Nader Vote = 1/2 Bush Vote

In the 2000 Election, which was primarily a duel between the two four-letter candidates Bush and Gore, the warning “A vote for Nader is a vote for Bush!” was often bounced around.

Of course, in one sense it’s kind of right. Nader tallied 97,421 votes in Florida. Most of these voters probably would have voted for Gore if they had known the actual outcome in advance, and our world would be quite different today. Of course, if they knew the actual outcome in advance, then it wouldn’t have been an actual outcome, but now I’m arguing myself down an infinite wormhole in space.

There are problems with this aphorism. The electoral college voting system, where “winner takes all” in a state meant that voting for Nader in a state that was statistically safe for Gore (or Bush!) was irrelevant to the outcome. In most states, you could send-your-message/throw-away-your-vote care-free. Not Florida, of course, but most states.

Most annoyingly, for me at least, is the numerical inaccuracy. A vote for Nader is not as destructive to Gore as a vote for Bush. It’s half as destructive.

Consider a simplified election with 15 voters: 9 left wing, who prefer Gore or Nader, and 6 right wing, who prefer Bush.

If all lefties vote Gore, he wins, 9-6.

If 4 of them to vote for Nader, Gore loses, 5-6 (and Nader gets 4).

If 2 of them vote for Nader, Gore wins 7-6 (and Nader gets 2).

If 2 of them vote for Bush, Gore loses 7-8.

So as you see, a vote for Nader is not the same as a vote for Bush. You need twice as many voters to switch to Nader for Bush to win, so it’s only half as damaging.

I guess “A vote for Nader is half a vote for Bush!” sounds a little weird. Nevertheless, it’s sad that once again, mathematically accurate details lose out to marketing poetry.

(Ah yes, a blog posting in 2009 about the election of 2000. Whattya think: Too soon?)