Arithmetic Errors

Sun Apr 7 01:13:46 PST 1996

1. Errata- In the first example there is
Also #(B>C) = 10 and
     #(C>B) = 48, so C beats B by 44-10 = 34.
The 48 should be 44 apparently.

2. Regarding Mr. Anderson's -- 
A New Example:

48:  A, (B & C tied)
 6:  B, (A & C tied)
46:  C, B, A

I suggest again that the least worst defeat should not be used for a tie
breaker. There seems to be a bias using the least worst defeat tie breaker
towards the candidate with the second highest number of first choice votes.

For a tie breaker I suggest that the candidate with the lowest number of
first choice votes lose (or first choice votes plus additional choice votes
if there were 4 or more candidates). Thus B should lose the tie breaker and A
should win 48 to 46 (or 51 to 49 if a half vote for each of the 6 A & C tied
votes from B as second choices is transferred).

More information about the Election-Methods mailing list