Social Choice is a field that studies how to make choices that apply to the whole society based on individual preferences. Here is an example which contradicts Arrow's Impossibility Theorem which claims that social choices are impossible.


Consider Arrow’s example on p. 27 of  Social Choice and Individual Values where he shows a deficiency of the Borda count:

“CONDITION 3: Let R1, . . .,  Rn and R1' , . . .,   Rn' be two sets of individual orderings and let C(S) and C'(S) be the corresponding social choice func­tions. If, for all individuals i and all x and y in a given environment S, xRi y  if and only if xRi' y  , then C(S) and C'(S) are the same (independ­ence of irrelevant alternatives).

The reasonableness of this condition can be seen by consideration of the possible results in a method of choice which does not satisfy Condi­tion 3, the rank-order method of voting frequently used in clubs. With a finite number of candidates, let each individual rank all the candidates, i.e., designate his first-choice candidate, second-choice candidate, etc. Let preassigned weights be given to the first, second, etc., choices, the higher weight to the higher choice, and then let the candidate with the highest weighted sum of votes be elected. In particular, suppose that there are three voters and four candidates, x, y ,z, and w. Let the weights for the first, second, third, and fourth choices be 4, 3, 2, and 1, respectively. Suppose that individuals 1 and 2 rank the candidates in the order x, y, z and w, while individual 3 ranks them in the order z, w, x, and y. Under the given electoral system, x is chosen. Then, certainly, if y is deleted from the ranks of the candidates, the system applied to the remaining candidates should yield the same result, espe­cially since, in this case, y is inferior to x according to the tastes of every individual; but, if y is in fact deleted, the indicated. electoral system would yield a tie between x and z.”


However, if you use a corresponding economic example, there is no such problem.


Let x be 1 unit of work eg 1 hour at task X.

Let y be 1 unit of work eg 1 hour at task Y.

Let  z be 1 unit of work eg 1 hour at task Z.

Let w be 1 unit of work eg 1 hour at task W.


Let’s stipulate that each individual must work the same number of hours and all the work must be done. The general idea is to assign the tasks in such a way as to maximize social utility where utility is defined as sum over individual utilities and individual utility is defined as 4 x (time spent on most preferred task) + 3 x (time spent on second most preferred task) etc. ie a general Borda ranking. Let individuals 1 and 2 have the preference ranking xyzw and individual 3 have the ranking zwxy. Since 1 and 2 have the same preferences, we let them spend half their time on their first preference x and half their time on their second preference y. Let 3 do z since it’s his first choice. That leaves w. Let each individual do one third of w. Then the individual utility for 1 and 2 is 4.(½) + 3. (½) + 1. (1/3) = 3 5/6. The individual utility for 3 is 4.1 + 3. (1/3) = 5. Social utility is 12 2/3. The maximum possible utility is 16.


Now, if task y is deleted as in Arrow’s example, we have the preferences for 1 and 2 as xzw and 3 as zwx. Now let 1 and 2 each do (½)x and (½)w and 3 do all of z. Each person does 1 hour of work. Total possible utility = 9. For 1 and 2, utility = 3. (1/2) + 1. (1/2) = 2. Utility for 3 = 3. Total utility = 7. Alternatively, let 1 and 2 do (½)x and (1/4)w and (1/4)z. Utility = 3. (1/2) + 2.(1/4) + 1. (1/4) = 9/4 = 2 (1/4). Let 3 do (1/2)z and (1/2)w. Utility = 3. (1/2) + 2. (1/2) = 5/2 = 2 (½). Total utility = 7.


The  point is that eliminating y does not upset  the work schedules in the same way that it upsets political  rankings because the work can be divided among the workers in many possible ways whereas in the political  case there can only be one winner. Hence Arrow’s Impossibility Theorem doesn’t apply. See Social Choice and Beyond for more.


A second example could be the following. There are four baskets of goods w, x, y and z. Each of three individuals ranks the baskets in the order of preference. Let the preference orders be the same as in the previous example. Let individuals 1 and 2 have the preference ranking xyzw and individual 3 have the ranking zwxy. If the baskets are divisible then the solutions will be the same as in the previous example. If they’re not then the baskets will be assigned based on the preference inputs. Let’s assume they’re not divisible. Then 1 would get x. 2 would get y and 3 would get z for individual utilities of  4, 3 and 4 and a social utility of 11. If the baskets are divisible, 1 and 2 would each get

4.(1/2) + 3. (½) = 3 (½) and 3 would get 4 for a total utility of  11. Notice that this example differs from the work example in not all baskets need be consumed whereas in the work example all the work needed to be done.


Now we can revisit the same examples and use range instead of Borda to rate the preferences.


The next example might be each one submits his work and consumption preferences. For example assume there are 4 tasks w, x, y, and z and 4 baskets of goods a, b, c and d. Let’s assume the tasks and the baskets are not dividible. Then each individual would rate w and a, w and b, w and c, w and d, x and a, x and b … z and a, z and b, z and c and z and d or 16 work/consumption possibilities. Let’s say the borda ratings are 1 to 16. Then the problem is the same as before.