FiveThirtyEight.com: Politics Done Right: 51,199,463,116,367 (Shout-Outs)

Wednesday, June 11, 2008

51,199,463,116,367 (Shout-Outs)

Congratulations to Isabel Lugo, a third-year PhD student at the University of Pennsylvania, who was the first person to come up with a solution to the number of unique winning scenarios in the electoral college, as well all the people who helped Isabel to get there and confirm the result. To my surprise, this became the most commented-upon thread ever at FiveThirtyEight.com, and is quite possibly my favorite.

...

In unrelated news that probably isn't quite worth a thread unto itself, I will be a guest tomorrow morning on CNN's American Morning with John Roberts. The segment will air live at roughly 8:15 AM Eastern. Postponed because of CNN's tornado coverage. This is why one should always be very cautious about touting a live TV gig. But we're trying to get rescheduled for tomorrow.

26 comments

Drew Miller said...: Can we do Maine and Nebraska now, too? :-); June 11, 2008 10:53 PM
Lukeness said...: Now you can keep busy between now and November with a post for each of the possibilities and an analysis of its likelihood!; June 11, 2008 10:58 PM
Another Mike said...: No recognition for Andras, who posted the correct answer at 5:15, 7 minutes before Isabel?; June 11, 2008 11:03 PM
Juris said...: Andras may have been Isabel, for all we know. I'd like to know who he is. That may remain a mystery.

I think it's reasonable for Isabel to be named the winner because she not only came up with an answer that others later verified using alternative methods but it was a nice solution and she "showed her work."; June 11, 2008 11:19 PM
Anonymous said...: For anyone interested in calculating with Maine and Nebraska split:

Nebraska has the following distinct possibilities for its districts and popular vote:

1|01001101001101
2|00101010101011
3|00010110010111
P|00000002222222
T|01112223334445

Maine has the following distinct possiblities for its districts and popular vote:

1|010101
2|001011
P|000222
T|011334

Neither of these have an equivalent set of smaller contests which result in the same possiblity space, so if you are going to attack this problem, you're going to have to use a different approach than most were taking in the other thread (or at least significantly modify one of them).; June 11, 2008 11:27 PM
Clemens said...: any chance you were talking to George Will? from his column in WaPo

5: That is the number of commas in the number of possible combinations of jurisdictions that can give a candidate 270 or more electoral votes. The votes disposed by the jurisdictions range from one (the Maine and Nebraska congressional districts) to three (7 states and D.C.) to California's 55, with 17 different numbers between three and 55.; June 11, 2008 11:45 PM
Anonymous said...: One thought I had on this topic is that if we define the win as discussed here as "maximally fragile", then we can obtain a political lemma: in real life, no party can win both DC and Wyoming without a landslide which (obviously) is not maximally fragile. Define "narrow victory" as 272 EV or less. It then follows that any maximally fragile victory must in real life be also narrow. That's disappointing as a maximally fragile victory would be interesting if you could point out at the victory and say: "you know, even though the margin appears solid, the win is actually maximally fragile"; but in real life maximal fragility would always be narrow and trivially obvious.; June 12, 2008 12:00 AM
Peter B Fitzgerald said...: In addition to what Clemens said, that whole Will piece looked like it was written by a 538 devotee ;); June 12, 2008 12:00 AM
Alex said...: Congrats to Isabel. Now that the algorithm is available it wouldn't be too hard to mix in the Nebraska/Maine results...simply add in however many districts by their EV at one EV apiece, and do it again.; June 12, 2008 12:19 AM
Alex said...: By the way Nate, I sent your problem to a computer science professor at Cornell that I TA'd for, and he said he'd would love to use it in class/as homework. Improving the world in more ways than one, you are =).; June 12, 2008 12:20 AM
Juris said...: A quantity that would probably be easier to calculate than fragility is that of disproportionality between the EV total and the popular vote (or the voting age population).

Given that small states get a disproportionately large allocation of electoral votes, it's likely that the winning EV combinations that involve the largest number of 3-EV states will also be the most disproportional.

I would conjecture that they're among the combinations that would also be most likely to lead to a situation in which the popular vote majority or plurality goes to one candidate and the EV majority goes to the other.

So a lot of those winning combinations could turn out to be nasty from the perspective of what's generally regarded as fair.

(This leaves aside Nate's calculation of the natural gerrymand that's occuring due to population change since the last redistricting in 2002.); June 12, 2008 12:21 AM
Drew Miller said...: His article says "4" right now for the number of commas, which I think is wrong (I get slightly over 1 quadrillion not even counting the Maine and Nebraska issues, although they certainly wouldn't push it to the quintillions.; June 12, 2008 12:34 AM
Brian said...: Damn, it certainly would've saved us all some time and effort if we'd known that the number of commas was all we had to calculate.; June 12, 2008 12:54 AM
Isabel Lugo said...: To answer the question that's been raised multiple times: Andras is not me.; June 12, 2008 7:13 AM
Mark said...: Congrats to Isabel, but I really think Andras is being short-changed here!; June 12, 2008 8:41 AM
Juris said...: Andras is not as rare a name among mathematicians as one might at first assume. Then again, it's far from being "Robert." I ran across one who is a specialist in combinatorics -- but what are the odds that he wandered over here to 538.com during an idle hour in Grenoble? http://www-leibniz.imag.fr/DMD/dmd.html; June 12, 2008 10:18 AM
Dave said...: To solve the Main/Nebraska problem, break it down into cases and use Isabel's solution.

For the case with M/N both having 0 votes, the solution is the same with just the remaining 49 states (yes, I know D.C, isn't a state).

For the cases with M or N receiving non 0 votes, the solution for the remaining 49 must be exactly 270 minus M and N votes. So use Isabel's solution on each with only the x^(270-M-N) coefficients.

You can also have solved the original problem by induction. List the states by increasing size. With:

E(n)=State n's votes allocated
C(n,v) = combinations for first n states to get v votes w/o excess

Then:

v>0: C(n,v)=C(n-1,v) + C(n-1,v-E(n))
v<=0: C(n,v)=1

C(1,v)= (1 if v<=3; 0 otherwise}

The first term in the recursion is where the n'th state votes no. The second is where it votes yes. These are obviously seperate solutions sets. The first is valid by definition.

The second term clearly has enough votes, and the first n-1 states are not excessive, and the n'th can not be because then a smaller state would be. Except if there were no earlier states, then the n'th would be excessive if v<0, that's why we need a special case.; June 12, 2008 11:39 AM

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Wednesday, June 11, 2008

51,199,463,116,367 (Shout-Outs)

26 comments