Election 2018: Vote Anyway.

I have some intense opinions about politics and generally, I’m happy to engage.  But I don’t want to make this a blog about politics.  If someone stumbles on this blog wanting to read about comics or mathematics or whatever they may not be interested in my opinions about candidate X or birthright citizenship or the current occupant of the Oval Office.  And that should be fine.  Some politics may sneak in from time to time but I’d like this to be a place that’s free from the most divisive arguments we’ve seen in my lifetime.

On the other hand, I’ve been fascinated with elections since I was 12.  I’ve done some work in voting theory and I’ve tried my hand at prognostication.  It’s been my  intention to eventually write about elections on this site.  But the problem, then, was what to write about?  We know the broad strokes of the 2018 election.  The democrats are doing remarkably well in the Generic Congressional Ballot and appear to be poised to retake the House.  That’s pretty remarkable given how heavily gerrymandered a lot of states are. Some of that has to do with the intensity of emotion engendered by President Trump.  It also helps that some of the most egregious gerrymandering we saw after the 2010 election has been overturned in the courts.

In the Senate, it’s a very different story. This is the class of senators that was elected in 2006, a Democratic wave that gave them the majority for the first time in four years. In 2012, despite defending more than 2/3 of the seats up for election, the Democrats actually increased their majority by two. So the Democrats are faced with what fivethirtyeight.com calls “the most unfavorable Senate mapthat any party has ever faced in any election.” Of the 35 senate elections being held this year, only 9 are held by Republicans and only one of those is in a state that’s bluish, namely Nevada. Meanwhile a lot of the seats being defended by democrats are in deep red states like North Dakota and Missouri. Despite being ahead on the Generic Congressional Ballot, it’s entirely possible that the Democrats will lose seats in the Senate.

Aside from the National stage, the most important elections that are happening this year are, in my opinion, the races for State Legislature. We don’t see much national coverage on these elections, but they’re crucially important. This is our first opportunity to elect some of the people who will be drawing the political maps in the wake of the 2020 Census. The candidates we elect now could determine control of the House of Representatives and of State Legislatures for a decade or more.

But all of this is known and it hasn’t shifted much. I could have written the last three paragraphs a month ago. Or two.  But the thing that motivates this post is that I stumbled across this.

A lot of folks pay attention to polls.  The polls influence their tendency to vote.

Democrats in Texas or Republicans in New York might decry their need to go to the polls because the opposition is going to “win anyway.”  But here’s the thing: according to this article (originally published in 2014) the average House poll has, since 1998, been off the final result by 6.2 percentage points.  Polls in senate races and gubernatorial elections have fared somewhat better, missing the final result by 5.1% and 5.2% respectively.  And polling is getting harder.  Response rates are declining making polls more expensive.  The decline in the prevalence of landlines along with laws about contacting people on cell phones are making it harder to get a representative sample.  You might think your Senate candidate is behind by three points, but the race could be a dead heat.

I see this graphic on Twitter a lot in Nate Silver’s feed.  The implication being made that Silver “predicted” that Clinton would win the White House and so, 538 “got it wrong.”  That’s not what this says at all.  This is a probability.  What this says is that, if you could repeat the election a bunch of times, Clinton would only win about 71.4% of the time.  In 28.6% of the “elections” Trump would be elected.  A Trump election isn’t surprising.

Imagine tossing a coin twice.  Would you be surprised if you got two tails?  You shouldn’t be.  The probability of that outcome is 25%.  Sure, it’s more likely that one of the other three outcome will happen, but it isn’t surprising at all.

The Trump victory, according to this analysis, is slightly less surprising than throwing two tails. The difference is that most people are not emotionally invested in the coins toss.

So, what’s the point?  Vote anyway.

Do you want the Democrats to win the senate?  Current estimates say there’s only a 1 in 6 chance of that happening.  Vote anyway.

Do you want the Republicans to retain control of the house?  Fivethirtyeight says they’ll “need a systematic polling error” for that to happen.  We’ve seen those before.  Vote anyway.

Do you want Heitcamp to get reelected in North Dakota, but you’re afraid she’s fallen too far behind?  Vote anyway.

Do you want DeSantis to win the Governorship in Florida but you think Gillum has pulled too far ahead? Vote anyway.

Not interested in the winner of the marquee race in your state?  The down ballot races and the initiatives are at least as important.  Vote anyway.

Can’t bring yourself to vote for either of the major party candidates?  You don’t have to use your vote to help determine the winner.  For example, here in New York the results of the Governor’s election determine which parties get dedicated ballot access.  You could vote to help the Working Families Party or the  Conservative Party or the Green Party or the “The Rent is Too Damn High” Party get on the ballot.  Vote anyway.

Elections are important.  We’d be a profoundly different country if everyone who could vote did vote.  But to quote Arron Sorkin or Benjamin Franklin or any number of people, “Decisions are made by those who show up.”  This one is really important.  No matter what you think is likely to happen, vote anyway.

Writing a Combinatorial Proof

This started life as a post on Quora, answering the question:

For any positive integer $n$, how do you write a combinatorial proof of the identity

$\displaystyle {2n \choose n} = \sum_{i = 0}^{n} {n \choose i} {n \choose n-i}$?

To write a combinatorial proof, the idea is to describe how each side of the equation is actually counting the same set of objects.

Now, if you have a set of $2n$ objects, $\displaystyle {2n \choose n}$ (that is $2n$ choose $n$) is the number of subsets of that set which contain exactly $n$ of them.

Now pretend that you’ve taken your $2n$ objects and put them into two boxes with $n$ objects in each. Think about what each of your terms on the other side of the equation, $\displaystyle {n \choose i}{n \choose n-i}$ represent. Then consider what you get when you add all of these together. You should be able to explain that this really does count the same thing as $\displaystyle {2n \choose n}$.

Now when I wrote the above, I wondered if this was a homework question someone had posted to Quora.  I didn’t fill in all the details since, while I’m happy to help someone with their homework, I don’t want to do it for them.

But in case anyone wants to see a worked out example, here’s the standard initial example of a combinatorial proof.  This is the identity that makes Pascal’s Triangle work as nicely as it does.

Theorem: $\displaystyle {n \choose k} = {n-1 \choose k} + {n-1 \choose k-1}$.

Proof: Let’s start by thinking about the expression on the left.  If we have a set with $n$ objects in it, $\displaystyle {n \choose k}$ is the number of ways we can select a subset of $k$ objects.  To say that another way, it’s the number of ways we can pick $k$ objects out of our set without caring what order they’re in.

Now suppose one of our objects has decided to wear a hat.  If we’re looking to select a subset with $k$ objects, we can decide to include or exclude the one wearing the hat.

Say we don’t want to include the guy with the hat.  In that case, all $k$ objects, have to be selected from the $n-1$ objects that aren’t wearing hats.  We can do this in $\displaystyle {n-1 \choose k}$ different ways.

Now suppose we decide to include the one with the hat.  Well then, to get $k$ objects altogether, we need to select $k-1$ more from the set.  There’s $\displaystyle {n-1 \choose k-1}$ ways to get the rest of the objects that you need.

Putting these two together we see that $\displaystyle {n-1 \choose k} + {n-1 \choose k-1}$ is also the number of ways to select a subset of $k$ objects from a set containing $n$ things.

Therefore, our identity must be true.

If you’d like to try one on your own, $\displaystyle 2^n = \sum_{i = 0}^{n} {n \choose i}$ is another nice example.

How do you prove a conjecture is false?

First posted to Quora on Friday, 7 September 2018

That depends on the nature of the statement.

If you have a universal statement, which is to say a statement that all of the things in some category share some property, you merely have to provide a counter-example.

So if you wanted to disprove the statement, “all prime numbers are odd” you’d merely have to point out that 2 is even and the statement cannot be true.

Disproving an existential statement is usually more work. These statements say that there is at least one thing that has a particular property. To disprove an existential statement, you need a general argument that that property can never happen.

So to prove that the statement “There is a pair of even integers whose sum is odd” is false, you must prove that the sum of any two even integers must be even.

Those are the cases “all” and “some.” The cases “none” and “some are not” are similar.

To disprove a statement like “None of the items in set A have property B” you simply have to find one that does. If you want to show a statement like “Some of the items in set A do not have property B” is false you need a general argument that everything in A has property B.

In any case, disproving a statement is equivalent to proving its negation.