### the numbers game

Tangental to another conversation, my friend N asserted that while a candidate could win the presidency with less than 50% of the popular vote, s/he would still need "close to half" in order to get the requisite 270 electoral votes. I set out to prove her wrong. For those of you who are not familiar with my country's electoral system (and since most of my readers are U.S. citizens, I hope that's not too many of you), here's the quick explanation: Each of the fifty states are assigned a number of electoral votes equal to the number of the state's U.S. Senators (always two per state) plus the number of the state's U.S. Representatives (apportioned by population, minimum one per state, total 435). The District of Columbia, which has no representation in Congress (a rant for another day) receives 3 electoral votes as well, bringing the total number of votes in the Electoral College to 538. A candidate needs 270 electoral votes in order to win the presidential election. Each state's electoral votes are granted as a block to the winner of that state's popular vote.

That doesn't sound too bad, until you consider that electoral votes in some states (generally speaking, the least populous) are worth more than those in others.

How so? Well, for starters, two of the electoral votes in each state are granted regardless of that state's population, meaning that in Vermont (1 Representative) these votes triple the state's electoral voting power, whereas in California (53 representatives) they add less than 4% more voting power to the state. Furthermore, since every state must have at least one U.S. Representative, and of course no state can send half a Representative to Washington (pity, that), in some states (such as Rhode Island) there are a mere half-million people for each Representative, whereas in other states (like Delaware or Montana) there are closer to 800,000 or more. It all boils down to an electoral vote in Wyoming weighing more than four times as much (in terms of population behind the vote) as one in California.

Don't believe me? I did the math and it's all in this nifty little spreadsheet.

"Ah, but wait--" you say, "you used total population! Obviously not all these people are registered to vote!" OK, then. You give me the number of registered voters in each state, and I'll use those. In the meantime, I'll settle for assuming that the proportional amount of registered voters to total population in each state is roughly the same.

Anyway, the second sheet of that Excel file is set up so you can play some electoral math games. It is currently sorted so the states with the highest weight electoral votes (fewest voters behind each vote) are at the top, and the lowest-weight states are at the bottom. In the scenario I have set up in the saved file, Party A has garnered 271 electoral votes (more than the required 270) with less than 22% of the popular vote! Of course, I finagled this by having Party A win by one or two votes in the 39 most heavily-weighed states, and having Party B sweep the other 12. I agree that this scenario is highly unlikely (given the nature of elections in general and the groupings of states' politcal leanings in particular), but it does show just how misrepresentative our electoral system really is. I suspect (though I haven't played this out yet) that in an election with three candidates, with states that give their electoral votes to the candidate who garners a plurality of votes in that state (rather than a majority), the popular vote percentage necessary to win would be close to 15%.

Feel free to play with that set-up and come up with more realistic projections that still have one party winning with far less than 50% of the popular vote. You should enter only the votes for Party A (column E); everything else will calculate itself out.

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