Optimal group composition for efficient division of labor

Abstract

This study examines a group performing multiple tasks, with each subgroup performing each task expressed as a binary choice problem. Each subgroup uses the simple majority rule; a correct decision benefits the subgroup. This study demonstrates that, assuming all individuals’ equal competence for all tasks and a sufficiently large group size, when each individual’s probability of making a correct decision exceeds one-half, the optimal group composition is an equal number of individuals per subgroup. Conversely, it is less than one-half, the assignment produces the lowest benefit. We also find that when decision-making costs exist, if the competence is greater than one-half, the possibility that the performance of division of labor outweighs that of plenary voting increases as the cost increases. On the other hand, if the competence is less than one-half, division of labor is always more beneficial than plenary voting. The optimal group compositions for the cases where the group size is not sufficiently large are also discussed.

Appendix: Proof of Fact 1

Although this fact itself may have already been known, we could not find the proof of it. Thus, we here show the way to prove this fact using straightforward algebra. To do that, we refer to the method in Appendix of Tomiyama (1991), which is originally used to prove Eq. (18) in the present paper.

First, let us consider a group consisting of $n\equiv 2k+1$ voters, where k is an integer. Then, the majority vote accuracy of this group is expressed as

$$\begin{aligned} P_n&=\sum _{h=k+1}^{2k+1} {\left( {\begin{array}{c} 2k+1 \\ h \\ \end{array}} \right) } p^{h}(1-p)^{2k+1-h} \nonumber \\&=p^{k+1}\left[ {\left( {\begin{array}{c} 2k+1 \\ k+1 \\ \end{array}} \right) (1-p)^{k}+\left( {\begin{array}{c} 2k+1 \\ k+2 \\ \end{array}} \right) p(1-p)^{k-1}+\left( {\begin{array}{c} 2k+1 \\ k+3 \\ \end{array}} \right) p^{2}(1-p)^{k-2}} \right. \nonumber \\&\quad \quad \cdots \cdots \nonumber \\&\quad +\left. {\left( {\begin{array}{c} 2k+1 \\ 2k-1 \\ \end{array}} \right) p^{k-2}(1-p)^{2}+\left( {\begin{array}{c} 2k+1 \\ 2k \\ \end{array}} \right) p^{k-1}(1-p)+\left( {\begin{array}{c} 2k+1 \\ 2k+1 \\ \end{array}} \right) p^{k}} \right] \nonumber \\ \end{aligned}$$

(33)

Similarly, the majority vote accuracy of a group consisting of n + 1 voters is

$$\begin{aligned} P_{n+1}&=\sum _{h=k+2}^{2k+2} {\left( {\begin{array}{c} 2k+2 \\ h \\ \end{array}} \right) } p^{h}(1-p)^{2k+2-h}+\frac{1}{2}\left( {\begin{array}{c} 2k+2 \\ k+1 \\ \end{array}} \right) p^{k+1}(1-p)^{k+1} \nonumber \\&=p^{k+1}\left[ \frac{1}{2}\left( {\begin{array}{c} 2k+2 \\ k+1 \\ \end{array}} \right) (1-p)^{k+1}+\left( {\begin{array}{c} 2k+2 \\ k+2 \\ \end{array}} \right) p(1-p)^{k}\right. \nonumber \\&\quad \left. +\left( {\begin{array}{c} 2k+2 \\ k+3 \\ \end{array}} \right) p^{2}(1-p)^{k-1} \right. \nonumber \\&\quad \quad {\cdots \cdots } \nonumber \\&\quad +\left. {\left( {\begin{array}{c} 2k+2 \\ 2k \\ \end{array}} \right) p^{k-1}(1-p)^{2}+\left( {\begin{array}{c} 2k+2 \\ 2k+1 \\ \end{array}} \right) p^{k}(1-p)+\left( {\begin{array}{c} 2k+2 \\ 2k+2 \\ \end{array}} \right) p^{k+1}} \right] \nonumber \\ \end{aligned}$$

(34)

Applying the well-known formula $\left( {\begin{array}{c} x \\ y \\ \end{array}} \right) =\left( {\begin{array}{c} x-1 \\ y-1 \\ \end{array}} \right) +\left( {\begin{array}{c} x-1 \\ y \\ \end{array}} \right)$ to every term in the square brackets of (34), we have

$$\begin{aligned} P_{n+1}&=\sum _{h=k+2}^{2k+2} {\left( {\begin{array}{c} 2k+2 \\ h \\ \end{array}} \right) } p^{h}(1-p)^{2k+2-h}+\frac{1}{2}\left( {\begin{array}{c} 2k+2 \\ k+1 \\ \end{array}} \right) p^{k+1}(1-p)^{k+1} \nonumber \\&=p^{k+1}\left[ \frac{1}{2}\left( {\begin{array}{c} 2k+1 \\ k \\ \end{array}} \right) (1-p)^{k+1}+\frac{1}{2}\left( {\begin{array}{c} 2k+1 \\ k+1 \\ \end{array}} \right) (1-p)^{k+1} \right. \nonumber \\&\quad +\left( {\begin{array}{c} 2k+1 \\ k+1 \\ \end{array}} \right) p(1-p)^{k}+\left( {\begin{array}{c} 2k+1 \\ k+2 \\ \end{array}} \right) p(1-p)^{k}\nonumber \\&\quad +\left( {\begin{array}{c} 2k+1 \\ k+2 \\ \end{array}} \right) p^{2}(1-p)^{k-1}+\left( {\begin{array}{c} 2k+1 \\ k+3 \\ \end{array}} \right) p^{2}(1-p)^{k-1} \nonumber \\&\quad +\left( {\begin{array}{c} 2k+1 \\ k+3 \\ \end{array}} \right) p^{3}(1-p)^{k-2}+\left( {\begin{array}{c} 2k+1 \\ k+4 \\ \end{array}} \right) p^{3}(1-p)^{k-2}\nonumber \\&\quad \quad \cdots \cdots \nonumber \\&\quad +\left( {\begin{array}{c} 2k+1 \\ 2k \\ \end{array}} \right) p^{k}(1-p)+\left( {\begin{array}{c} 2k+1 \\ 2k+1 \\ \end{array}} \right) p^{k}(1-p) \nonumber \\&\quad +\left. {\left( {\begin{array}{c} 2k+1 \\ 2k+1 \\ \end{array}} \right) p^{k+1}} \right] . \end{aligned}$$

(35)

Here, the sum of the first and second terms inside the square brackets is $\left( {\begin{array}{c} 2k+1 \\ k+1 \\ \end{array}} \right) (1-p)^{k+1}$ because $\left( {\begin{array}{c} 2k+1 \\ k \\ \end{array}} \right) =\left( {\begin{array}{c} 2k+1 \\ k+1 \\ \end{array}} \right)$. Then, the difference between the two majority vote accuracies,

$$\begin{aligned}&P_n -P_{n+1}\\\nonumber&\quad =p^{k+1}\left[ {\sum _{j=1}^{k+1} {\left( {\begin{array}{c} 2k+1 \\ k+j \\ \end{array}} \right) \big \{p^{j-1}(1-p)^{k-j+1}-p^{j-1}(1-p)^{k-j+2}-p^{j}(1-p)^{k-j+1}\big \}} } \right] , \end{aligned}$$

(36)

is 0 because the inside of the curly brackets is 0 for every j. This ends the proof. $\square$