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Session 5, Part B: Unfair Allocations
 
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Session 5, Part B:
Unfair Allocations

In This Part: Fair and Unfair Allocations | Measuring the Degree of Fairness
Looking at Excesses and Deficits

We've seen how to judge the relative fairness of an allocation. A related question asks how you might determine the number of moves required to make an unfair allocation fair, without making the actual moves. Note 4

Below is Allocation A from Problem B2, arranged in ascending order. We've already determined that only two moves are required to make this allocation fair. Let's look more closely at why this is true:

The notations below each stack indicate the number of coins that each stack is above or below the mean of 5. In other words:

a. 

Two stacks are above the mean. Each of these has 6 coins, an excess of 1 coin (+1) above the average. (These excesses are noted in the figure above.) The total excess of coins above the average is 2.

b. 

Two stacks are below the mean. Each of these has 4 coins, a deficit of 1 coin (-1) below the average. (These deficits are also noted in the figure above.) The total deficit of coins below the average is 2.

c. 

Five stacks are exactly average and have no excess or deficit (0).

Problem B4

Solution  

Perform the calculations above to show that 20 moves are required to obtain a fair allocation for Allocation B, which is arranged in ascending order below:


 

Problem B5

Solution  

Perform the same calculations to find the number of moves required to obtain a fair allocation for Allocation C, which is shown in ascending order below:


Next > Part C: Using Line Plots

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