Coombs' Method

Hugh Tobin htobin at redstone.net
Sat Dec 28 18:19:04 PST 1996


Perhaps Coombs (Davison) is better than IRO, but why should we care when
Smith//Condorcet is clearly better than both?

Coombs, as described in recent postings, suffers from the same basic
problem as IRO: it may disqualify the beats-all winner.  Coombs will do
this sometimes when IRO will not.  Consider the following example:

24 ABC
24 ACB
5 BAC
9 CAB
18 CBA
20 BCA

First place votes: A 48, B 25, C 27
Last place votes: A 38, B 33, C 29
Pairwise: A 57 > B 43; A 53 > C 47; C 51 > B 49

Condorcet: A wins outright.  And why not, when a majority prefers A over
each alternative?

IRO: Drop B and A wins, 53 - 47 over C.

Coombs: Drop A (!) and C wins.  Note that if this started out as a race
between A and C, by joining the contest B changed the result, even
though B loses to each of them.  Note also that the result is the same
if the A voters truncate (under the "half-demerit" rule posited on this
list), as may be likely in a public election where both B and C are
intolerable to A voters.  

Of course, the A voters could have organized themselves to vote the same
candidate in last, and thereby could have won.  But in Condorcet no such
insincere voting would be necessary in order for A to prevail.

	Coombs may not present as severe strategic problems as IRO, but it
seems the temptations to order-reversal would be at least as great as in
Condorcet.  (One ranks one's second favorite last, not to create a
circular tie, but so that one's favorite will benefit from the
second-place votes of those who support one's second favorite.)  So what
advantage does it have, such that we should accept results such as shown
above?  Do electoral reformers want to try to explain to the public why
A is not in the runoff in a case such as shown above?

-- Hugh Tobin




More information about the Election-Methods mailing list