Matt -- Adam says adding 1/2 votes leads winning votes to become margins. I do not know, I am trying to check. But the result seems to corroborate what he says. Using your matrix, I do not find that A still wins. I find back the result I had with margins (X=1):
>By the way, >with x=0, margins finds a triple equality. >With x>0 it locks B>C, then gets stuck between the two last pairwise >comparison. >Winning votes used as tie-breaker would give C>A. Thus B wins. >Relative margins used as tie-breaker would give A>B. Thus A wins. Check by yourself. Steph. [EMAIL PROTECTED] a �crit : > On 24 Nov 2002 at 14:41, Elisabeth Varin/Stephane Rouillon wrote: > > > 3: A > > 2: A > B > C > > 2: B > A > C > > 2: B > C > A > > 4: C > > Ranked pairs with winning votes produces: > > A (7) > C (6) , B (6) > C (4) and A (5) > B (4). > > A is the Condorcet winner and wins. > > Margins and relative margins produce of course the same result. > > If I am one of the two B > A > C voter, my 2nd (A) > > choice harms my favorite 1st choice (B). > > The proof is, if I and my co-thinker vote B only: > > 3: A > > 2: A > B > C > > 2: B (truncated !) > > 2: B > C > A > > 4: C > > Ranked pairs with winning votes produces: > > B (6) > C (4), C (6) > A (5) and A (5) > B (4) can't lock. > > B wins now. > > This appears to be an example that illustrates a more stable outcome is achievable > by counting equal ranked options 1/2 vote each. With an additional 1/2 vote for > each non-voted pair the option pair tally matrix after > 2: B (truncated !) looks >like: > # A B C > 7.0 7.0 6.0 > 6.0 6.0 7.5 > 7.0 5.5 8.0 > and A still wins (RP and SSD). You guys are (mis?)computing the tally matrix like: > # A B C > 7.0 5.0 5.0 > 4.0 6.0 6.0 > 6.0 4.0 8.0 > (diagonal represents "approval count"). > > ---- > For more information about this list (subscribe, unsubscribe, FAQ, etc), > please see http://www.eskimo.com/~robla/em ---- For more information about this list (subscribe, unsubscribe, FAQ, etc), please see http://www.eskimo.com/~robla/em
