Comments on: The chunky numbers that underlie electoral statistics

By: Eric Lund

Eric Lund — Tue, 06 Nov 2012 18:50:52 +0000

It should be noted that there are two states (Maine and Nebraska) which apportion their electoral votes by congressional district. But it’s rare for congressional districts in those two states to split–the first time it happened was 2008, when Obama carried NE-02 (basically metro Omaha). Obama gets to 332 EVs by carrying all of the swing states except North Carolina. To get to 331 or 333 would require a district in ME or NE, respectively, to flip. I don’t think Sam Wang’s model accounts for this possibility (he also missed NE-02 in 2008), but Nate Silver’s does. Silver gives Obama a 12% chance of carrying NE-02. The other two Nebraska districts are safely red, and both Maine districts are safely blue.

By: Jamie

Jamie — Tue, 06 Nov 2012 17:25:06 +0000

Perhaps the easiest way to think about it is that states come in uneven packets, ranging from 3 for less populous states to 55 for California. If a candidate wins a state, he gets that whole packet. To get a histogram where every possible vote count were possible (i.e., one that would be smooth — or as smooth as it can be for a fundamentally discrete thing like a count of votes), the packets would have to be all of size = 1.

The coin-flipping is Greg’s metaphor for the all-or-nothing nature of the electoral college. If it comes up D, the state’s electoral college votes go to President Obama. If it comes up R, they go to Mr. Romney and because of the chunkiness of the packets we are left with funny-looking histograms where some vote combinations are essentially impossible.

By: Greg Laden

Greg Laden — Tue, 06 Nov 2012 16:05:13 +0000

Well, it was kind of a joke.

OK, so here’s how you do it: First consider the simplified version. You take each state that is definitely for Obama or Romney and add them up to get one number (in this case, EV’s for Obama, but it could be either way). Then you take all the other state electoral vote counts that are up for grabs and assign them to one of the two candidates in all possible combinations, and come up with that number of counts and put them on the histogram. If Obama is way ahead on the “knowns” this will mean that only a small number of randomly assigned combinations would fail to put him over the top. Since the electoral college is mostly winner take all by state, the number of possible places along the “x axis” of the histogram is limited, there are many impossible numbers and some numbers that are more likely to occur with multiple possibilities.

The actual histogram uses a slightly different algorithm.

By: Dale

Dale — Tue, 06 Nov 2012 15:53:36 +0000

You lost me at flipping a coin with two heads or two tails. Why bother flipping? It’s the proportion of heads and tails that counts. Is there one coin for each elector? In short, I don’t understand how this histogram was generated. Could you explain for dullards like myself?