just how nakedly undemocratic are we gonna get?

I suddenly was in a number-crunching mood, so…

I listed all fifty states by the number of their electoral votes. I also found a chart of the states’ total populations (as of 2001).

Then I added up the high-EV states until I got to 268 (I had to leap out of order at the end), i.e., just barely losing. I further pretended that each state’s popular vote for its winning candidate was 50% plus 1 (the lowest possible). I further tweaked the undemocratic nature of this exercise by assuming that voter turnout in low-EV states was very low (20% of eligible voters) and very high in high-EV states (80% of eligible voters). I assumed the percentage of eligible voters was the same in each state.

What are the results? Let’s say Candidate A wins the following high-EV states (with, as I said, 50%-plus-one votes, with 80% voter turnout):

California, Texas, New York, Florida, Illinois, Pennsylvania, Ohio, Michigan, Georgia, North Carolina, Massachusetts

Candidate B wins every other state, with 50%-plus-one votes and 20% voter turnout.

Who wins the electoral vote? Candidate B, with 270 electoral votes.

Who loses the electoral vote? Candidate A, with 268 electoral votes…and 84.3% of the popular vote.

Of course, such a concatenation of just-barelies is extremely unlikely…but the point is that such a result is theoretically possible. Even with a more equitable assumption (that every state has the same percentage of voter turnout), our losing candidate wins 57.3% of the popular vote.

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2 Comments

Filed under geek, politricks

2 responses to “just how nakedly undemocratic are we gonna get?

  1. Aaron

    For maximum undemocraticness, candidate A should actually get 100% of the vote in the states he or she wins. It’s like racist gerrymandering– where it happened, we ended up with overwhelmingly black districts and narrowly (but unshiftably) majority-white districts.

  2. Aaron

    Wait, did I get that backward?

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