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

2. A statement contrary to received opinion. - from www.dictionary.com
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

3. 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.

4. 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.

5. 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.

6. 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.)

7. 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).

8. 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
Kendall
Wolfson
Total
Enrollment
10,170
9,150
680
20,000
Suppose there are 40 full-time faculty members to be distributed among these campuses according to their enrollment.

9. Example #1 – the Alabama Paradox
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…
Campus
Enrollment
North
10,170
Kendall
9,150
Wolfson
680
Total
20,000

10. Example #1 – the Alabama Paradox
Campus
Enrollment
Quota
North
10,170
10,170/500
20.34
Kendall
9,150
9150/500
18.3
Wolfson
680
680/500
1.36
Total
20,000 40

11. Example #1 – the Alabama Paradox
Campus
Enrollment
Quota
Lower Quota
Add remaining seat
North
10,170
10,170/500
20.34 20
Kendall
9,150
9150/500
18.3 18
Wolfson
680
680/500
1.36 1 +1
Total
20,000 40 39

12. Example #1 – the Alabama Paradox
Campus
Enrollment
Quota
Lower Quota
Add remaining seat
Final Apportionment
North
10,170
20.34 20
Kendall
9,150
18.3 18
Wolfson
680
1.36 1 +1 2
Total
20,000 40 39

13. Example #1 – the Alabama Paradox
Campus
Enrollment
North
10,170
Kendall
9,150
Wolfson
680
Total
20,000
We conclude 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

14. Example #1 – the Alabama Paradox
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?
Campus
Enrollment
North
10,170
Kendall
9,150
Wolfson
680
Total
20,000

15. Example #1 – the Alabama Paradox
Campus
Enrollment
North
10,170
Kendall
9,150
Wolfson
680
Total
20,000
The enrollments will remain the same.
The only change is to increase the number of available full-time faculty.
Now the standard divisor becomes
(total population)/(number of items) =
20,000/41 = 487.8

16. 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
680/487.8
1.39
Total
20,000 41

17. 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
Wolfson
680
680/487.8
1.39 1
Total
20,000 41 39

18. 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 19
Wolfson
680
1.39 1
Total
20,000 41 39

19. Example #1 – the Alabama Paradox
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
Campus
Enrollment
North
10,170
Kendall
9,150
Wolfson
680
Total
20,000

20. Example #1 – the Alabama Paradox
With 40 full-time faculty:
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.
Campus
Enrollment
North
10,170
Kendall
9,150
Wolfson
680
Total
20,000