Let’s say there are three rooms with 40 identical boxes in each room. Each box contains some number of cards inside. The number of cards in each box is different. It may be as many as 60 cards in one box and as few as one card in one box.
Although the amounts of cards in each box of the same room are different, the sets of boxes in rooms are identical. That means that if in the room A there is a box containing, say, 39 cards, then there should also be a similar box with 39 cards in the room B, and another such one in the room C.
The set of boxes in each room is compiled as follows: the first box has only one card, the second one has two, the third one has three and so on until the 40th box (inclusive). The 40th box, thus, has 40 cards in it. So, the total number of cards in each room is 820 (1+2+3+4+5+6+7+…+37+38+39+40 = 820). However, the contestants don’t know the total number of cards in each room.
The boxes are placed on the floor in each room in a random order.
Now, three contestants are given the task – each one of them is asked to enter into his room and, using his intuition, within one minute to try to get a bigger number of cards than his contenders by opening the smaller number of boxes than his contenders.
Therefore, each contestant will have two results: the “box performance” – let’s denote it by “B”, and the “card performance” – let’s denote it by “C”
So, let's say they have done their job and their results are as follows:
Contestant 1 opened 3 boxes and gathered 49 cards from them;
Contestant 2 opened 4 boxes and gathered 52 cards from them;
Contestant 3 opened 6 boxes and gathered 54 cards from them;
This can be represented in the following way:
B1 = 40-3 = 37, C1 = 49
B2 = 40-4 = 36, C2 = 52
B3 = 40-6 = 34, C3 = 54
Because each contestant has two numbers as a result (instead of just one number like that would be, say, in racing), I was a bit puzzled here as to how to decide the winner.
I have thought of a number of ways:
1) Dividing B by C in each case and then comparing the resultant quotients:
B1/C1 ~ B2/C2 ~ B3/C3
2) Dividing C by B in each case and then comparing the quotients:
C1/B1 ~ C2/B2 ~ C3/B3
3) Comparing the averages of C and the offset of B:
(C1+B1)/2 ~ (C2+B2)/2 ~ (C3+B3)/2
4) Calculating the “percentage indices” (I don't know how to name them in English properly) and then comparing them. Here is how I calculated them:
Since we know the minimum and the maximum possible numbers of boxes that can be opened by one contestant, which are 1 and 40 respectively, and since we know the minimum and the maximum possible numbers of cards that one contestant can gather, which are 1 and 820 respectively, we can try calculating the percentage:
820/100 = 1%, which is 1% of the quality of the contestant’s “card performance”.
40/100 = 1%, which is 1% of the quality of the contestant’s “box performance”.
Thus, in case with contestant 1, the percentage indices will look as follows:
%C1 = C1/(820/100) = 49/8.2 = 5.98%
%B1 = B1/(40/100) = 37/0.4 = 92.5%
Now we can calculate the average of these two indices:
(%C1 + %B1)/2 = (5.98%/92.5%)/2 = 49.24%
The first two ways didn’t strike me as correct, but the last two seemed fine.
However, when I calculated the averages and the percentage indices for the three contestants, I got the following results:
As it can be seen from this table, if I take the “Average” way, then the second and third contestants will share the first place, while the first one will win silver. But if I take the “Percentage Index” way, then the first contestant will get gold, the second one silver, and the third one bronze!
So, the final results of the “Average” way are quite different from the “Percentage Index”. Why is this so?
Which way here is more correct, the “Average” or the “Percentage Index”?
There is no "correct" answer here - it depends on which factors you think are more important. I will expand below.
First of all, methods 1 and 2 are not good for measuring what you want to measure. The contestant's goals are to maximize both $B$ (the number of unopened boxes) and $C$ (the number of cards obtained). The ratios $B/C$ or $C/B$ would be good measures if $B$ and $C$ were competing quantities - i.e., we want to maximize one and minimize the other. Thus you could instead use the score of $C/(40-B)$, the ratio of cards obtained to boxes opened. With this scoring method, your contestant's scores would be:
As for the third and fourth scoring methods, they are both weighted averages. The third is a simple average where $B$ and $C$ are weighted equally: $\frac{1}{2}B+\frac{1}{2}C$. The fourth is a weighted average where the weights are unequal. You can check that this score is equal to $\frac{1}{40}\cdot \frac{1}{2}B + \frac{1}{820}\cdot \frac{1}{2}C = \frac{1}{80}B + \frac{1}{1640}C$. So this scoring method places more importance on the value $B$ than on the value $C$. This might be desirable since $C$ can take much larger values than $B$, so this score "normalizes" the quantities. You could obtain other scoring methods by choosing any constants $\alpha,\beta>0$ and using the score $\alpha B + \beta C$. These scores differ in the importance they place on maximizing $B$ vs maximizing $C$.
The ratio score $C/(40-B)$ can't be so simply compared to the score $\alpha B+ \beta C$, since the ratio doesn't place a constant amount of weight on $C$ and $B$. When the number of boxes opened is small, $C/(40-B)$ places a lot of weight on increasing the number of cards, whereas if the number of boxes opened is large, the score doesn't increase much when you open more cards. From another perspective, if the number of cards is large, it places a lot of weight on decreasing the number of boxes opened. But if the number of cards is small, the score doesn't increase much by decreasing the number of boxes opened.
All in all, there is no "right" way to score this contest. You have to consider which factors are most important to you in determining who wins. It might help you to create more sample contestants, score them by various methods, and see how much you agree with the rankings.