Monday, July 18, 2016

Mathematical Possibilities

Up front disclaimer: No, I'm not looking to pick on Gary Johnson in particular with this post. He's far from the first or only person to say something similar to what I'm going to quote him saying. He just happens to be some combination of the most recent/most prominent, having said it in the New York Times, and having said it this year, and being on of the principals in a lawsuit related to it. Here it is:

The contention is on our part that if you're on the ballot in enough states to mathematically be elected, then you should be included in the presidential debate.

Q: How many states does a candidate have to be on the ballot in for it to become mathematically possible for that candidate to be elected president?

A: None.

Here's a scenario featuring a way that Johnson himself could be elected:

This November, Gary Johnson carries one state. Let's just assume that that state is New Mexico, which comes with five electoral votes.

Now, let's say that Hillary Clinton carries California (55), New York (29), Florida (29), Michigan (16), Ohio (18), Pennsylvania (20), Washington (12), Virginia (13), Massachusetts (11), Maryland (10), New Jersey (14), Texas (38), and Vermont (3). No, those specific states aren't likely; they were just the ones I picked offhand to demonstrate the math. They come with a total of 268 electoral votes.

That leaves the remaining states, which come with 265 electoral votes, for Donald Trump.

If no candidate receives 270 votes in the Electoral College, the US House of Representatives picks the next president from the three candidates with the most electoral votes. Which means that Gary Johnson could conceivably become president.

But, then, so could Jill Stein, who will be on the ballot in a number of states. Maybe not enough states to win in the electoral college, but as long as she carries at least one state and only Hillary Clinton and Donald Trump end up with more electoral votes than she does, she is still eligible for consideration by the House.

For that matter, if a write-in candidate (in states that allow them) carried a single state while holding the major party candidates below 270 electoral votes each, ditto.

Mathematically, a candidate doesn't need to be on the ballot in a single state for it to be possible for that candidate to be elected president.

[Update, 07/19/16: As Shawn L points out in comments, a candidate wouldn't even have to carry a state to be eligible for election by the House -- Maine and Nebraska apportion their electoral votes rather than assigning them "winner take all." So if (for example) a candidate got one electoral vote, and the other two candidates got 269 and 268 respectively,  all three would be eligible for election by the US House of Representatives. And now that I think about it, a "carried no states victory" could also occur under the auspices of one or more "faithless electors" - TLK]

Bonus question: How many states must a vice-presidential candidate win in order for it to be mathematically possible for that vice-presidential candidate to be elected?

Hint: It's a lot more complicated than the other question/answer set.

Addendum: This turned into a bit of a series. Check out Part 2 and Part 3.

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