Showing posts with label Electoral College. Show all posts
Showing posts with label Electoral College. Show all posts

Friday, December 16, 2016

CoupCoup Clock


Cuckoo Clock by rones from https://openclipart.org/detail/219786/cuckoo-clock

* Electors meet in their respective state capitals on Monday, December 19, at various times, beginning with Vermont and perhaps Maryland at 10am Eastern (here's the list).

The final set of electors don't meet until 7pm in the Hawaii–Aleutian Time Zone (three guesses which state we're talking about there) -- midnight Eastern.

For obvious reasons it will be some time Tuesday before all elector votes have been cast and the totals are known. But we'll probably have a pretty good idea much earlier as to whether or not there's a coup attempt under way in the electoral college.

Monday, July 25, 2016

And Again With the Math ...


So, I've covered how many electoral votes, or states or congressional districts carried, it would require to possibly become president or vice-president of the United States via the "bust the electoral college and send the election to Congress" path. How many individual votes would it take?

The answer is in fact indeterminate, but it's possible to generate a reasonable estimate of the minimum.

We know that a presidential candidate could kick the election into the House, and become eligible for election to the presidency, with as little as one electoral vote.

And since Maine and Nebraska apportion their electoral votes by congressional district, it would be possible to pick one of those electoral votes up with as few as 1/3 + 1 of the individual votes cast in a congressional district.

No, I am not going to go through all of those two states' congressional districts to find the one with the lowest historical turnout. I'm just going to choose Nebraska's 1st US House District, and its presidential vote totals from 2012, as a reasonable proxy.

In 2012, a total of 264,712 presidential votes were recorded in that district. In a three-way race, 88,238 votes would be 1/3 + 1.

Of course, the number could go even lower than that -- lower turnout, or more candidates in the race (in 2012 there were four in the district: Mitt Romney, Barack Obama, Gary Johnson and Randall A. Terry), or whatever.

But it seems plausible to put the minimum number of popular votes required to become president via a House election at "fewer than 100,000."

Vice-president is, again, more complex. Remember, it's going to take 135 electoral votes and a second place finish to get to possible election via the US Senate ... and we can't just assume bare pluralities in the three of the four most populous states plus one state with 13 or more electoral votes.

Why? Because electoral votes are proportionally weighted toward the smaller states. EVERY state gets AT LEAST three electoral votes (two base electors plus one per US House district).

For example, the three least populous states -- Wyoming, Vermont and Alaska, with a combined population of not quite two million -- dispose of a total of nine electoral votes, while the four states disposing of nine electoral votes each (Tennessee, Indiana, Arizona and Massachusetts) have populations in the 6.5-6.8 million range.

I suspect a Ph.D.-level mathematician or statistician could model the problem and produce a number. But I'm not a Ph.D.-level mathematician or statistician. I've reached the level of my own mathematical incompetence.



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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