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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). |
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