Whose votes count the least in the Electoral College?
Dale R. Durran, University of Washington
In the days following the 2016 presidential election, many pundits and voters alike were stunned by the disparity between the popular vote, which went for Hillary Clinton, and the Electoral College, which favored Donald Trump.
If the president were elected by popular vote, every voter’s ballot would have been given equal weight, or influence, over the outcome, and Hillary Clinton would have won. But, as evidenced by Donald Trump’s victory, the Electoral College gives different weights to votes cast in different states. What are these weights, and how can we best compare them?
Most people believe the Electoral College weighs ballots in states with large populations much less than those in small states. For example, as the Washington Post noted shortly after the election, Wyoming has three electoral votes and a population of 586,107, while California has 55 electoral votes and 39,144,818 residents. Distributing the electoral vote evenly among each state’s residents suggests that individual votes from Wyoming carry 3.6 times more influence, or weight, than those from California.
The electoral vote total for each state is determined by its population relative to other states, plus two more votes equal to its representation in the Senate. Yet focusing on state population is not the most useful way to determine the relative weight accorded each state’s ballots. It does not help us understand how the weights assigned to voters by the Electoral College differ from the equal weights given to all voters in a popular vote. That’s because the popular vote weighs each vote according to the total turnout, not the total population.
As a professor who studies how mathematics can be used to model weather using computers, I was curious to make an apples-to-apples comparison between the Electoral College and the popular vote. I did this by using the number of ballots cast, rather than population, to compare the weight given to voters in each state by the Electoral College.
Large states such as California, Texas and New York do comparatively well under this analysis; it is the midsized states that fare the worst. These unexpected results help us understand whose votes carry the least weight in U.S. presidential elections.
Crunching the numbers
Roughly 136 million people voted in the 2016 presidential election. If we divide the total number of electoral votes – 538 – by the total number of voters, we can determine how much an individual vote counted toward an Electoral College vote. It turns out that, on a national average, each individual’s vote counted for about four millionths of one full Electoral College vote.
To find the relative weight of a vote in each state, I divided each state’s electoral vote total by the total number of ballots cast in that state, and then divided again by the exact fraction of an Electoral College vote accorded the average American voter (roughly four millionths). Let’s call this the “vote weight,” or simply the weight.
Note that, in a hypothetical system where the total electoral vote for each state equals the precise fraction of the total nationwide ballots cast in that state, votes in all states would be assigned weights of one, the same as in a national popular vote.
My calculations show voters in Wyoming did indeed receive the most weight, 2.97, for their votes. Voters in Florida came out on the bottom, with a voting weight of just 0.78. The weight given to the votes in Louisiana exactly matched the national average of one.
Two of the largest states, California and New York, came out only slightly below the national average. Votes from Texas, the second most populous state, actually received an above-average weight of 1.07.
Except for Florida, the states with the smallest weights are midsized, with between seven and 20 electoral votes each. For these states, there is no systematic relationship between vote weight and each state’s electoral vote total.
The surprisingly low weights carried by votes in many midsized states are partly explained by the difference between the population – which determines the number of electoral votes – and the actual number of eligible voters in each state. Eligible voters do not include those who are too young to vote, noncitizens and, in some states, prisoners or former prisoners.
But it is voter turnout that primarily explains the low vote weights in states with seven or more electoral votes. In fact, the state-to-state difference in voter turnout was the most important factor in determining the variation of vote weights in midsized and large states in the 2016 presidential election.
In contrast to the weak relationship between a state’s weight and its electoral vote total, the weight attached to each vote clearly tends to decrease as voter turnout within a state increases.
It is hardly surprising that higher turnout within a state decreases the weight accorded to each ballot, because the fixed number of electoral votes for any given state must be shared among the total number of ballots cast. But it does seem remarkable that the link between turnout and weight is so much stronger than the link between the number of electoral votes and weight.
For example, consider the difference between Oklahoma and Oregon. Both states have seven electoral votes, but their weights, 1.22 and 0.89, are quite different. That’s primarily because only 52 percent of eligible voters turned out in Oklahoma – much less than the 66.6 percent turnout in Oregon. That gave Oklahomans a greater weight per vote in the Electoral College than their fellow citizens in Oregon.
In another example, South Carolina saw a relatively low 56.8 percent turnout, versus the much higher 69.8 percent turnout in Colorado. Both states have nine electoral votes. But, as a consequence of voter turnout, ballots in South Carolina received a weight of 1.09, while those in Colorado were given a much lower weight of 0.82.
As a final example, consider the pair of states with 29 electoral votes each: New York and Florida. New York’s weight of 0.97 exceeds Florida’s weight of 0.78, mostly because turnout in New York was 55.7 percent, while that in Florida was 64.5 percent.
Untangling the Electoral College
My analysis does not completely capture the many ways in which the Electoral College modifies the influence attached to individual votes.
For example, it does not take into account our winner-take-all system, where all of the electoral votes from each state are awarded to whoever wins the majority of the popular vote. (The congressional district method used in Maine and Nebraska has the same effect, just aggregated into smaller units.)
Consider the election’s “battleground states,” where the election was decided by about one percentage point or less. The relative weight of votes in four out of five of these states was less than 0.85.
The low weights for these states are largely due to high turnout, which was likely increased by the residents’ awareness of the significance of their votes. While this increase in turnout did lower the weight of each vote in a battleground state, few would assert that the voters in these states had less opportunity to influence the presidential election than voters living in less competitive states with a high vote weight.
The way the Electoral College rewires American presidential elections in comparison to a simple popular vote is clearly complex. The Electoral College does add extra weight to votes cast in the least populated states. But the way this system treats voters in the remaining states is not well-understood. In states with seven or more electoral votes, it tends to weigh votes based on that state’s voter turnout, rather than its number of electoral votes.
Whatever one’s political affiliation, it is hard to be enthusiastic about a system that penalizes voters in high-turnout states.Comment on this article
Dale R. Durran is a registered Democrat.
University of Washington provides funding as a member of The Conversation US.