Given the following preference schedule, find the winner under the different methods.
18 | 12 | 10 | 9 | 4 | 2 |
A | B | C | D | E | E |
D | E | B | C | B | C |
E | D | E | E | D | D |
C | C | D | B | C | B |
B | A | A | A | A | A |
A - B;--> B wins
A - C;--> C wins
A - D;--> D wins
A - E;--> E wins
B - C;--> C wins
B - D;--> D wins
B - E;--> E wins
C - D;--> D wins
C - E;--> E wins
D - E;--> E wins
This means:
A has 0 wins, B has 1 win, C has 2 wins, D has 3 wins, and E has 4
wins. Therefore E (with the most number of wins) is
the winner. In
this particular case E is also a Condorcet
winner. Because he has won every head-to-head (one-to-one) contest.
In general this is not the case with the pairwise comparison
A | 5(18) + 4(0) + 3(0) + 2(0) + 1(37) | = 127 |
B | 5(12) + 4(14) + 3(0) + 2(11) + 1(18) | = 156 |
C | 5(10) + 4(11) + 3(0) + 2(34) + 1(0) | = 162 |
D | 5(9) + 4(18) + 3(18) + 2(10) + 1(0) | = 191 |
E | 5(6) + 4(12) + 3(37) + 2(0) + 1(0) | = 189 |
This means D is the Borda Count winner.
18 | 12 | 10 | 9 | 4 | 2 |
A | B | C | D | E | E |
D | E | B | C | B | C |
E | D | E | E | D | D |
C | C | D | B | C | B |
B | A | A | A | A | A |
This leaves us with
18 | 37 |
A | B |
B | A |
Thus B is the Runoff winner.
18 | 12 | 10 | 9 | 4 | 2 |
A | B | C | D | E | E |
D | E | B | C | B | C |
E | D | E | E | D | D |
C | C | D | B | C | B |
B | A | A | A | A | A |
18 | 16 | 10 | 9 | 2 |
A | B | C | D | C |
D | D | B | C | D |
C | C | D | B | B |
B | A | A | A | A |
18 | 16 | 21 |
A | B | C |
C | C | B |
B | A | A |
18 | 37 |
A | C |
C | A |