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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 while opening the smaller number of boxes than his contenders.

So, let's say they have done their jobs 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;

Here each contestant comes up with at least two different results: the number of boxes he has opened – let’s denote it by “B”, and the number of cards he has gathered – let’s denote it by “C”.

Because each contestant has two numbers as a result (instead of just one number like in racing), I am a bit puzzled here as to how to decide the winner. I have thought of a number of ways:

1) Dividing the offset of B (which is 40-B) by C in each case and then comparing the resultant quotients:

(40-B1)/C1 ~ (40-B2)/C2 ~ (40-B3)/C3.

2) Dividing C by the offset of B in each case and then comparing the quotients:

C1/(40-B1) ~ C2/(40-B2) ~ C3/(40-B3).

3) Comparing the averages of C and the offset of B:

(C1+40-B1)/2 ~ (C2+40-B2)/2 ~ (C3+40-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 calculate 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 “card performance” (the more cards are gathered, the better)

40/100 = 1%, which is 1% of the “anti-quality” of the contestant “box performance” (the more boxes are opened, the worse; “anti-quality” is the offset of quality: the anti-quality result “7 opened boxes” corresponds to the quality result “33 unopened boxes”, “8 opened boxes” corresponds to “32 unopened boxes”, and so on)

Thus, in case with contestant 1, whose B value equals to 3 (3 boxes opened) and whose C value equals to 49 (49 cards gathered), his percentage indices will look as follows:

%C1 = C1/(820/100) = 49/8.2 = 5.98%

%B1 = (40-B1)/(40/100) = (40-3)/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 ones, 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:

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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”?

brilliant
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  • Please add the `[self-study]` tag & read its [wiki](https://stats.stackexchange.com/tags/self-study/info). Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. – gung - Reinstate Monica Dec 19 '19 at 16:11
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    This question isn't answerable, because the criteria for choosing boxes given to the contestants aren't actionable: it is not possible simultaneously to obtain "as many cards as possible by opening as few boxes as possible." Some instructions are needed concerning how to trade off between getting more cards and opening more boxes. Thus, there is no definite basis for choosing winners. – whuber Dec 19 '19 at 20:08
  • @whuber - “it is not possible simultaneously to obtain "as many cards as possible by opening as few boxes as possible."” - Why is it not possible? Imagine one contestant opens the first three boxes: the box 1 with only one card, the box 2 with two cards and the box 4 with 4 cards inside, and the other contestant opens only the box number 40, which has 60 cards inside. In this case, the first contestant, the who has opened 3 boxes, will have gathered only 7 cards (1+2+4=7), while the second contestant, the one who opened only one box, will have 60 cards. – brilliant Dec 20 '19 at 08:41
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    @brilliant the phrase "as many cards as possible by opening as few boxes as possible." is ambiguous (you seem to be thinking about something like a ratio, but that does not clearly follow). The most possible cards is 840, but this requires to open 40 boxes. The fewest possible boxes is zero, but this ends in only zero cards. Therefore it is not possible to have both the most cards (840) and as few boxes possible (0). – Sextus Empiricus Dec 20 '19 at 14:19
  • @SextusEmpiricus - So, can you, please, help me phrase it properly? I don't know how to express it in English well (I am not a native English speaker). In my first language the phrase "each contestant within one minute has to get as many cards as possible by opening as few boxes as possible" wouldn't mean that a contestant is expected to get all 840 cards, nor would that mean that he would be expected not to open any box at all. That phrase would only mean (in my language) "as many/few as possible in order to be the winner" or "in order to beat the other contestant". – brilliant Dec 20 '19 at 14:45
  • @brilliant if I drop part of the sentence "each contestant within one minute has to get as many cards as possible by opening as few boxes as possible" and change it into "each contestant has to get as many cards as possible" then what does it mean in your language? It does not mean that a contestant with 840 cards wins? – Sextus Empiricus Dec 20 '19 at 15:15
  • @SextusEmpiricus - The phrase "each contestant has to get as many cards as possible" would still not mean in my language that a contestant would necessarily have to get all 840 cards to be the winner (especially in the situation described in my question - when contestants don't know the total number of cards in one room). The phrase "as many cards as possible by opening as few boxes as possible" in this context in my language would be taken as "do your best to get more cards than your opponent gets, and do your best not to open as many boxes as your opponent opens" – brilliant Dec 20 '19 at 15:52
  • @brilliant 'sure as many cards as possible' may not need to be 840 (because 840 might not be practically possible). You are missing the point, you can substitute whatever number you believe is the most possible number instead of 840. – Sextus Empiricus Dec 20 '19 at 16:33
  • @SextusEmpiricus - (1) “You are missing the point, you can substitute whatever number you believe is the most possible number instead of 840” – I don’t quite understand why you say that I am missing this point. Yes, I do realize that you can substitute whatever number as the most possible instead of 840. And for the contestant that would be any number that would make him the winner. I never said that the most possible number meant exactly 480. That’s, in fact, what I said in my previous comment. – brilliant Dec 21 '19 at 00:33
  • @SextusEmpiricus - (2) Contestants in my question don’t even know how many numbers are in the room, so the number 840 is not even on their mind. The only thing that they know is that as long as they get a number of cards bigger than the number of cards of their competitors, they can win. – brilliant Dec 21 '19 at 00:33
  • *"a number of cards bigger than the number of cards of their competitors"* They may not to know the number is 840, but they know that the best possible result that they can achieve is opening all the boxes. – Sextus Empiricus Dec 21 '19 at 08:56
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    Did you already explain why you have this question? The background, more precise reasons what you are gonna do with the answer and where the question came from? – Sextus Empiricus Dec 21 '19 at 08:58
  • @SextusEmpiricus - "*they know that the best possible result that they can achieve is opening all the boxes*" – No! Rather, they know that the best possible result that they can achieve is opening fewer boxes than their opponents while getting more cards than their opponents, in which case they will definitely be the winner. If they open all the boxes, it is still possible that their opponents can also open all the boxes, in which case it will be a draw. – brilliant Dec 21 '19 at 10:49
  • @SextusEmpiricus - “*Did you already explain why you have this question? The background… and where the question came from?*” – Not yet. I am still thinking of how to explain that properly in English. And for that I firstly need to clarify this misunderstanding with you, which was caused by my poor English. – brilliant Dec 21 '19 at 10:50
  • @brilliant you have to go back to my post where I write "if I drop part of the sentence", if it is not about three different parameters, but instead about only one, what does it mean in *that* case? – Sextus Empiricus Dec 21 '19 at 10:58
  • @SextusEmpiricus - Of course, in *that* case it would mean “the more boxes you open, the better”. However, that case is not my case. In my case the phrase "each contestant has to get as many cards as possible" is right away modified by “by opening as few boxes as possible”, in which case it’s already not “the-more-the-better” meaning. It is more now like “the more, the better” in one parameter (cards) and “the-fewer-the-better” in the other parameter (opened boxes). Actually, I kind of don’t understand why you needed to drop that part of a sentence. – brilliant Dec 21 '19 at 11:11
  • I dropped that part of the sentence in order to verify what it means in a simpler case. What I believe now is that what the fuller/complex sentence means has nothing to do with language. Otherwise you could tell *us* who according to your language interpretation would be the winner, instead of posting a question about it. – Sextus Empiricus Dec 21 '19 at 15:11
  • @SextusEmpiricus - I see. Thank you. OK, so now it's the time for me to re-write my question. I'll posted it as soon as I'm done. Thanks again fro your time. – brilliant Dec 21 '19 at 16:38
  • @moderators - I have just re-written my question. Please, re-open it. – brilliant Dec 22 '19 at 16:10
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    @SextusEmpiricus - It's done now. – brilliant Dec 22 '19 at 16:20
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    @brilliant your problem with the students seems much more viable for being a good question. The problem with the box problem is that there is no single/unique way to turn three numbers into one number, and the problem is too abstract (missing context) to come up with a way that is best. The box problem could be usefull to help to write the answer, but you would do better to explain in more detail your problem with the students and explain why it corresponds to the boxes in the room problem. – Sextus Empiricus Dec 22 '19 at 20:01
  • *"What is the best statistical way to calculate the performance of each contestant in such cases and express it by only one value?"* There is no such thing as 'a statistical way to calculate'. (although it might be possible to involve statistics into the problem, for instance when you want to use data in order to create a performance measure, for instance if you have knowledge of correlation between your three values and some single relevant measure like exam marks or something like that) – Sextus Empiricus Dec 22 '19 at 20:02
  • @SextusEmpiricus - I see. Thank you. I'll drop "*What is the best statistical way*" from my question then. – brilliant Dec 22 '19 at 22:38
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    See https://stats.stackexchange.com/a/9361/17230. – Scortchi - Reinstate Monica Dec 22 '19 at 22:58
  • @Scortchi-ReinstateMonica - Thank you for that link. It has a lot of discussion there, but not the definite answer. Perhaps, there's no one definite answer here, which makes me consider narrowing down my question once again. – brilliant Dec 23 '19 at 09:34
  • @brilliant: You're welcome. There's certainly no one definite answer. If you're able, however, to judge rank order in given cases, there are more or less systematic approaches to defining a common scale, one of which is recommended there. – Scortchi - Reinstate Monica Dec 23 '19 at 10:36
  • @Xi'an Please, re-open my question. I have already re-written it. – brilliant Dec 23 '19 at 15:11
  • @Peter Flom - Reinstate Monica Please, re-open my question. I have already re-written it. – brilliant Dec 23 '19 at 15:11

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