A recurrence for the Condorcet Paradox with Three Candidates where the probability distribion on the ranking is , [x, 1/2 - 2 x, x, 1/2 - 2 x, x, x] By Shalosh B. Ekhad Suppose that there are 2n-1 voters each completely ranking three candidates,\ let's call them 1,2,3. Also assume that each voter independtly picks a\ randking with the following probability distribution Prob. of 123 is, x Prob. of 132 is, 1/2 - 2 x Prob. of 213 is, x Prob. of 231 is, 1/2 - 2 x Prob. of 312 is, x Prob. of 321 is, x Let b(n) be the probability of a Condorcet scenario when there are 2n-1 vote\ rs Using Wilf-Zeilberger proof theory it is easy to get the following fifth-ord\ er recurrence for b(n) 2 2 n (8 x - 1) (4 x - 1) (2 n + 5) (2 n + 3) (2 n + 1) (2 n + 7) b(n) -1/16 -------------------------------------------------------------------- + 2 2 (n + 2) (n + 4) (n + 3) 2 4 2 3 4 1/16 (2 n + 5) (2 n + 3) (2 n + 7) (6144 n x - 4608 n x + 15360 n x 2 2 3 4 2 2 3 + 1408 n x - 11520 n x + 10240 x - 192 n x + 3360 n x - 7680 x 2 2 / + 10 n - 432 n x + 2160 x + 21 n - 264 x + 12) b(n + 1) / ((n + 2) / 2 2 3 4 3 3 (n + 4) (n + 3) ) - 1/4 (2 n + 7) (2 n + 5) (3072 n x - 2304 n x 2 4 3 2 2 3 4 3 + 18432 n x + 864 n x - 13824 n x + 37120 n x - 144 n x 2 2 3 4 3 2 2 + 4944 n x - 27840 n x + 25600 x + 10 n - 792 n x + 9620 n x 3 2 2 / - 19200 x + 52 n - 1494 n x + 6420 x + 93 n - 966 x + 57) b(n + 2) / / 2 2 3 4 3 3 ((n + 2) (n + 4) (n + 3) ) + 1/2 (2 n + 7) (1024 n x - 768 n x 2 4 3 2 2 3 4 3 2 2 + 8704 n x + 448 n x - 6528 n x + 24320 n x - 96 n x + 3568 n x 3 4 3 2 2 3 2 - 18240 n x + 22400 x + 10 n - 744 n x + 9520 n x - 16800 x + 73 n 2 / 2 - 1944 n x + 8520 x + 180 n - 1716 x + 150) b(n + 3) / ((n + 4) / 2 (n + 3) ) - ( 2 2 2 2 2 2 80 n x - 24 n x + 560 n x + 5 n - 168 n x + 980 x + 32 n - 294 x + 51 / 2 ) b(n + 4) / (n + 4) + b(n + 5) = 0 / and in Maple format -1/16*n*(8*x-1)^2*(4*x-1)^2*(2*n+5)*(2*n+3)*(2*n+1)*(2*n+7)/(n+2)/(n+4)^2/(n+3) ^2*b(n)+1/16*(2*n+5)*(2*n+3)*(2*n+7)*(6144*n^2*x^4-4608*n^2*x^3+15360*n*x^4+ 1408*n^2*x^2-11520*n*x^3+10240*x^4-192*n^2*x+3360*n*x^2-7680*x^3+10*n^2-432*n*x +2160*x^2+21*n-264*x+12)/(n+2)/(n+4)^2/(n+3)^2*b(n+1)-1/4*(2*n+7)*(2*n+5)*(3072 *n^3*x^4-2304*n^3*x^3+18432*n^2*x^4+864*n^3*x^2-13824*n^2*x^3+37120*n*x^4-144*n ^3*x+4944*n^2*x^2-27840*n*x^3+25600*x^4+10*n^3-792*n^2*x+9620*n*x^2-19200*x^3+ 52*n^2-1494*n*x+6420*x^2+93*n-966*x+57)/(n+2)/(n+4)^2/(n+3)^2*b(n+2)+1/2*(2*n+7 )*(1024*n^3*x^4-768*n^3*x^3+8704*n^2*x^4+448*n^3*x^2-6528*n^2*x^3+24320*n*x^4-\ 96*n^3*x+3568*n^2*x^2-18240*n*x^3+22400*x^4+10*n^3-744*n^2*x+9520*n*x^2-16800*x ^3+73*n^2-1944*n*x+8520*x^2+180*n-1716*x+150)/(n+4)^2/(n+3)^2*b(n+3)-(80*n^2*x^ 2-24*n^2*x+560*n*x^2+5*n^2-168*n*x+980*x^2+32*n-294*x+51)/(n+4)^2*b(n+4)+b(n+5) = 0 Let's try out an exapmple with x=1/10 and n=100, i.e. 199 voters The exact probabilty of a Condorcet scenario is, 1533884152218570348511895718\ 735762537888209928042461624140987269509978623875731398346818549353633590\ 665863959648275855754005182896922217409726355904148903619382506670947799\ / 027934766024908592333 / 2000000000000000000000000000000000000000000000\ / 000000000000000000000000000000000000000000000000000000000000000000000000\ 000000000000000000000000000000000000000000000000000000000000000000000000\ 0000, and in decimals it is, 0.07669420761 Let's test is with simulations, with, 5000, tries We get that the estimated probability is, 0.080400000000000000000000000000000\ 00000000000000000000000000000000000000000000000000000000000000000000 Let's try out an exapmple with x=1/5 and n=100, i.e. 199 voters The exact probabilty of a Condorcet scenario is, 4151248910465821484403500060\ 892723237170924623712868062281404182057241633709331616361100373258424930\ 040761578982799108916484585574990900000623853458332727659994932635680100\ / 30494686932778688011 / 50000000000000000000000000000000000000000000000\ / 000000000000000000000000000000000000000000000000000000000000000000000000\ 000000000000000000000000000000000000000000000000000000000000000000000000\ 00, and in decimals it is, 0.08302497821 Let's test is with simulations, with, 5000, tries We get that the estimated probability is, 0.075800000000000000000000000000000\ 00000000000000000000000000000000000000000000000000000000000000000000 Let's try out an exapmple with x=23/25 and n=100, i.e. 199 voters The exact probabilty of a Condorcet scenario is, 4068465296370716454523269213\ 659764239088971712643223144380061897987354810512323701654904986778051461\ 980179855480074697428760752436945233731685635877801644649516105854840663\ 949185204883323526585816013069469253565219839660547998499446069791363787\ 267953456398264307057979294257558916677161840567724394537105397176379001\ 321999281467452234487015733905392202398288219478873786289646208766840280\ / 13 / 62500000000000000000000000000000000000000000000000000000000000000\ / 000000000000000000000000000000000000000000000000000000000000000000000000\ 000000000000000000000000000000000000000000000000000000000000000000000000\ 000000000000000000000000000000000000000000000000000000000000000000000000\ 000000000000000000000000000000000000000000000000000000000000000000000000\ 00000000000000000000000000000000000000, and in decimals it is, 0.06509544474 Let's test is with simulations, with, 5000, tries We get that the estimated probability is, 0.065000000000000000000000000000000\ 00000000000000000000000000000000000000000000000000000000000000000000 Finally let's see how the probability of theCondorcet paradox changes when t\ here are 99 voters and the prob. dist. is , [x, 1/2 - 2 x, x, 1/2 - 2 x, x, x], from x=0 to x=1/4 + HHHHHHHHHHHHHHHHH + HHHHHHHH HHHHHH 0.08 HHHHHH HHHH + HHHHH HH + HHHH HH + HHHH HH 0.06 HHH HH + HHH HH + HHH HH + HHH H + HH H 0.04 HH H + HH HH + HH H + H H 0.02 H H + H HH + H H +H H +H H *--+--+--+--+--+--+--+--+--+--+--+--+---+--+--+--+--+--+--+--+--+--+--+--+--* 0 0.05 0.1 0.15 0.2 0.25