[EM] More easily hand-counting three-slot Condorcet

Jobst Heitzig heitzig-j at web.de
Fri Nov 5 04:00:53 PST 2004

Dear Kevin!

you wrote:
> If there are three ranks, and R1 is the number of candidates
> placed in the first rank, R2 for second, and R3 the number of
> unmarked candidates, then using the above method the manual
> counter will have to mark
> (R1*R2) + (R1*R3) + (R2*R3) tallies.
> For a given ballot, mark a vote at X,X for every candidate X
> placed in the first or second slot.  Mark a vote at X,Y for
> every candidate X placed in the first slot above a candidate
> Y placed in the middle slot.
> This means that for each ballot, we will only make:
> R1 + R2 + (R1*R2) tallies.
> A beats B pairwise if A,A - B,A > B,B - A,B.

Very nice! I think it will simplify the tally considerably.

> I looked long and hard, but I don't believe it is possible
> to recover the magnitudes of victory.  

That's true: When we denote the number of voters who have A in slot i
and B in slot j by Vij, then we know the following five linear
combinations of the nine numbers V11,V12,...,V33:
  V11+V12+V13+V21+V22+V23+V31+V32+V33 = total number of voters
  V11+V12+V13+V21+V22+V23             = entry at position A,A
      V12                             = entry at position A,B
  V11+V12    +V21+V22    +V31+V32     = entry at position B,B
              V21                     = entry at position B,A

When we search for winning votes A>B, we need to determine
      V12+V13        +V23,

But the vector
is linearly independent from
  (1,1,0,1,1,0,1,1,0), and

> I think, though, that
> the approval measure is almost as good.  I also believe that
> when the approval of the pairwise winner is used as the
> defeat strength, then all methods I know of will elect the
> same candidate (RP, Schulze, Raynaud, MinMax, probably River).
> I believe all rules are then equivalent to "Elect the least-
> approved candidate who pairwise beats every candidate with
> greater approval."

That's true, but it would erroneously count those having A in slot 2 and
B in slot 1 towards the strength of defeat A>B. In accordance with my
recent "grand compromise" proposal, I suggest to use instead the number
of voters who have A in slot 1 or 2 and B in slot 3, which corresponds
to V13+V23.

Although also this number cannot be determined from the numbers you
suggest to tally, it is not too expensive to make it possible by also
counting all voters who have both A and B in slot 1 or both A and B in
slot 2. This corresponds to counting V11+V22 also. Defeat strength can
then be defined as

  strength of A>B
    := number of voters having A but not B in 1 or 2
     = V13+V23
     = (V11+V12+V13+V21+V22+V23) - V12 - V21 - (V11+V22)
     = entry at position A,A
       minus voters having both in 1 or both in 2

This would increase the number of tallies from
  R1 + R2 + (R1*R2)
  R1 + R2 + (R1*R2) + (R1*R1) + (R2*R2)
which can still be expected to be much smaller than
            (R1*R2) + (R1*R3) + (R2*R3)
since in general, R1+1 and R2+1 will be considerably smaller than R3.

In all, I think this could be a nice further simplification of the
"grand compromise" proposal!

Yours, Jobst

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