the three triangular inequalities are all backward, which implies that x,y,z are negative.  In all other cases, there is no paradox  (which is to say that the collective preferences of the voters are consistent).

How frequent is the paradox ?

Sometimes, ranking three options really boils down to a simple choice between two options  (whenever the third choice is either clearly inferior or clearly superior).  The paradox will occur with vanishing probability in such cases, since a vote between two options is never paradoxical.

To evaluate the situation when the three options offered to the voters are  a priori  on the same footing,  we shall determine the probability of the above paradoxical situation when all preferences are assumed to be equiprobable...

The number of triples of integers forming the sides of a triangle of perimeter  n  is either   (n+2)(n+4)/8   (if n is even)  or   (n-1)(n+1)/8   (if n is odd).

Curiously, that's also exactly the number of ordered triples of integers forming the sides of a triangle of nonzero area and perimeter n+3.

Number of triples  (x,y,z)  forming a triangle of perimeter  n = x+y+z

n	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17
A008795	1	0	3	1	6	3	10	6	15	28	21	36	28	45	36	45	66	55

The paradoxical situation  A>B>C>A  among v voters is obtained by choosing a positive  n  of the same parity as  v  and three nonnegative integers  a,b,c  which add up to  m = (v-n)/2.