Arithmetic Errors
Bruce Anderson
landerso at ida.org
Sat Apr 6 00:47:00 PST 1996
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
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