[EM] Strong FBC
Forest Simmons
fsimmons at pcc.edu
Fri Feb 14 14:30:18 PST 2003
It seems to me that any neutral method that gives a three way tie to a
reverse order pair (like the following ballot pair) cannot satisfy both
Pareto and the strong FBC:
1 A>B>C
1 C>B>A .
Here's my reasoning. Suppose that there are only two voters and one has
already voted A>B>C. Suppose further that your utilities for candidates
A, B, and C are zero, 80, and 100, respectively.
If you vote sincerely, then a tie results, and your expected utility is 60
= (0+80+100)/3 . If you vote CAB (insincerely but still loyal to
favorite), then Pareto rules out B, so the result is (1) a win for A, (2)
a win for C, or (3) a tie for A and C.
(1) has utility zero, (2) has utility 100, and (3) has expected utility of
50.
So the only loyal utility expectations are 0, 50, 60, and 100. If we can
surpass 60 by betrayal of favorite, then the FBC implies that we can also
surpass 60 by being loyal. So let's see if we can surpass 60 without
voting favorite first:
How about just staying home from the polls or voting a spoiled ballot?
The result will be a win by B with utility 80, which is greater than 60.
So we see that we can surpass 60 without ranking favorite first, therefore
by the FBC, we can surpass 60 while ranking favorite first. So the ballot
set ABC+CAB must yield C as winner.
However, we will see that this yields a contradiction:
Suppose that (in a different but similar election under the same rules) my
utilities are 100, 80, and zero for A, B, and C, respectively, and that
the only other ballot cast in the election is CAB.
If I vote sincerely (ABC) then according to the above argument the method
winner has to be C with utility zero.
If I remain loyal to my favorite but bury compromise, the ballot set
becomes (ACB+CAB), which by Pareto and neutrality must yield a tie between
A and C with an expected utility of (100+0)/2 = 50, for me.
So the best I can do without betraying favorite is 50.
But if I betray favorite and vote BAC, the ballot set becomes (BAC+CAB),
i.e. a reverse pair, which (according to our original assumption) is a
three way tie, with an expected utility of 60.
Therefore it is to my advantage to betray favorite C, i.e. the FBC does
not hold after all.
This contradiction, shows that the original assumptions are incompatible.
These kinds of arguments can be extended to the other possibilities for
reverse pairs, as well, (i.e. the middle guy wins as in Bucklin or the
outside guys tie.)
To be continued ...
Forest
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