**Like Life, Distribution Ain’t Fair**

## Gems in STEM: Discussion of Dean’s Method

Let’s time-travel back to kindergarten. You and I have found 4 cute rocks during recess, but we have to split them between us to take home to our rock collections. Easy enough, we each get two rocks–that’s only fair!

But what if there are five rocks? Who decides which of us gets the extra rock and who gets the short end of the stick (or should I say rock)? What if I found more rocks (because I rock) or you care more about them? If we can’t figure this out, we’ll need to battle it out…race you to the swings! Last one there is a rotten egg.

While this decision doesn’t really matter, we all learn the concept of fair distribution pretty early on. The same concept comes up in one of the most important parts of the United States government: the US House of Representatives. In the House, there are 435 delegates that must be distributed amongst 50 states as fairly as possible so that the citizens’ votes will be accurately represented in the Electoral College to determine our president…a decision slightly more important than sharing rocks. (I want to quickly note that the Electoral College is a very flawed system, but that’s besides the point for right now.)

These representatives are usually elected to give a voice to the constituencies in certain geographic regions. **How do we assign representatives to states so that it is proportional to their respective populations?**

Wait, why do we want it to be proportional? Well, think about this. California (the best state) has a population of over 39 million people, whereas Wyoming has a population of around 580,000 people. If California and Wyoming both got one representative, a Californian’s vote would be worth like 1/39,000,000 of a vote while a Wyomingite’s vote would be worth about 1/580,000 of a vote — about 70 times more! Not super fair to us Californians.

Unfortunately, it’s impossible to *perfectly* distribute delegates proportionally since that would give most states a non-integer number of delegates (and we can’t exactly have .73 of a person). So, we have to somehow decide which states get more or less delegates than they are mathematically entitled to. #Favoritism?

To help us make this decision, we have a variety of **apportionment methods**, which are basically just algorithms to decide where these extra delegates go.

An example of one of these methods is **Dean’s Method**, first proposed by Professor James Dean at Dartmouth College in 1832. The basis of this method is something called the **harmonic mean**, which is a popular kind of numerical average!

## Harmonic Mean

Harmonic mean is one of the three so-called Pythagorean means; the other two means are the more popular arithmetic mean and geometric mean. For example, the arithmetic mean of a and b is just the normal average of them: (a+b)/2. The geometric mean is defined to be √(ab). Fun fact, the value of the harmonic mean is always smaller than the other two Pythagorean means!

Harmonic mean is calculated by taking the reciprocal of the arithmetic mean of the reciprocals of the given values…phew, that was a mouthful. Let’s look at an example to make sense of it!

Say we have two numbers: 4 and 6. The reciprocals of the 4 and 6 are ¼ and ⅙, by definition. Then, the arithmetic mean, which is just the average, of ¼ and ⅙ is (¼ + ⅙)/2 = (5/12)/2 = 5/24. Finally, the reciprocal of their arithmetic mean is 1/(5/24) = 24/5 = 4.8. So, the harmonic mean of 4 and 6 is 4.8. Weird…who even uses this?

Well, this unique numerical average comes up in situations where we want to find some sort of average rate. While the harmonic mean is not as well-known as the other Pythagorean means, this unique numerical average comes up in situations that have to do with some sort of average rate. Once again, let’s do an example!

## How fast does Rainborca Bike?

Suppose Rainborca is biking to the ocean from home. On her way there, she averages 4 miles per hour (mph), and on her way back, she averages 6 mph. Because we’re nosy, curious mathematicians, we ask her what her overall average speed for the entire trip was! At first, she makes a common mistake and says her average speed was (4 + 6)/2 = 5 mph, applying the arithmetic mean to this question. Unfortunately, this doesn’t work because Rainborca didn’t travel for the same amount of time on the two trips.

The *correct* average speed can be found using harmonic mean! From before, we know that the harmonic mean of 4 and 6 is 4.8, so her average speed is 4.8 miles per hour.

How can we be sure this is true?

Well, we know the average speed/rate is equal to the total distance over the total time, using the common equation* distance=rate*time*. If we let the distance between the ocean and her home be *d*, the total distance is *d+d = 2d*, because she goes there and back. Since *distance=rate*time*, we have that *time = distance/rate*. So, the time Rainborca takes to bike to the ocean at 4 mph is *d/4* and the time she takes to bike back home at 6 mph is *d/6*, so the total time she took on her trip is *d/4 + d/6*.

Now, we can calculate that Rainborca’s average *rate* = *distance/time = 2d/(d/4 + d/6) = *2/(1/4 + 1/6) = 2/(5/12) = 4.8 miles per hour, aka the harmonic mean of 4 and 6. Voila! It’s like magic…or is it just math? ❤

The harmonic mean is also often used in circuit analysis to find the equivalent resistance of resistors in parallel, capacitors in series, or inductors in parallel, to name a few examples! There are also many applications for something called the **weighted harmonic mean**, which associates corresponding weights to each value in the mean. However, since Dean’s Apportionment Method is only concerned with the standard harmonic mean (where all weights are equal), we won’t get into the details of the weighted version.

Speaking of Dean, let’s get back to our original discussion of apportionment! (I guess the conversation always somehow returns to politics…)

## Dean’s Method

Dean’s Method shares similarities with the more popular Webster’s Method and Hill’s Method, except Dean’s Method uses harmonic mean as the cutoff for rounding whereas Webster’s method uses arithmetic mean — the conventional average — as the cutoff point and Hill’s Method uses geometric mean.

Dean’s Method is actually rarely used because it tends to favor small states more than other methods. Remember our fun fact? Let me remind you: the value of the harmonic mean is always smaller than the other two Pythagorean means! So, Dean’s Method lowers the threshold needed for small states to round up to another delegate, meaning, in the name of fairness, we can’t use it in real life. (Turns out, the US uses Hill’s Method to distribute delegates!)

Okay, enough talk. How do we actually run Dean’s Apportionment Method?

Well, we first need a few definitions. (Heads up, things might get a little more detailed/technical here on out, so hang in there!)

We say that the **standard divisor **of a population is the fraction of the total population divided by the number of representatives, which gives the number of people represented by the average representative. Then, a state’s **standard quota** is the population of the state divided by the standard divisor.

The first step of Dean’s method is to calculate the standard quota for each state. This number is almost definitely not going to be an integer (AKA whole number), so we calculate the harmonic mean of the two closest integers to the standard quote and round it according to this mean. If the standard quota is above the harmonic mean, round up by one, and if the standard quota is below the harmonic mean, round down by one. To make this make sense, let’s do a quick example! Say the standard quota is 5.37. Then, the two closest integers are 5 and 6 and the harmonic mean of 5 and 6 is 1/((⅕ + ⅙)/2) = 60/11 = 5.45. Since 5.37 is less than 5.45, we round down to 5, and that is the number of delegates for the state!

Cool, so let’s just do this for every state. Easy, right? Nope. When has politics ever been easy? In this case, when we sum up the delegates for all fifty states, it *should* come out to a total of 435, but it doesn’t always. If it sums up to less than 435, that means there are *surplus* delegates, i.e., extra delegates that aren’t assigned to any state. To fix this, we need to decrease the standard divisor to modify all the quotas until it totals up to 435. On the flip side, if the total number of delegates is over 435, we increase the divisor to modify the quotas until the total goes down to 435.

Weird. Why are we doing this? Does it actually fix things?

Well, the reason this works is because if we go back to our definitions, we can see that the standard divisor is the denominator of the standard quota. This means that as its value increases, the fraction decreases, and vice versa. (For example, 2/5 < 2/4 < 2/3.) So, using trial and error, we adjust the standard divisor until we reach the correct total of 435 delegates, then use those delegate numbers as the apportionment. Ta-da!

## Welcome to Hogwarts

To see Dean’s Method in action, let’s look at an example! Boy, do we love examples around here.

Suppose we have 20 new students to apportion to four houses Gryffindor, Slytherin, Hufflepuff, and Ravenclaw. (I’m going to pretend I’m one of these students to fulfill my childhood dream of going to Hogwarts.) Now, Gryffindor has a population of 2,964 people, Slytherin has a population of 3,220 people, Hufflepuff has a population of 829 people, and Ravenclaw has a population of 4,987 people. So, the total number of students in the 4 houses is 12,000, and the standard divisor is 12000/20 = 600.

Recall that the house’s **standard quota** is the population of the house divided by the standard divisor. So, we calculate that the standard quota is 2,964/600 = 4.94 for Gryffindor, 3,220/600 = 5.37 for Slytherin, 829/600 = 1.38 for Hufflepuff, and 4,987/600 = 8.31 for Ravenclaw.

Now, using Dean’s Apportionment Method, the harmonic mean cutoffs are 4.44 (the harmonic mean of 4 and 5) for Gryffindor, 5.45 (the harmonic mean of 5 and 6) for Slytherin, 1.33 (the harmonic mean of 1 and 2) for Hufflepuff, and 8.47 (the harmonic mean of 8 and 9) for Ravenclaw. This means the final apportionment of the twenty students is 5 students for Gryffindor (because 4.94 > 4.44), 5 students for Slytherin (since 5.37 < 5.45), 2 students for Hufflepuff (since 1.38 > 1.33), and 8 students for Ravenclaw (because 8.31 < 8.47). Since 5+5+2+8 = 20 students, we don’t have to make any adjustments to the divisor — hallelujah! Our twenty students have found a house (and I hope I’m one of the Slytherin five).

The rounding step is where the divisor apportionment methods vary, depending on which kind of rounding they use. In Dean’s Method, we’re using the harmonic mean to round, whereas Webster’s Method uses conventional rounding, i.e., rounding using the arithmetic mean and Hill’s Method uses geometric mean as the rounding cutoff. So, since harmonic mean is always less than or equal to arithmetic mean (according to our fun fact), Dean’s Method favors smaller states with harmonic mean while Webster’s method is neutral to state size.

Dean’s Apportionment Method, like other divisor methods, has the advantage of avoiding typical paradoxes that apportionment methods run into, but it’s not perfect. Dean’s Method violates something called the** quota rule**. The quota rule states that each state should get the number of representatives that is a rounding up or down of their quota. But, when using a divisor method like Dean’s Method, some states can get more delegates than their quota while others get less as the standard divisor is modified. In fact, Dean’s Method often violates this rule in favor of smaller states since the harmonic mean cutoff is lower when the standard quota is smaller, which explains why it is not used in real life.

In fact, it is proven that no apportionment method is without its problems. The Balinsky-Young Theorem says that no apportionment method can satisfy the quota rule *and* avoid paradoxes. That’s kind of interestingly disheartening, isn’t it? *There is no perfect method*. All we can do is weigh the consequences of each method and choose one with the least drawbacks.

Turns out the consensus is that avoiding paradoxes is more important than violating the quota rule. So, the United States government uses a divisor method, the Huntington-Hill Method, to officially apportion the House of Representatives, using geometric mean as the rounding convention. This method, along with Webster’s Method, are neutral, while Dean’s Method is not.

Though Dean’s Method has its flaws like any other apportionment method, it’s an interesting example of how divisor methods can be created simply by finding a new measure for rounding. This idea of trying to find a fair distribution when no objectively perfect one exists is a crucial question to ask, for government, politics, and life now *and* in the future.

Before you go, I have one last question for you: Are you Dean’s Apportionment Method? Because your beauty ain’t *fair*. ❤

Until next time! If you found this interesting, make sure to follow me to stay updated on the next ones.

In the meantime, check out other articles in my column **here**! If you have any questions or comments, please email me at apoorvapwrites@gmail.com.

*To be the first one to hear about all my new articles, recent events, and latest projects, make sure to subscribe to my newsletter: ***Letter? I Hardly Know Her!**

*As a reminder: this **column**, Gems in STEM, is a place to learn about various STEM topics that I find exciting, and that I hope will excite you too! It will always be written to be fairly accessible, so you don’t have to worry about not having background knowledge. However, it does occasionally get more advanced towards the end. Thanks for reading!*