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9.3: Apportionment Paradoxes

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    Each of the apportionment methods has at least one weakness. Some potentially violate the quota rule and some are subject to one of the three paradoxes.

    The quota rule says that each state should be given either its upper quota of seats or its lower quota of seats.

    Example \(\PageIndex{1}\): Quota Rule Violation

    A small college has three departments. Department A has 98 faculty, Department B has 689 faculty, and Department C has 212 faculty. The college has a faculty senate with 100 representatives. Use Jefferson’s method with a modified divisor of d = 9.83 to apportion the 100 representatives among the departments.

    Table \(\PageIndex{1}\): Quota Rule Violation
    State A B C Total
    Population 98 689 212 999
    Standard Quota 9.810 68.969 21.221 100.000
    d = 9.83 9.969 70.092 21.567
    quota 9.000 70.000 21.000 100

    District B has a standard quota of 68.969 so it should get either its lower quota, 68, or its upper quota, 69, seats. Using this method, District B received 70 seats, one more than its upper quota. This is a Quota Rule violation.

    The population paradox occurs when a state’s population increases but its allocated number of seats decreases.

    Example \(\PageIndex{2}\): Population Paradox

    A mom decides to split 11 candy bars among three children based on the number of minutes they spend on chores this week. Abby spends 54 minutes, Bobby spends 243 minutes and Charley spends 703 minutes. Near the end of the week, Mom reminds the children of the deal and they each do some extra work. Abby does an extra two minutes, Bobby an extra 12 minutes and Charley an extra 86 minutes. Use Hamilton’s method to apportion the candy bars both before and after the extra work.

    Table \(\PageIndex{2}\): Candy Bars Before the Extra Work
    State Abby Bobby Charley Total
    Population 54 243 703 1,000
    Standard Quota 0.594 2.673 7.734 11.000
    Lower Quota 0 2 7 9
    Apportionment 0 3 8 11

    With the extra work:
    Abby now has 54 + 2 = 56 minutes
    Bobby has 243 + 12 = 255
    Charley has 703 + 86 = 789 minutes

    Table \(\PageIndex{3}\): Candy Bars After the Extra Work
    State Abby Bobby Charley Total
    Population 56 255 789 1,100
    Standard Quota 0.560 2.550 7.890 11.000
    Lower Quota 0 2 7 9
    Apportionment 1 2 8 11

    Abby’s time only increased by 3.7% while Bobby’s time increased by 4.9%. However, Abby gained a candy bar while Bobby lost one. This is an example of the Population Paradox.

    The new states paradox occurs when a new state is added along with additional seats and existing states lose seats.

    Example \(\PageIndex{3}\): New-States Paradox

    A small city is made up of three districts and governed by a committee with 100 members. District A has a population of 5310, District B has a population of 1330, and District C has a population of 3308. The city annexes a small area, District D with a population of 500. At the same time the number of committee members is increased by five. Use Hamilton’s method to find the apportionment before and after the annexation.

    Table \(\PageIndex{4}\): Apportionment Before the Annexation
    State A B C Total
    Population 5,310 1,330 3,308 9,948
    Standard Quota 53.378 13.370 33.253 100.000
    Lower Quota 53 13 33 99
    Apportionment 54 13 33 100

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    Table \(\PageIndex{5}\): Apportionment After the Annexation
    State A B C D Total
    Population 5,310 1,330 3,308 500 10,448
    Standard Quota 53.364 13.366 33.245 5.025 105.000
    Lower Quota 53 13 33 5 104
    Apportionment 53 14 33 5 105

    District D has a population of 500 so it should get five seats. When District D is added with its five seats, District A loses a seat and District B gains a seat. This is an example of the New-States Paradox.

    In 1980, Michael Balinski (State University of New York at Stony Brook) and H. Peyton Young (Johns Hopkins University) proved that all apportionment methods either violate the quota rule or suffer from one of the paradoxes. This means that it is impossible to find the “perfect” apportionment method. The methods and their potential flaws are listed in the following table.

    Table \(\PageIndex{6}\): Methods, Quota Rule Violations, and Paradoxes
    Paradoxes
    Method Quota Rule Alabama Population New-States
    Hamilton No violations Yes Yes Yes
    Jefferson Upper-quota violations No No No
    Adams Lower-quota violations No No No
    Webster Lower- and upper-quota violations No No No
    Huntington-Hill Lower- and upper-quota violations No No No

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