### 12.25.2007

## Merry Christmas

Have you ever wondered what it would sound like if U2 dropped into a bar on karaoke night and Bono took the stage after having a few pints of Guinness? It'd probably sound something like this.

### 12.21.2007

## Letter grades bad. Numerical grades good.

If universities around the world (or at least UCLA) would adopt numerical grades of 2 or 3 digits of precision instead of letter grades, I wouldn't have to deal with emails from 10 students per quarter who think they got hosed.

That is all.

That is all.

### 12.19.2007

## From the archives: the mathematics of the electoral college

*This post was originally written on September 29, 2004.*

Many folks seem to think that in Presidential elections, voters in smaller states have more power than voters in larger states to influence the Presidential election. Actually, that's not really true. The two extra senators and possible roundoff errors in population don't come close to compensating for the fact that bigger states can more easily swing the election, thanks to electoral college bloc voting.

In general, in a state of population n, the probability of the vote being split equally is then

(n! / (n/2)! (n/2)!) * (1/2)^n.

We use Stirling's approximation to evaluate large factorials:

n! = (approx.) (n/e)^n sqrt(2 pi n)

Hence, the probability of the vote being equally split is approximately

((n/e)^n * sqrt (2 pi n) / (n/2e)^n * (pi n)) * (1/2)^n

= sqrt (2 / pi n).

If you live in a state of population n, this is the probability that you hold a swing vote in your state. (Yes, n should really be the number of voters, not the population. Oh well.)

Calculating the probability that your state is a swing state is a bit trickier. Suppose your state has k electoral votes; we're then interested in finding the probability that the other states give between 269-k and 269 electoral votes to Bush. Again it's hopeless to try to compute this probability explicitly for every state, but we can make a few simplifications.

Each state votes as a bloc; it gives either j or 0 votes to Bush. The variance (standard deviation squared) in this quantity is j^2/4. A basic theorem of probability is that if you have a bunch of independent random variables, the variance in the sum is the sum of the variances.

Looking at the current electoral vote map and summing j^2 over all states, we get that the sum of j^2 is about 10000. So the variance in the number of votes that the other states deliver to Bush is about (10000 - k^2)/4, where k again is the number of votes in your state. The standard deviation is the square root of the variance, so the standard deviation would range from about 43 if you're in California to almost 50 if you're in a small state. Call the standard deviation sigma, as is customary.

The basic simplifying assumption we can make is that the distribution of electoral votes in other states going to Bush is a normal distribution (bell curve) with standard deviation as calculated above, centered at (538-k)/2. The probability that your state is a swing state is the area of the probability function lying over the interval (269-k, 269); this interval lies in the center of the probability distribution and has radius k/2.

Recall that sigma is in the 43-50 range. The interval we're looking at has radius at most 28 (if you live in California); this is less than two-thirds of sigma. The bell curve is actually pretty close to flat if you only go two-thirds of sigma away from the center. So the probability that your state is a swing state is pretty close to a being linear function of k. The bell curve has height 1/(sqrt(2 pi) sigma) at the center, so the probability that your state is the swing state is about 1/sqrt(2 pi) (k/sigma).

Recall that the probability that your vote is the swing vote in your state is about sqrt (2/pi n). Hence the probability that your vote decides the election is about

sqrt (2 / pi n) (1/sqrt(2 pi)) (k/sigma) = 1/pi sqrt(1/n) (k/sigma).

Recall that k = (n / 660,000) + 2, or something like that (plus roundoff error). For n large, we can neglect the 2. The probability that a voter in a big state decides the election is thus proportional to sqrt(n), i.e. voters in big states have more power. If you do the math, the extra votes in states with 3 electoral votes help a bit; it's the voters in states with 4 or 5 votes that get screwed.

Plugging in some numbers:

California: n = 35,000,000

k = 55

sigma = 43

Probability of your vote being a swing vote in state= 1.35 x 10^-4

Probability of California being a swing state = 0.51

Probability of your vote deciding the Presidential election = 6.88 x 10^-5

Pennsylvania: n = 12,000,000

k = 21

sigma = 49

Probability of your vote being a swing vote in state = 2.30 x 10^-4

Probability of Pennsylvania being a swing state = 0.171

Probability of your vote deciding the Presidential election = 3.93 x 10^-5

Colorado: n = 4,500,000

k = 9

sigma = 50

Probability of your vote being a swing vote in state = 3.76 x 10^-4

Probability of Colorado being a swing state = 7.18 x 10^-2

Probability of your vote deciding the presidential election = 2.70 x 10^-5

Nevada: n = 2,200,000

k = 5

sigma = 50

Probability of your vote being a swing vote in state = 5.38 x 10^-4

Probability of Nevada being a swing state = 3.99 x 10^-2

Probability of your vote deciding the presidential election = 2.14 x 10^-5

Wyoming: n = 500,000

k = 3

sigma = 50

Probability of your vote being a swing vote in state = 1.13 x 10^-3

Probability of Wyoming being a swing state = 2.39 x 10^-2

Probability of your vote deciding the presidential election = 2.70 x 10^-5

A caveat: The "probability of your vote deciding the presidential election" is a conditional probability: it is the probability that your vote decides the election, conditioned on the event that everyone else votes randomly. We can't add up these probabilities, since they're conditioned on different events. That's how we get probabilities of the order 10^-5 in a nation of 300,000,000. But they are valid for comparing voter power in different states.

### 12.03.2007

## Oh, dear God

Apparently Michigan will play Florida in the Capital One Bowl. Since the last three games Michigan played against spread offenses with mobile QBs went so well, this should be a dandy.