Largest Possible Percentage of the Popular Vote for a Losing Presidential Candidate?

(Not sure if this topic should be in Politics or GQ - It’s a mathematical question about a set of political rules.)
We all know that the electoral votes determine the the next President and it’s possible for the losing candidate to have received substantially more popular votes than the victor. I’m curious to know the mathematical limit of the percentage that the mismatch can achieve. The solution starts out ‘relatively’ simple when considering the winner-takes-all states (i.e. 100% of votes in all states won by the loser and [ [# of registered voters in state] / 2] - 1 for the states lost). Then it becomes more complicated.

Technically, a winning candidate can win by a vote of 1 vote (just one single solitary voter, nobody else showing up in the entire state) in the top twelve most populous states, plus one vote apiece for the Maine and Nebraska districts. So, 14 votes in total for the loser. That gives him 270 electoral votes in total.

Then the losing candidate can carry every single other state by a 100% total vote turnout, unanimous victory, every single voter going for him. That gives him 268 electoral votes in total.

I can’t do detailed math (I’m at work,) but it must surely come out to something like 100 million for the loser, vs. 14 votes for the winner. So, a winning percentage of 99.9999% in the popular vote.

Or, if you want full participation to rule out that case, you can win all of those twelve states by a margin of 1 vote, with all of the rest still being blowouts the other way.

That approach gives the smallest number of votes to win 270 EC votes.
To get the largest possible % of the popular vote for the losing presidential candidate as asked by the OP the winner needs a single vote in the smallest states sufficient to win 270 and cede 100% of the largest.
I think that equates to 41 states with New Jersey being the swing state.

Ah, gotcha, I see.

But the difference is miniscule. We’re talking about the loser winning the popular vote by 99.99 percent vs. winning by 99.99999 percent. Something like that. Either way, the loser wins the popular vote by what essentially amounts to unanimity, yet still loses the Electoral College.

But the difference is miniscule.


If one wants to be more realistic, you’d deal with the actual number of voters, and not dream about just one voter in the entire state. In such a case, you would actually want the LEAST populous states.

For example, in Nov 2020, California’s 55 electors were decided by 17,500,881 voters. That’s more than 318,000 voters per elector. But if you would focus on Wyoming, you could get three electors for only 276,765 voters!

About 10-20 years ago, I calculated that if you got 270 electoral votes from low-population states where Candidate A got 51% of the vote, and the other 268 electors were from high-population states that went 100% for Candidate B, then A could win with as little as about21% of the popular vote.

I’m too lazy to figure out the exact numbers again.

And what if you add a third party? They get enough for two electoral votes. Neither of the other two get 270. It goes to the House, and they elect the 2 EV person.