[EM] voter strat & 2-party domination under Condorcet voting

[Warren Smith]

On the probability that insincerely ranking the two frontrunners max and min, is
optimal voter-strategy in a Condorcet election.
----------Warren D. Smith Aug 2005----------------------------------------------

MATHEMATICAL MODEL: 3-candidate V-voter Condorcet elections
with random voters (all 3!=6 permutations=votes equally likely).

QUESTION: Is there a subset of identically-voting voters, who, by changing their vote
to rank the two "perceived frontrunners" max and min ("betraying" their 
true favorite "third party" candidate) can make their least-worst frontrunner win
(whereas, their true favorite cannot be made to win no matter what they do)?

THEOREM: The answer to the above question is "yes" with probability
at least  25% = 1/4  in the V=large limit.

PROOF SKETCH:
1. The probability of a condorcet cycle is exactly 25% if we 
ignore situations that include exact ties.  
(Since: Assume A>B wlog.  Then B>C with prob=50%, then with prob=50% C>A.)

2. The probability in the V=large limit tends to 1 that all the pairwise
victory margins are of order approximately sqrt(V), and that all the 6 kinds of voters 
occur with counts approximately V/6 each, i.e. much larger than sqrt(V).

3. So assume there is a condorcet cycle, wlog it is A>B>C>A, and wlog the smallest
margin of victory is C>A so that A is the winner (according to all the usual
Condorcet methods, since they all are the same in the 3-candidate case).

4. Choose a subset, of cardinality of order sqrt(V), of the voters of type "B>C>A".
(More precisely, we must choose the cardinality*2 to lie above the
previously-mentioned victory margin.)
If they betray their favorite B by insincerely switching to "C>B>A", 
then C becomes the Condorcet winner,
which from their point of view is a better outcome.
Q.E.D.

STRENGTHENING.
Note our "B>C>A"-type voter subset can argue that obviously, nothing they can do
will elect B, since when they rank B top honestly that fails to do it. Therefore,
their only chance for an improvement is to go for electing C.  And the only
way they can try is to raise C in the rankings.  As we've seen, this reasoning
yields success for them, 25% of the time.  However, given their preconception
that B has essentially no chance of victory, it actually makes sense
for them to rank C top 100%, not 25%, of the time, even though we know
this will only be successful for them with probability 25%.  Because given
their belief B has no chance, this cannot hurt them, and they know there is a
25% chance it will help them.  So we conclude from this that in fact, the "betray B"
strategy is plausibly better than honesty, 100% of the time.

Summary.
Adam Tarr in previous posts had questioned my claim that this this plurality-like voter
strategy could ever be optimal in Condorcet elections.  He said
"I can't easily imagine a scenario where it is useful in Condorcet."
He demanded that I "make some simulations that demonstrate
this, or at least show some examples."  He apparently had not noticed the
fact I had already exhibited an example on the CRV web site, 
   http://math.temple.edu/~wds/crv/RangeVoting.html
"Why range is better than Condorcet" discussion, perhaps because said example was in the
subpage  http://math.temple.edu/~wds/crv/IncentToExagg.html.
But now the present discussion shows that Tarr was maximally wrong: this strategy
is ALWAYS the right move.

Given that this is the case, we now can take it to be 100% certain that 
Condorcet voting methods will lead to 2-party domination, just like the flawed 
plurality system those methods were supposed to "fix", and just like experiemntlly
is true with IRV.  So anybody who is interested in third parties ever having
a chance, would be advised NOT to foolishly advocate either IRV or Condorcet,
but insetad would be advised to advocate RANGE VOTING (which experimentally
favors third parties far more than either plurality or approval, incidentally,
see the CRV web site).
-wds