1. Math for Liberal Studies

2.  The US Senate has 100 members: two for each state  In the US House of Representatives, states are represented based on population  PA has 19 representatives  Delaware has 1

3.  The process by which seats are assigned based on population is called apportionment  The number of seats each state gets is also called that state’s apportionment

4.  Consider a fictional country with four states and a representative legislature with 50 seats StatePopulation Angria80,000 Bretonnia60,000 Curaguay40,000 Dennenberg20,000 Total200,000

5.  Each state should get a proportion of the seats that is equal to its proportion of the total population StatePopulation Angria80,000 Bretonnia60,000 Curaguay40,000 Dennenberg20,000 Total200,000

6.  Divide each state’s population by the total population to get the % population StatePopulation% Pop. Angria80,000 40% Bretonnia60,000 30% Curaguay40,000 20% Dennenberg20,000 10% Total200,000 100%

7.  Multiply that percentage by the total number of seats (in this case 50) to get each state’s fair share of seats StatePopulation% Pop.Fair Share Angria80,000 40%20 Bretonnia60,000 30%15 Curaguay40,000 20%10 Dennenberg20,000 10%5 Total200,000 100%50

8.  Real world examples rarely work out as nicely as the previous example did  Let’s use more realistic population numbers and see what happens

9.  We will start the problem in the same way StatePopulation Angria83,424 Bretonnia67,791 Curaguay45,102 Dennenberg17,249 Total213,566

10.  Divide each state’s population by the total population to get the % population StatePopulation% Pop. Angria83,424 39.06% Bretonnia67,791 31.74% Curaguay45,102 21.12% Dennenberg17,249 8.08% Total213,566 100%

11.  Multiply that percentage by the total number of seats (in this case 50) to get each state’s fair share of seats StatePopulation% Pop.Fair Share Angria83,424 39.06%19.53 Bretonnia67,791 31.74%15.87 Curaguay45,102 21.12%10.56 Dennenberg17,249 8.08%4.04 Total213,566 100%50

12.  The problem is that we can’t assign a state 19.53 seats… each apportionment must be a whole number! StatePopulation% Pop.Fair Share Angria83,424 39.06%19.53 Bretonnia67,791 31.74%15.87 Curaguay45,102 21.12%10.56 Dennenberg17,249 8.08%4.04 Total213,566 100%50

13.  We could try rounding each fair share to the nearest whole number, but we end up with 51 seats, which is more than we have! StatePopulation% Pop.Fair ShareSeats Angria83,424 39.06%19.5320 Bretonnia67,791 31.74%15.8716 Curaguay45,102 21.12%10.5611 Dennenberg17,249 8.08%4.044 Total213,566 100%5051

14.  Named for Alexander Hamilton, who developed the method for apportioning the US House of Representatives  With Hamilton’s method, we start like we did before, and compute the fair shares

15.  Here are the fair shares we computed before  Round each fair share down StatePopulation% Pop.Fair Share Angria83,424 39.06%19.53 Bretonnia67,791 31.74%15.87 Curaguay45,102 21.12%10.56 Dennenberg17,249 8.08%4.04 Total213,566 100%50

16.  The rounded-down fair shares are called lower quotas  Each state should receive at least this number of seats StatePopulation% Pop.Fair Share Lower Quota Angria83,424 39.06%19.5319 Bretonnia67,791 31.74%15.8715 Curaguay45,102 21.12%10.5610 Dennenberg17,249 8.08%4.044 Total213,566 100%5048

17.  Notice that now we have 2 “leftover” seats that need to be assigned StatePopulation% Pop.Fair Share Lower Quota Angria83,424 39.06%19.5319 Bretonnia67,791 31.74%15.8715 Curaguay45,102 21.12%10.5610 Dennenberg17,249 8.08%4.044 Total213,566 100%5048

18.  The states that had the highest decimal part in their fair share get priority for the leftover seats StatePopulation% Pop.Fair Share Lower Quota Angria83,424 39.06%19.5319 Bretonnia67,791 31.74%15.8715 Curaguay45,102 21.12%10.5610 Dennenberg17,249 8.08%4.044 Total213,566 100%5048

19.  The states that had the highest decimal part in their fair share get priority for the leftover seats StatePopulation% Pop.Fair Share Lower Quota Priority Angria83,424 39.06%19.5319 Bretonnia67,791 31.74%15.87151st Curaguay45,102 21.12%10.5610 Dennenberg17,249 8.08%4.044 Total213,566 100%5048

20.  The states that had the highest decimal part in their fair share get priority for the leftover seats StatePopulation% Pop.Fair Share Lower Quota Priority Angria83,424 39.06%19.5319 Bretonnia67,791 31.74%15.87151st Curaguay45,102 21.12%10.56102nd Dennenberg17,249 8.08%4.044 Total213,566 100%5048

21.  The states that had the highest decimal part in their fair share get priority for the leftover seats StatePopulation% Pop.Fair Share Lower Quota Priority Angria83,424 39.06%19.53193rd Bretonnia67,791 31.74%15.87151st Curaguay45,102 21.12%10.56102nd Dennenberg17,249 8.08%4.044 Total213,566 100%5048

22.  The states that had the highest decimal part in their fair share get priority for the leftover seats StatePopulation% Pop.Fair Share Lower Quota Priority Angria83,424 39.06%19.53193rd Bretonnia67,791 31.74%15.87151st Curaguay45,102 21.12%10.56102nd Dennenberg17,249 8.08%4.0444th Total213,566 100%5048

23.  Now, in priority order, we assign the extra seats StatePopulation% Pop.Fair Share Lower Quota PrioritySeats Angria83,424 39.06%19.53193rd Bretonnia67,791 31.74%15.87151st Curaguay45,102 21.12%10.56102nd Dennenberg17,249 8.08%4.0444th Total213,566 100%5048

24.  Now, in priority order, we assign the extra seats StatePopulation% Pop.Fair Share Lower Quota PrioritySeats Angria83,424 39.06%19.53193rd Bretonnia67,791 31.74%15.87151st16 Curaguay45,102 21.12%10.56102nd Dennenberg17,249 8.08%4.0444th Total213,566 100%5048

25.  Now, in priority order, we assign the extra seats StatePopulation% Pop.Fair Share Lower Quota PrioritySeats Angria83,424 39.06%19.53193rd Bretonnia67,791 31.74%15.87151st16 Curaguay45,102 21.12%10.56102nd11 Dennenberg17,249 8.08%4.0444th Total213,566 100%5048

26.  We only had two leftover seats to assign, so the other states get their lower quota StatePopulation% Pop.Fair Share Lower Quota PrioritySeats Angria83,424 39.06%19.53193rd Bretonnia67,791 31.74%15.87151st16 Curaguay45,102 21.12%10.56102nd11 Dennenberg17,249 8.08%4.0444th Total213,566 100%5048

27.  We only had two leftover seats to assign, so the other states get their lower quota StatePopulation% Pop.Fair Share Lower Quota PrioritySeats Angria83,424 39.06%19.53193rd19 Bretonnia67,791 31.74%15.87151st16 Curaguay45,102 21.12%10.56102nd11 Dennenberg17,249 8.08%4.0444th4 Total213,566 100%504850

28.  Notice that Bretonnia and Curaguay received their fair shares rounded up (their upper quota) StatePopulation% Pop.Fair Share Lower Quota PrioritySeats Angria83,424 39.06%19.53193rd19 Bretonnia67,791 31.74%15.87151st16 Curaguay45,102 21.12%10.56102nd11 Dennenberg17,249 8.08%4.0444th4 Total213,566 100%504850

29.  With Hamilton’s method, each state will always receive either their lower quota or their upper quota: this is the quota rule  If a state is apportioned a number of states that is either above its upper quota or below its lower quota, then that is a quota rule violation

30.  Use Hamilton’s Method to assign 60 seats to the following states  State A has population 4,105  State B has population 5,376  State C has population 2,629

31.  Here is the solution StatePopulation% Pop.Fair Share Lower Quota PrioritySeats A4,105 33.90%20.34202 nd 20 B5,376 44.39%26.64261 st 27 C2,629 21.71%13.03133 rd 13 Total12,110 100%605960

32.  Hamilton’s method is relatively simple to use, but it can lead to some strange paradoxes:  The Alabama paradox: Increasing the number of total seats causes a state to lose seats  The New States paradox: Introducing a new state causes an existing state to gain seats  The Population paradox: A state that gains population loses a seat to a state that does not

33.  When the population of each state stays the same but the number of seats is increased, intuitively each state’s apportionment should stay the same or go up  That doesn’t always happen

34.  Consider these state populations, with 80 seats available. Watch what happens when we increase to 81 seats StatePopulation% Pop.Fair Share Lower Quota PrioritySeats Angria83,42439.06%31.2531 Bretonnia67,79131.74%25.3925 Curaguay45,10221.12%16.89161 st 17 Dennenberg17,2498.08%6.4662 nd 7 Totals213,566100%807880

35. StatePopulation% Pop.Fair Share Lower Quota PrioritySeats Angria83,42439.06%31.2531 Bretonnia67,79131.74%25.3925 Curaguay45,10221.12%16.89161 st 17 Dennenberg17,2498.08%6.4662 nd 7 Totals213,566100%807880 StatePopulation% Pop.Fair Share Lower Quota PrioritySeats Angria83,42439.06%31.64312 nd 32 Bretonnia67,79131.74%25.71251 st 26 Curaguay45,10221.12%17.1117 Dennenberg17,2498.08%6.5466 Totals213,566100%817981

36.  After the 1880 Census, it was time to reapportion the House of Representatives.  The chief clerk of the Census Bureau computed apportionments for all numbers of seats from 275 to 350.  Alabama would receive 8 seats if there were 299 total seats, but only 7 seats if 300 were available.

37.  If a new state is added to our country, but the total number of seats remains the same, then we would expect that the apportionment for the existing states should stay the same or go down  This doesn’t always happen

38.  Consider this country with 4 states and 70 seats StatePopulation% Pop.Fair Share Lower Quota PrioritySeats Elkabar80,42439.80%27.86271 st 28 Florin59,90229.64%20.75202 nd 21 Gondor48,33823.92%16.75163 rd 17 Hyrkania13,4056.63%4.6444 Totals202,069100%706770

39.  Now suppose the small state of Ishtar is added to this country StatePopulation% Pop.Fair Share Lower Quota PrioritySeats Elkabar80,42439.80%27.86271 st 28 Florin59,90229.64%20.75202 nd 21 Gondor48,33823.92%16.75163 rd 17 Hyrkania13,4056.63%4.6444 Totals202,069100%706770

40. StatePopulation% Pop.Fair Share Lower Quota PrioritySeats Elkabar80,42439.80%27.86271 st 28 Florin59,90229.64%20.75202 nd 21 Gondor48,33823.92%16.75163 rd 17 Hyrkania13,4056.63%4.6444 Totals202,069100%706770 StatePopulation% Pop.Fair Share Lower Quota PrioritySeats Elkabar80,42438.72%27.1127 Florin59,90228.84%20.1920 Gondor48,33823.28%16.2916 Hyrkania13,4056.45%4.5242 nd 5 Ishtar5,6112.70%1.8911 st 2 Totals207,680100%706870

41.  In 1907, when Oklahoma was admitted to the Union, the total number of seats in Congress did not change  Oklahoma was assigned 5 seats, but this resulted in Maine gaining a seat!

42.  As time goes on, populations of states change  States that increase population rapidly should gain seats over those that do not  However, this doesn’t always happen

43.  Consider this country with 4 states and 100 seats StatePopulation% Pop.Fair Share Lower Quota PrioritySeats Javasu28,90018.09%18.0918 Karjastan76,20047.68%47.68471 st 48 Libria44,20027.66%27.66272 nd 28 Malbonia10,5006.57%6.5766 Totals159,800100%10098100

44.  Suppose that the populations of the states change StatePopulation% Pop.Fair Share Lower Quota PrioritySeats Javasu28,90018.09%18.0918 Karjastan76,20047.68%47.68471 st 48 Libria44,20027.66%27.66272 nd 28 Malbonia10,5006.57%6.5766 Totals159,800100%10098100

45. StatePopulation% Pop.Fair Share Lower Quota PrioritySeats Javasu28,90018.09%18.0918 Karjastan76,20047.68%47.68471 st 48 Libria44,20027.66%27.66272 nd 28 Malbonia10,5006.57%6.5766 Totals159,800100%10098100 StatePopulation% Pop.Fair Share Lower Quota PrioritySeats Javasu28,90017.97%17.97172 nd 18 Karjastan76,40047.51%47.5147 Libria45,00027.99%27.99271 st 28 Malbonia10,5006.53%6.5363 rd 7 Totals160,800100%10097100

46.  In the 1901 apportionment, Virginia lost a seat to Maine even though Virginia’s population grew at a faster rate!

47.  Several alternatives to Hamilton’s method have been developed that avoid these paradoxes  The first was developed by Thomas Jefferson, and was the method used to apportion the Congress from 1792 to 1832

48.  With this method, we return to Hamilton’s idea of rounding down the fair shares  The trick is to round down the fair shares, but not have any leftover seats

49.  Consider this example StatePopulation% Pop.Fair Share Lower Quota Angria 87,43829.15%14.5814 Bretonnia 82,51127.50%13.7513 Curaguay 66,94222.31%11.1611 Dennenberg 63,10921.04%10.5210 Total 300,000100%5048

50.  Here we have two extra seats, and each seat represents 300,000/50 = 6,000 people StatePopulation% Pop.Fair Share Lower Quota Angria 87,43829.15%14.5814 Bretonnia 82,51127.50%13.7513 Curaguay 66,94222.31%11.1611 Dennenberg 63,10921.04%10.5210 Total 300,000100%5048

51.  When we divide the total population by the number of seats, we get the standard divisor  Instead of figuring out the % population, we could just divide the population of each state by the standard divisor

52.  For example, Angria’s fair share is 87,438/6,000 = 14.58 StatePopulationFair Share Lower Quota Angria 87,43814.5814 Bretonnia 82,51113.7513 Curaguay 66,94211.1611 Dennenberg 63,10910.5210 Total 300,0005048

53.  Jefferson’s idea was to modify the standard divisor so that when the shares for each state are rounded down, there are no leftover seats  Right now we are only assigning 48 seats  Making the divisor smaller will increase each state’s share

54.  Here is the apportionment with the original divisor (6,000) StatePopulationFair Share Lower Quota Angria 87,43814.5814 Bretonnia 82,51113.7513 Curaguay 66,94211.1611 Dennenberg 63,10910.5210 Total 300,0005048

55.  Let’s make the divisor smaller  Here we have the divisor equal to 5,850 StatePopulation Modified Share Lower Quota Angria87,43814.9514 Bretonnia82,51114.1014 Curaguay66,94211.4411 Dennenberg63,10910.7910 Totals300,00049 Still not enough seats!

56.  Let’s make the divisor even smaller  Here we have the divisor equal to 5,700 StatePopulation Modified Share Lower Quota Angria87,43815.3415 Bretonnia82,51114.4814 Curaguay66,94211.7411 Dennenberg63,10911.0711 Totals300,00051 Now we have too many seats!

57.  With some trial and error we find a number that works (in this case 5,800) StatePopulation Modified Share Lower Quota Angria87,43815.0815 Bretonnia82,51114.2314 Curaguay66,94211.5411 Dennenberg63,10910.8810 Totals300,00050 Success!

58.  Jefferson’s method has the advantage that it doesn’t suffer from any of the paradoxes that Hamilton’s method did  However, there is still a problem with Jefferson’s method  Let’s consider another example with 4 states and 50 seats

59.  This time the standard divisor is 220,561/50 = 4,411.2 StatePopulationFair Share Lower Quota Elkabar96,97421.9821 Florin45,90210.4110 Gondor44,92110.1810 Hyrkania32,7647.437 Totals220,5615048

60.  We have leftover seats, so Jefferson’s method tells us to make the divisor smaller StatePopulationFair Share Lower Quota Elkabar96,97421.9821 Florin45,90210.4110 Gondor44,92110.1810 Hyrkania32,7647.437 Totals220,5615048

61.  After some trial and error, we find that a divisor of 4,200 works StatePopulation Modified Share Lower Quota Elkabar96,97423.0923 Florin45,90210.9310 Gondor44,92110.7010 Hyrkania32,7647.807 Totals220,56150

62.  The top chart uses the standard divisor (4,411)  The bottom chart is our Jefferson’s method solution  We have a quota rule violation! StatePopulationFair Share Lower Quota Elkabar96,97421.9821 Florin45,90210.4110 Gondor44,92110.1810 Hyrkania32,7647.437 Totals220,5615048 StatePopulation Modified Share Lower Quota Elkabar96,97423.0923 Florin45,90210.9310 Gondor44,92110.7010 Hyrkania32,7647.807 Totals220,56150

63.  As we saw, Jefferson’s method can violate the quota rule  This tends to favor larger states over smaller ones

64.  Variations of Jefferson’s method were proposed  Adams’ method (developed by the 6 th US President John Quincy Adams)  Webster’s method (developed by famed orator and lawyer Daniel Webster)

65.  This method is the same as Jefferson’s method, except you always round the shares up instead of down  This gives you too many seats, so you need to increase the divisor until you get the right number of seats

66.  In this method, we round the shares to the nearest whole number  It’s possible that this gives us the right number of seats, in which case we’re done  Otherwise, we need to increase (if we have too few seats) or decrease (if we have too many seats) the divisor

67.  Even these alternative methods can cause quota rule violations  In fact, it was proved in 1980 that it is impossible to find an apportionment system that both  avoids the paradoxes we discussed  doesn’t violate the quota rule