Probabilities of "favorite-betrayal lesser-evil" (FBLE) situations in
3-candidate
ranked-ballot elections
Definition of "FBLE situation": Call the election-winner A. An FBLE situation
then
occurs when some C>B>A voters, by switching to B>C>A ("betraying their
favorite" C)
can make B win (an outcome they prefer).
Probabilistic model: All elections equally likely. That is, with V voters,
there are 6^V
possible elections since each voter can vote in 6 ways:
A>B>C, A>C>B, B>A>C, B>C>A, C>A>B, C>B>A.
(We disallow "truncated ballots" and "ranking equalities.") We shall consider
the large-V limit.
Theorem 1:
Basic Condorcet's FBLE probability is 25%.
Theorem 2.
Instant Runoff's FBLE probability is arctan(1/sqrt(2)/pi = 19.5913...%.
Isn't that amazing? :) I had previously analysed this latter
incorrectly and thought 25%.
Note.
In both theorems the CBA voters are at least as well off switching to BCA
IF they believe C's winning chances are well below 25% and 19.6% - thus
the betrayal in some sense is strategically justified with 100% probability in
both cases.
The proofs are at http://math.temple.edu/~wds/crv/IRVStratPf.html .
wds
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