Winning an election with 22% of the popular vote
Monday, November 1st, 2004A presidential candidate could be elected with as a little as 21.8% of the popular vote by getting just over 50% of the votes in DC and each of 39 small states. This is true even when everyone votes and there are only two candidates. In other words, a candidate could lose with 78.2% of the popular vote by getting just under 50% in small states and 100% in large states.
The optimal set of states to take (the one that lets a candidate win with the smallest popular vote) is not the N states with the smallest population. It's also not the N states with the smallest value for (population/electors), which would be optimal if you could get exactly 270 electoral votes that way.
The optimal solution happens to get exactly 270 electoral votes. In this solution, the winner takes DC, the 37 smallest states, the 39th smallest state, and the 40th smallest state. (The winner takes Alabama, Alaska, Arizona, Arkansas, Colorado, Connecticut, Delaware, DC, Hawaii, Idaho, Indiana, Iowa, Kansas, Kentucky, Louisiana, Maine, Maryland, Minnesota, Mississippi, Missouri, Montana, Nebraska, Nevada, New Hampshire, New Mexico, North Carolina, North Dakota, Oklahoma, Oregon, Rhode Island, South Carolina, South Dakota, Tennessee, Utah, Vermont, Virginia, Washington, West Virginia, Wisconsin, and Wyoming.)
Read on for my assumptions and algorithm.
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