        
Sunday, October 26, 2008
this page last modified on Sunday, October 26, 2008
The closer each state is to a tie, the more likely the election will end with the popular candidate losing.
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The distribution of votes for George W. Bush in 2000. States fall along the horizontal axis, the vertical axis is percentage of votes. States such as Massachusetts and New York (where Bush did poorly) are on the far left. Utah, Wyoming and Idaho (where Bush did very well) are on the right. Notice the distribution is roughly a straight line. The straight line means a linear distribution.
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As we all know, a presidential candidate can win the popular vote but still lose the election. It happened in 2000, and it happened before that. Exactly how unusual is it?
I was curious, so I wrote a simulator. In each pass, the simulator designs a hypothetical United States with a random number of states (between 20 and 90), a random number of electoral college votes (between 300 and 700) and random state populations. The simulator also randomly picks the votes in each state (assuming a two-party election). Each state's electoral votes are calculated and totaled, and then compared to the popular vote.
The answer? In a tight race where every single vote is essentially a coin flip, there's about a 16% chance of a mismatch between the popular vote winner and the electoral college winner.
My simulator is primitive (for example, I devised a simple way of distributing electoral college votes among the states; in reality, this is quite complicated. Google the term Huntington-Hill to learn more). But I did factor in the non-linear distribution of state populations (we have few big states, for example, but many small ones). I also factored in non-linear voting results (few states will vote in a landslide; most states will give between 40% and 60% of their votes to either party).
I ran millions of simulations. I played around with all of the non-linear distributions and discovered something interesting. The distributions of the state populations make no difference. Also, the absolute range of votes between the winner and loser didn't matter either (in other words, I allowed some simulations to have landslides and restricted others to razor-thin margains). None of it made a difference; there was still about a 16% chance of the popular candidate losing.
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When the distribution looks like this -- almost every state is nearly in a 50-50 tie -- the odds of the popular candidate losing climb dramatically. For this distribution, the odds are more than 30 percent.
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The only variable that made any difference was the distribution of votes within each state. Look at the graph above. In the 2000 election, the distribution was approximately linear. I tested linear and non-linear distributions in my simulator. Linear distributions (such as in the 2000 election) put the odds at 16 percent. I found that when most states are nearly-tied, the odds that the popular candidate will lose begin to soar. But when the states split into two a bi-polar condition (that is, every state is a landslide), the odds drop.
I pushed the simulation to see if I could find some limits. It turns out that the odds of the popular candidate losing NEVER exceed 35 percent. I determined this through simulations. I haven't proved it mathematically (I'll do that some other day). Going in the other direction, the odds never go to zero. Even when the states are dramatically polarized, there is always a chance the popular candidate will lose.
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When the distribution looks like this -- every state votes overwhemingly in one direction with no near-ties -- the odds of the popular candidate losing drop to just 8.9 percent.
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