[EM] Fw: IBCM, Tideman, Schulze

Markus Schulze schulze at sol.physik.tu-berlin.de
Sun Jun 4 00:56:02 PDT 2000

Dear Steve,

you wrote (2 Jun 2000):
> We can infer from Norm's message that he does not agree with 
> Markus' objection that simulations based on random voting are 
> meaningless.  Markus said that the use of random voting in 
> simulations implies that my underlying model of voter behavior 
> is that voters behave randomly, but that's not a logical 
> conclusion.  People have at various times posted estimates of 
> the percentage of scenarios which would have a Condorcet winner, 
> using simulations involving randomness, but Markus has not 
> criticized those estimates.

It seems to me that you are talking about Craig Carey. Craig
claimed (A) that when the number of candidates is large then
the probability that there is a Condorcet winner is neglectable
and (B) that therefore the Condorcet Criterion is quite

There was no need for me to question Craig's claim A because
his conclusion B was incorrect as Blake pointed out (19 Nov 1999):
> In the same way, most people think that someone who gets a majority
> of first-place votes should win.  We could define the "Majority
> method" as the procedure that carries this out.  Of course, the
> procedure would sometimes not give us a winner, so we would advocate
> various majority-completion methods that gave a winner even in these
> cases.  Plurality, Condorcet, and plurality-elimination, are all
> majority-completion methods.  Anyone using any of these can be said to
> be using the Majority method.
> I think it clearly makes no sense to talk about the majority method
> as "going bad," or criticize the majority completion methods because
> there isn't always a majority winner.
> If you disagree with the majority method, the same argument can be
> made using a "unanimity method".  It is quite possible to say that
> someone who is unanimously chosen as the favourite should win, and
> still acknowledge that such a candidate may not exist.  The unanimity
> method doesn't "go bad", it just doesn't always give a conclusion.


You wrote (2 Jun 2000):
> Here's an example which, if I haven't erred, shows that SD is 
> not completely monotonic:
>    Definition of SD:
>    Repeat while no alternative is unbeaten:
>       Discard the non-discarded defeat(s) which is smallest
>       in some unbroken cycle.
>    Elect the unbeaten alternative(s).
>    Example of SD non-monotonicity:
>    Small majorities:  AC51,CB52,BA53,DB53,GA54
>    Larger majorities: AD,FC,CE,EF,AH,HG
>    (All other pairings are pairties.)
>    First AC51 is dropped.  Then BA53 and DB53 are dropped.
>    (CB52 is not dropped because the cycle it is in was broken
>    by the dropping of AC51.)  Then GA54 is dropped and A has
>    become unbeaten, so SD chooses A.
>    Assume that at least 3 voters ranked A just below C.
>    Suppose 3 of them uprank A over C, giving AC54.
>    Now AC is not the smallest majority of any cycle;
>    CB52 is the smallest majority of the A>C>B>A cycle.
>    First CB52 is dropped.  Then BA53 and DB53 are dropped
>    and B has become unbeaten, so SD chooses B.
>    (I didn't attempt to find an example having fewer
>    alternatives or fewer pairties.)

When we talk about Sequential Dropping (SD) then I have to say
that -to my opinion- the biggest problem of SD is its violation
of Independence from Clones when there are identical elements in
the pairwise matrix.


   Suppose that the Senate uses SD to elect its President pro
   tempore. Suppose that 50 Senators are Democrats and 50
   Senators are Republicans. Suppose that the Democrats
   nominate three candidates A1, A2 and A3 and that the
   Republicans nominate only one candidate B. Then a possible
   situation looks as follows:

      40 Senators vote A1 > A2 > A3 > B.
      35 Senators vote B > A2 > A3 > A1.
      15 Senators vote B > A3 > A1 > A2.
      10 Senators vote A3 > A1 > A2 > B.

   The pairwise matrix looks as follows:

       A1:B = 50:50
       A2:B = 50:50
       A3:B = 50:50
       A1:A2 = 65:35
       A1:A3 = 40:60
       A2:A3 = 75:25

   SD elects candidate B decisively.

   On the other side (A1,A2,A3) is a set of clones. And when this
   set of clones is substituted with a single makro candidate A
   then the situation above looks as follows:

      50 Senators vote A > B.
      50 Senators vote B > A.

   Therefore, Independence from Clones says that candidate B must
   be elected with a probability of 50% and that one of the
   candidates A1,A2,A3 must be elected with a probability of 50%.

Markus Schulze
schulze at sol.physik.tu-berlin.de
schulze at math.tu-berlin.de
markusschulze at planet-interkom.de

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