Democratic elections permit us to vote for whomever we perceive has
the highest expectation to do better with the issues we care about. Let’s
simplify and assume we can quantify how satisfied we are with an
elected official’s performance. Denote this quantity with *X*. Because
when we cast our vote we still don’t know for sure how the candidate
will perform, we base our decision on what we expect, denoted here with
*E(X)*. Thus we try to maximize *E(X)*. However, both political theory
and data tell us that in US presidential elections only two parties
have a non-negligible probability of winning. This implies that
*E(X)* is 0 for some candidates no matter how large *X* could
potentially be. So what we are really doing is deciding if *E(X-Y)* is
positive or negative with *X* representing one candidate and *Y* the
other.

In past elections some progressives have argued that the difference
between candidates is negligible and have therefore supported the Green Party
ticket. The 2000 election is a notable example. The
2000 election
was won by George W. Bush by just five electoral votes. In Florida,
which had 25 electoral votes, Bush beat Al
Gore by just 537 votes. Green Party candidate Ralph
Nader obtained 97,488 votes. Many progressive voters were OK with this
outcome because they perceived *E(X-Y)* to be practically 0.

In contrast, in 2016, I suspect few progressives think that
*E(X-Y)* is anywhere near 0. In the figures below I attempt to
quantify the progressive’s pre-election perception of consequences for
the last five contests. The first
figure shows *E(X)* and *E(Y)* and the second shows *E(X-Y)*. Note
despite *E(X)* being the lowest in the last past five elections,
*E(X-Y)* is by far the largest. So if these figures accurately depict
your perception and you think
like a statistician, it becomes clear that this is not the election to
vote third party.