Arithmetic Errors

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Arithmetic Errors

Craig Carey-2
As I said, I was tired.

Original example corrected (I hope):

46:  A, (B & C tied)
 6:  B, A, C
 4:  B, C, A
44:  C, B, A

Then #(A>B) = 46 and
     #(B>A) = 54, so B beats A by 54-46 =  8.
Also #(A>C) = 52 and
     #(C>A) = 48, so A beats C by 52-48 =  4.
Also #(B>C) = 10 and
     #(C>B) = 48, so C beats B by 44-10 = 34.

C says:  "I scored 48 in my only defeat, while A only got 46, and B only got 10.
  So my worst pairwise defeat was the smallest."

B says:  "My opponent only scored 44 in my one defeat, while C's opponent got  
  52, and A's opponent got 54.  So my worst pairwise defeat was the smallest."

C also says:  "I lost by only 4 in my one defeat, while A lost by 8, and B lost
  by 34.  So, again, my worst pairwise defeat was the smallest."

Now according to Steve:

<<"Try my notation:
  B beats A by  8                -46 +6 +4 +44
  A beats C by  4                +46 +6 -4 -44
  C beats B by 34                  0 -6 -4 +44

I hope you'll leave it to a method's proponents to teach it, and save
the complexities for the rebuttal arguments.  :-)">>

This seems to me that Steve is saying that C wins according to Condorcet's
method since C only loses by 4.  Is this correct Steve?

But Mike says:

<<"Bruce has surely seen my definition of Condorcet's method, and
he knows that A's & C's claims, as he writes them, have nothing
to do with Condorcet's method's scoring. If his point is that
people won't understand why Condorcet scores as it does
(instead of counting votes-for, or margins of defeat), I hope
I've demonstrated how briefly & simply that can be explained.

> This is simple?  At best, it's understandable only if it's very carefully

No that isn't simple. It's called "obfuscation". But I've answered it.

> explained.  And the need for such a vary careful explination is part of my
> argument for complexity here.

My explanation, given above, for why Condorcet counts as it does, isn't
complicated. It's brief & simple.">>

This seems to me that Mike is saying that B wins according to Condorcet's method
because "A's & C's claims, as he writes them, have nothing
to do with Condorcet's method's scoring. If his point is that
people won't understand why Condorcet scores as it does
(instead of counting votes-for, or margins of defeat), I hope
I've demonstrated how briefly & simply that can be explained."  Is this correct
Mike?

Either I am missing a whole lot here (which is certainly possible), or Steve and
Mike disagree on which of their clearly explained, well understood, and
obviously right results is, in fact, the right result--and all I am questioning
for now is is the clearly explained and well understood part.

A New Example:

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

Then #(A>B) = 48 and
     #(B>A) = 52, so B beats A by 52-48 =  4.
Also #(A>C) = 48 and
     #(C>A) = 46, so A beats C by 48-46 =  2.
Also #(B>C) =  6 and
     #(C>B) = 46, so C beats B by 46- 6 = 40.

A says:  "I scored 48 in my only defeat, while C only got 46, and B only got 6.
  So my worst pairwise defeat was the smallest."

B says:  "My opponent only scored 46 in my one defeat, while C's opponent got
  48, and A's opponent got 52.  So my worst pairwise defeat was the smallest."

C says:  "I lost by only 2 in my one defeat, while A lost by 4, and B lost by
  40.  So my worst pairwise defeat was the smallest."

Steve:  Does C win?
Mike:  Does B win?
DEMOREP and everyone else:  1) Using your best understanding of Condorcet's
method as discussed on this list, who wins?  2) Who do you think really ought to
win (or ought not to win)?

Bruce