Best two or all three?
Consider the following situation. A course has three assignments, each worth \(\displaystyle 100\) marks. While calculating the final assignment score, which of these two scoring formulas would be the most beneficial from a student’s point of view?
- Average of all three assignments
- Average of the best two out of the three assignments
What happens if there are \(n\) assignments and we consider best \(k\) out of \(n\) assignments, with \(k < n\)? Will the policy chosen in the previous step remain the same?
Let the scores of a student in all three assignments in decreasing order be \(\displaystyle a\geqslant b\geqslant c\). Then, according to the first criterion, the average score is \(\displaystyle \frac{a+b+c}{3}\) and according to the second criterion, the average score is \(\displaystyle \frac{a+b}{2}\). Let us now consider their difference:
\[ \begin{equation*} \begin{aligned} \frac{a+b}{2} -\frac{a+b+c}{3} & =\frac{a+b-2c}{6}\\ & =\frac{( a-c) +( b-c)}{6}\\ & \geqslant 0 \end{aligned} \end{equation*} \]
We see that the policy of computing the average of the best two out of three assignment scores always results in a higher final score. Hence, this is the better alternative.
The generalization of this also holds. If we let \(a_1 \geqslant \cdots \geqslant a_n\) be the scores of a student in \(n\) assigments, then we can show by a similar argument that for any \(k < n\):
\[ \begin{equation*} \frac{a_{1} +\cdots +a_{k}}{k} \geqslant \frac{a_{1} +\cdots +a_{n}}{n} \end{equation*} \]