Opps – Correction
I made a mistake in an earlier post on this thread:
I stated that in a BTR-Score election with a 3-way cycle, the highest seed in the cycle would have a 50% chance of winning, and the two lower seeds would each have a 25% chance of winning.
This is incorrect. The highest seed in the cycle would always win.
Consider this BTR-Score tournament with a 3-way cycle:
Candidate A is the highest seed in the cycle.
Candidate B is the 2nd highest seed in the cycle.
Candidate C is the 3rd highest seed in the cycle.
In every such election, B and C will face off in the tournament; the winner will then face A in the deciding contest.
In a 3-way cycle, each of the candidates wins one match and loses one match to the other two candidates. The winner of the B vs C contest uses its only win to prevail; therefore, that winner has no chance of defeating A.
2 Possibilities:
A > B > C > A - B beats C then B loses to A
C > B > A > C - C beats B then C loses to A
A BTR-Score election with a Condorcet winner rewards majority.
A BTR-Score without a Condorcet winner rewards utility.
It would be possible, but extremely weird, for a BTR-Score election to have a 3-way cycle that does not include the top seed, the candidate with the highest total score.
