Our Electoral College System is Broken

A numerical Illustration

Originally posted: Nov. 9, 2004 
Last updated: Sep. 29, 2005

The Electoral College system in the U.S. relies on each state to vote for president elect. Those votes come in the form of electoral votes, and each state is allocated some number of electoral votes — from 3 to over 30 — based roughly on its population. Most states have a policy that whoever wins the largest percentage of the state’s popular vote (ballots cast by registered voters) will receive all of the state’s electoral votes. It is rare that the popular winner does not receive all electoral votes, but it has happened.

So, what kinds of problems can arise from this system? Many. Not only do you get an unfair concentration of attention on so-called “battleground “states” (those with large numbers of electoral votes and close contests), it turns out that the loser of the national popular vote can still win a majority (270) of the electoral college’s 538 possible votes. We saw this happen in 2000, when Al Gore received more of the national popular vote but lost the election.

My curiosity was piqued — exactly how much of the national popular vote could someone win and still lose in the electoral college? So, I decided to create a mathematical model to help answer this question. What would be the fewest number of actual voters someone would need to get to vote for him/her, and where would those voters need to live, in order to earn 270 electoral college votes. This page shows the result of that experiment.

RESULT: It turns out that a candidate for President need only win about 21.7% of the national popular vote in order to win the electoral college if those votes are in the right places. The converse of this is equally shocking: a candidate could win 78.3% of the popular vote and still not win the election.

Impossible, you say? Actually, it’s quite possible due to the structure of our electoral college system. I’ll use actual numbers from the 2000 election to illustrate my points.

The clue is that different states have different ratios of voters to electoral votes…sometimes wildly different. For example, in 2000, Wyoming had only 358,000 registered voters, but got 3 electoral votes (the minimum). So, Wyoming has only 73,390 registered voters per electoral vote. In contrast, Michigan had roughly 7,358,000 registered voters and 18 electoral college votes. That means there were 381,390 registered voters per electoral vote — over 5 TIMES as many as in Wyoming. In other words, one Wyoming resident’s vote is over 5 times more influential than one Michigan resident’s vote is on the election outcome.

So, how does that help my original question? Well, it turns out that if you do the following, you get the scary answer above:

1. Figure out the ratio of voters (people) to electoral votes for each state, and then rank the states from lowest to highest.

2. Start at the top of the list (Wyoming) and allocate just a smidgen over 50% of the popular vote to candidate A and the rest to candidate B. Each state for which candidate A gets over 50% of the popular vote then allocates its electoral college votes to candidate A.

3. Do that until you rack up exactly 270 electoral college votes for candidate A.

This table (see below) provides an illustrated example of these steps, or download the Excel spreadsheet and play with the model yourself.

The results are shocking:

— Only 21.7% of registered voters need to vote for candidate A in order for him/her to win 270 electoral college votes.

— The winner (candidate A) needs only win the majority of the popular vote in these 36 states/districts: Wyoming, D.C., North Dakota, Vermont, South Dakota, Alaska, Hawaii, Rhode Island, Delaware, Idaho, New Mexico, West Virginia, New Hamsphire, Nebraska, Maine, Nevada, Utah Montana, Connecticut, Mississippi, Arkansas, Iowa, South Carolina, Kansas, Arizona, Maryland, Oregon, Oklahoma, Alabama, Colorado, Tennessee, Wisconsin, Virginia, California, Georgia, and Kentucky. Notice how some of 2004’s “battleground states,” such as Ohio, Pennsylvania, and Minnesota, aren’t on the list?

— The loser (candidate B) could wind up winning 49.99% of the popular vote in those 36 states/districts and 100% of the popular vote in the other 15 states and still lose by two electoral college votes.

— Perhaps most useful is the realization that NATIONAL POLLS MEAN NOTHING. Just because candidate B is winning 55% to 44% doesn’t mean that candidate A couldn’t still win the electoral college handily. So, national polls be damned…I’m not listening to them any more. State polls, on the other hand, can be slightly more useful.

Assumptions: I had to make several assumptions in order to model this in any relatively straightforward manner. 
 — Only two candidates are running in the election 
 — Everyone registered in the state votes for one of the two candidates (100% turnout) 
 — Each state grants 100% of its electoral votes to winner of popular majority in state 
 — Registered voter percentages were imputed for Wisconsin and North Dakota (I had no data on registration counts for those states)

Since this is a heuristic-based solution, it may not be optimal, so the actual minimum % of the popular vote needed may actually be lower.

I’m not suggesting that we should rely purely on a popular vote for electing a president — that’s not a good solution either — but I do think we need to re-evaluate and revise our election system.

Sources: 
All data used for this analysis were obtained from the 2000 US Census (here) or the Federal Election Commission (which had pulled its 2000 voter turnout data, but thankfully Google’s cache still had it…you can download a PDF of the cached page here).



Originally published at craigfroehle.com.

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