[EM] Condorcet and Reversal Symmetry are incompatible with DMTBR

[Kristofer Munsterhjelm]

Here's a proof that we can't have all of Condorcet, reversal symmetry,
and dominant mutual third burial resistance.

It's a proof by exhaustion, and I see what Wikipedia means when it says
"A proof with a large number of cases leaves an impression that the
theorem is only true by coincidence, and not because of some underlying
principle or connection. "

I handle ties by saying that "X wins with certainty" if there is no tie
and "X wins with positive probability" if the probability of X being
winning after whatever tiebreaker is greater than zero.

-

DMTBR is incompatible with Reversal Symmetry and Condorcet:

Start with (A):

4: A>B>C
4: B>A>C
1: C>A>B

A is the CW and by Condorcet, wins with certainty.
Bury A on the 4 BACs to get (B):

4: A>B>C
4: B>C>A
1: C>A>B

If B wins with positive probability, we're done: (A)->(B) is a DMTBR
failure. Otherwise, either C wins with certainty or A wins with positive
probability. See the end for the case where C wins with certainty.

Suppose A wins with positive probability on (B). Reverse the ballots to
get (D):

4: A>C>B
1: B>A>C
4: C>B>A

By reversal symmetry, at least one of B and C must win with positive
probability on (D).
Suppose C wins with positive probability. Then flipping 4 CBA to CAB
gives (E):

4: A>C>B
1: B>A>C
4: C>A>B

where A is the CW, and so (E)->(D) is a DMTBR failure.

On the other hand, if B wins, then flipping 1 BAC to BCA gives us (F)

4: A>C>B
1: B>C>A
4: C>B>A

where C is the CW, so (F)->(D) is a DMTBR failure.

==================================
The case where C wins in (B):

Suppose that C wins with certainty. Consider the reversed ballots (D) again:

4: A>C>B
1: B>A>C
4: C>B>A

By reversal symmetry and our assumption that C won on (B), C can't win.
First suppose A wins with positive probability.
Relabel the candidates according to ACB -> abc. We get:

4: b>c>a
4: a>b>c
1: c>a>b

By relabeling and assumption that A won, a must win with positive
probability.
But this is just (B), where c must win with certainty by our initial
assumption, so we have a contradiction.

Hence B must win with certainty, and (F)->(D) gives us a DMTBR failure
as above.
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[Richard Lung]

Reversal symmetry is a property of Binomial STV. That is to say it
consists of both an election count and an exclusion count, the latter is
conducted in exactly the same way as the former, except the preferences
are counted in reverse, so that the quota becomes exclusive instead of
elective. The two counts are averaged using the geometric mean.
My new book on FAB STV introduces three more averages into the count,
each contributing to a more representative result.
About a week after initial publication, I added a chapter on Condorcet
method.

From
Richard Lung.

Post by Kristofer Munsterhjelm
Here's a proof that we can't have all of Condorcet, reversal symmetry,
and dominant mutual third burial resistance.
It's a proof by exhaustion, and I see what Wikipedia means when it
says "A proof with a large number of cases leaves an impression that
the theorem is only true by coincidence, and not because of some
underlying principle or connection. "
I handle ties by saying that "X wins with certainty" if there is no
tie and "X wins with positive probability" if the probability of X
being winning after whatever tiebreaker is greater than zero.
-
4: A>B>C
4: B>A>C
1: C>A>B
A is the CW and by Condorcet, wins with certainty.
4: A>B>C
4: B>C>A
1: C>A>B
If B wins with positive probability, we're done: (A)->(B) is a DMTBR
failure. Otherwise, either C wins with certainty or A wins with
positive probability. See the end for the case where C wins with
certainty.
Suppose A wins with positive probability on (B). Reverse the ballots
4: A>C>B
1: B>A>C
4: C>B>A
By reversal symmetry, at least one of B and C must win with positive
probability on (D).
Suppose C wins with positive probability. Then flipping 4 CBA to CAB
4: A>C>B
1: B>A>C
4: C>A>B
where A is the CW, and so (E)->(D) is a DMTBR failure.
On the other hand, if B wins, then flipping 1 BAC to BCA gives us (F)
4: A>C>B
1: B>C>A
4: C>B>A
where C is the CW, so (F)->(D) is a DMTBR failure.
==================================
4: A>C>B
1: B>A>C
4: C>B>A
By reversal symmetry and our assumption that C won on (B), C can't win.
First suppose A wins with positive probability.
4: b>c>a
4: a>b>c
1: c>a>b
By relabeling and assumption that A won, a must win with positive
probability.
But this is just (B), where c must win with certainty by our initial
assumption, so we have a contradiction.
Hence B must win with certainty, and (F)->(D) gives us a DMTBR failure
as above.
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