Paradoxes of Apportionment The Alabama Paradox Population Paradox

Paradoxes of Apportionment The Alabama Paradox, Population Paradox, and the New State’s Paradox

What is a Paradox? • • A seemingly contradictory statement that may nonetheless be true: the paradox that standing is more tiring than walking. Something that exhibits inexplicable or contradictory aspects: “The silence of midnight … rung in my ears” (Mary Shelley). An assertion that is essentially self-contradictory, though based on a valid deduction from acceptable premises. A statement contrary to received opinion. - from www. dictionary. com

What is a Paradox? • We identify paradoxes of apportionment when we follow logical steps and discover a result that may seem surprising because in some way the result violates what we expect to be logical. • That is, the paradoxes arise when we follow logical steps and derive a result that doesn’t seem logical. • Of course, the results we derive are perfectly logical in the sense that they are the correct results of our deductions and it’s important to recognize that in higher mathematics paradoxes are often found in many areas and resolving these paradoxes opens doors to new understanding.

Paradoxes of Apportionment • Alabama Paradox – An increase in the number of available items causes a group to lose an item (even though populations remain the same). • Population Paradox – Group A can lose an item to group B even when the rate of growth of the population of group A is greater than in group B. • New States Paradox – The addition of a new group, with a corresponding increase in the number of available items, can cause a change in the apportionment of items of among the other groups.

The Alabama Paradox • Alabama Paradox – An increase in the number of available items causes a group to lose an item (even though populations remain the same). • The Alabama Paradox first surfaced after the 1870 census. With 270 members in the House of Representatives, Rhode Island had 2 representatives but when the House size was increased to 280, Rhode Island lost a seat. After the 1880 census, C. W. Seaton (chief clerk of U. S. Census Office) computed apportionments for all House sizes between 275 and 350 members. He then wrote a letter to Congress pointing out that if the House of Representatives had 299 seats, Alabama would get 8 seats but if the House of Representatives had 300 seats, Alabama would only get 7 seats.

The Population Paradox • Population Paradox – Group A can lose an item to group B even when the rate of growth of the population of group A is greater than in group B. • The Population Paradox was discovered around 1900, when it was shown that a state could lose seats in the House of Representatives in spite of a rapidly growing population. (Virginia was growing much faster than Maine--about 60% faster--but Virginia lost a seat in the House while Maine gained a seat. )

The New State’s Paradox • New States Paradox – The addition of a new group, with a corresponding increase in the number of available items, can cause a change in the apportionment of items among the other groups. • The New States Paradox was discovered in 1907 when Oklahoma became a state. Before Oklahoma became a state, the House of Representatives had 386 seats. Comparing Oklahoma's population to other states, it was clear that Oklahoma should have 5 seats so the House size was increased by five to 391 seats. The intent was to leave the number of seats unchanged for the other states. However, when the apportionment was recalculated, Maine gained a seat (4 instead of 3) and New York lost a seat (from 38 to 37).

Example #1 – the Alabama Paradox • Suppose Miami Dade College has 3 campuses and enrollments at each campus are as given in the table below. Campus North Enrollment 10, 170 Kendall Wolfson Total 9, 150 680 20, 000 • Suppose there are 40 full-time faculty members to be distributed among these campuses according to their enrollment.

Example #1 – the Alabama Paradox Campus Enrollment North 10, 170 Kendall 9, 150 Wolfson 680 Total 20, 000 There are 3 groups (the campuses) and 40 items (the full-time faculty positions) to be divided among them. We will use Hamilton’s method to divide the items so that each group gets a proportional integer number of faculty. (Faculty are less effective when cut into pieces!) The standard divisor is (total enrollment)/(number of items) = 20, 000/40 = 500. Next, we calculate the quotas for each campus…

Example #1 – the Alabama Paradox Campus Enrollment Quota North 10, 170/500 20. 34 Kendall 9, 150 9150/500 18. 3 Wolfson 680/500 1. 36 Total 20, 000 40

Example #1 – the Alabama Paradox Campus Enrollment Quota Lower Quota North 10, 170/50 0 20. 34 20 Kendall 9, 150 9150/500 18. 3 18 Wolfson 680/500 1. 36 1 Total 20, 000 40 39 Add remaining seat +1

Example #1 – the Alabama Paradox Campus Enrollment Quota Lower Quota Add remaining seat North 10, 170 20. 34 20 20 Kendall 9, 150 18. 3 18 18 Wolfson 680 1. 36 1 Total 20, 000 40 39 +1 Final Apportionment 2 40

Example #1 – the Alabama Paradox Campus Enrollment North 10, 170 We conclude that, with 40 full-time faculty, the College should make the following apportionment: North – 20 full-time faculty Kendall 9, 150 Wolfson 680 Total 20, 000 Kendall – 18 full-time faculty Wolfson – 2 full-time faculty

Example #1 – the Alabama Paradox Campus Enrollment North 10, 170 Kendall 9, 150 Wolfson 680 Total 20, 000 We concluded that, with 40 full-time faculty, the College should make the following apportionment: North – 20 full-time faculty Kendall – 18 full-time faculty Wolfson – 2 full-time faculty Now, suppose the College opens a new line and decides to hire a new full-time faculty member. If the number of full-time faculty is increased to 41, where will the new faculty member be assigned? At which campus will he or she teach?

Example #1 – the Alabama Paradox Campus Enrollment The enrollments will remain the same. North 10, 170 The only change is to increase the number of available full-time faculty. Kendall 9, 150 Now the standard divisor becomes (total population)/(number of items) = Wolfson 680 20, 000/41 = 487. 8 Total 20, 000

Example #1 – the Alabama Paradox Campus Enrollment Quota North 10, 170 10170/487. 8 20. 85 Kendall 9, 150 9150/487. 8 18. 76 Wolfson 680/487. 8 1. 39 Total 20, 000 41

Example #1 – the Alabama Paradox Campus Enrollment Quota Lower Quota Add remaining seats North 10, 170 10170/487. 8 20. 85 20 +1 Kendall 9, 150 9150/487. 8 18. 76 18 +1 Wolfson 680/487. 8 1. 39 1 Total 20, 000 41 39

Example #1 – the Alabama Paradox Campus Enrollment Quota Lower Quota Add remaining seats Final Apportionment North 10, 170 20. 85 20 +1 21 Kendall 9, 150 18. 76 18 +1 19 Wolfson 680 1. 39 1 1 Total 20, 000 41 39 41

Example #1 – the Alabama Paradox Campus Enrollment North 10, 170 Kendall 9, 150 Wolfson 680 Total 20, 000 We concluded that, with 40 full-time faculty, the College should make the following apportionment: North – 20 full-time faculty Kendall – 18 full-time faculty Wolfson – 2 full-time faculty Now, with 41 full-time faculty, we conclude that the College should make the following apportionment: North – 21 full-time faculty Kendall – 19 full-time faculty Wolfson – 1 full-time faculty

Example #1 – the Alabama Paradox With 40 full-time faculty: Campus Enrollment North 10, 170 Kendall 9, 150 Wolfson 680 Total 20, 000 North – 20 full-time faculty Kendall – 18 full-time faculty Wolfson – 2 full-time faculty With 41 full-time faculty: North – 21 full-time faculty Kendall – 19 full-time faculty Wolfson – 1 full-time faculty The Alabama Paradox has occurred! With the addition of a new faculty member, Wolfson loses a faculty member while North and Kendall both gain one – all this in spite of the fact that there were no changes in the enrollments at any campus.