Instigator / Pro
4
1596
rating
9
debates
88.89%
won
Topic
#2131

It is Impossible For 100% of People in a group to be Above That Group Average at Anything

Status
Finished

The debate is finished. The distribution of the voting points and the winner are presented below.

Winner & statistics
Better arguments
0
3
Better sources
2
2
Better legibility
1
1
Better conduct
1
1

After 1 vote and with 3 points ahead, the winner is...

RationalMadman
Parameters
Publication date
Last updated date
Type
Standard
Number of rounds
3
Time for argument
Three days
Max argument characters
5,000
Voting period
Two weeks
Point system
Multiple criterions
Voting system
Open
Contender / Con
7
1702
rating
574
debates
67.86%
won
Description

BoP on pro.

Round 1
Pro
#1
We can begin the proof by assigning an arbitrary system of valuing a person or group's "ability" at any particular "thing" (loosely described in the resolution intentionally so as to allow con more room for potential counterarguments I may not have thought of). This is most easily done by assigning numerical values, either in a ranking system (best, second best, etc.) or a hard measurement (139 points, 45 points, 6 points, etc.). In either case the following demonstration holds true. Note this also applies to any combination of ties, as if anything is "tied" to the average it is not "above" the average as worded in the resolution.

The average of any particular group of numbers can be defined by finding that groups mean, median, or mode. Whichever of these definitions is used the statement in the resolution holds true for any numerical group (and hence any group that can be translated to be represented numerically).

Mean

The mean is the typical and most common definition of the word average in day-to-day conversation, though not strictly speaking the only acceptable one. If one interprets the word average in the resolution to refer a particular group's mean then the resolution is upheld. Any group where all the members are equal will have a mean equal to all the members of that group (therefore none are above average in that case). As soon as there are two different numbers in the group the mean will be somewhere between those two numbers meaning that one will be above average and the other bellow average. Adding more members to the group that are different numbers does not help. If the new members are higher than all the old members then the mean is increased and therefore the lowest member is still below average. If the new member is lower than the previous two the mean is decreased but the new value of the mean is still held above the value of the new member by the higher numbers. There is always at least one value below the mean, therefore less than 100% are above it.

Median or Mode

The resolution is perhaps most easily upheld if one uses the median or mode of a group as the definition for the average of that group. By definition the median or mode must be exactly equal to at least one of the members of a group and being exactly equal we can at least rule out that member being above the mode, meaning no matter the status of the other members ere is at least one (and therefore less than 100%) that is equal to (and therefore t above) the average.
Con
#2
Forfeited
Round 2
Pro
#3
Unfortunately con has made no arguments last round. I have nothing to rebut and nothing to defend my own arguments from and therefore will simply pass this round.
Con
#4
My case is extremely simple and since my opponent didn't use sources, I will keep if very theoretical too.

This debate is won by Con due to three things:

  1. The term 'anything' meaning if even one, or a few, scenarios can make it possible, the vast majority are irrelevant to Pro winning the debate.
  2. The difference and vagueness in time/tense between when the average is calculated and when the original measurements are taken vs what the 'true measurement' may be at the time when we analyse if the group has surpassed the average.
  3. Scenarios of 'identical' groups only favour Pro unless every single member surpasses their average as time goes by (linked to point 2)
  4. (Linked to point 1) subjective-analysis leading to qualitative outcomes absolutely annihilates the Pro side of the debate thanks to the 'anything' factor.
So, to state the obvious and not abuse 'sources' points (I'll bring them in the later Round if need be), let's admit that 'anything' means if I can prove that in some (or even just one) situation(s) that 100% of people in a group can surpass the group's average, I win this debate. This is important to address as I am conceding that if the resolution said 'everything' or even 'most things', I'd lose in the former scenario and need a ton of proof and complexity to my case in the latter scenario where I still think I'd lose.

Basically, this resolution can be technically nitpicked via points 2 and 3 of what I listed, I want to instead go straight to point 4 so that we first destroy Pro's case in a swift move before nitpicking at technicalities to further our victory.

Let's say we have a group of men and have a selection of heterosexual women, bisexual women and/or homosexual men to judge their sexual appeal. The key thing to bear in mind here is that we are having a scenario where the 'value' of the subjects is completely and utterly dynamic, analogue and thus ambiguous. No matter what 'metric' you used, even if they had to rank them out of 10, the key aspect of this is that at any given point in time it would be possible for absolutely all of those men, equally to each other, to be sexier than the mean average of the group (or even median). Mode is irrelevant to what this debate is trying to explore, neither side wants this debate to be about modal average so let's sit that aside.

When we are discussing subjectively evaluated things, Pro will retort that even though it is subjective, the measurements out of 10 taken at the particular time still can't all be greater than the mean/median average. Here is where Con need to bring in points 2 and 3 to nitpick technicalities and truly secure the win. Whether it's sexiness, art-appeal, funniness or anything subjectively analysed, we need to understand that this resolution is extremely vague as to 'when' and 'where' the average and the people in the group's measurement/value are located in our logical algorithm.

You see, semantically if we have a group of 3 year olds, 100% of the people in that group has surpassed their measured group average once they have all reached age 4. Furthermore, the debate's topic is written as 'above' and it is the people that are above. Let me reiterate this to you, so that you understand:

The debate's title does not say 'the measured data relating to people' and while this is a sensible way to interpret the thing being resolved here, I would like to nitpick the semantics a little more to completely win this debate from all angles:

It says 'people in a group'. The people in a group themselves do not have a value, so their value can be called 0 of any unit. What I mean is that a person is not the temperature of the person that is taken for an experiment, that person is simply a being that is there and which you are taking a measurement from. Thus, if 100% of the people in a group are taken as a whole, we realise that they are impervious and valueless to our actual metric.

In other words, can Donald Trump, Joe Biden and Barrack Obama all three be above the group average of milliletres cellulose in a plant? Well no, they can't, as that value will always be positive or 0. How about if they all three can be above the temperature, in Celcius, of the Arctic? I think they can, as that will be negative and Donald Trump isn't a temperature value at all and will be rendered functionally 0 since he's not a value and neither are the other 2 people. Do you understand what I am saying? The 'people' are not the values taken and even if the values taken are what you interpret it as, there's the time-taken aspect that makes me win this debate no matter what.

To further elaborate on the original point-4 extension, something like 'how appealing or disgusting' something is has the pandora's box element to it if we take 'impossible' to apply to us saying so before the measurements are taken. It's always possible beforehand.
Round 3
Pro
#5
Well first thing first, yes the wording of the resolution does dictate that only a single counter-example is needed to prove con's case.

That being said I do not think the counter examples presented are sufficient, and I shall explain why. As the voter has seen from con's previous round their argument relies on the resolution being - in their words - "extremely vague as to 'when' and 'where' the average and the people in the group's measurement/value are located in our logical algorithm."

I would briefly argue that this is not entirely the case. The fact that the resolution and description make no mention of time is not indicative of any vagueness but instead is indicative of an implicit assumption that time is not a relevant factor.  However, I also believe that even if time is made to be considered a relevant factor the resolution can still be upheld. I will demonstrate why this is so by examining what I believe to be the stronger of con's two example test cases. Here con says:

You see, semantically if we have a group of 3 year olds, 100% of the people in that group has surpassed their measured group average once they have all reached age 4.

Con correctly anticipates my response when he says "The debate's title does not say 'the measured data relating to people'" but I will take a moment to articulate it clearly for the audience before addressing his rebuttal. So, let us suppose that the hypothetical group above consists of 3 children each 3 years old. the average age in this case is 3 (3+3+3 = 9... 9/3 = 3). One year later if their age is 4 then you may suppose that since 4 is greater than 3 all of them have surpassed their average age and thus the resolution is defeated. However, the average is only less than 4 if you keep the data points from when the ages were measured a year earlier. In that case the average is 3.5 (3+3+3+4+4+4 = 21... 21/6 = 3.5). However if you keep the 3 year old versions of the people in the data set you are looking at a group where half the members are 3 years in age, therefore not all in the group are above average 

Anyway that is the response that con anticipated and and attempted to preemptively counter. 

The problem is that his attempted pre-empt depends upon trying to "have his cake and eat it too". For instance in the above example there are six entities, 3 kids age 3 and 3 kids age 4. Con wants to say that the age 3 entities being below the average does not count because those age 3 entities no longer exist, yet at the same time he wants to say that they do count as being in the group because he wants them to drag down the average.

What it comes down to is this, you can count the 3 year old version of the kids as separate entities that are part of the measured group, in which case you get an average age of 3.5 with half of the members below the average age as shown in the maths above. Alternately you can say that the younger versions are not part of the group and therefore the average is 4 (4+4+4 = 12... 12/4 = 4), which makes all the group members equal to (and therefore not greater than) the group average age.

The reason this interpretation works from a semantics point of view is the part of the resolution that says "to be Above That Group's Average at Anything". The first thing we need to do, therefore, is define the group we are looking at. Say the kids are named Abe, Betty, and Claire (A B and C)... we can look at the resolution and say "The group we are looking at is 3 year old A, B, and C" or "We are looking at 3 year old A, B, and C along with 4 year old A, B, and C" and any of these views have results detailed above which support the resolution. What con wants to do is say "the group we are looking at is Abe, Betty, and Claire but not at any particular age" then try to misdirect into actually looking at the 4 year old versions in conjunction with younger versions, but not include the younger versions... but also include the younger versions... kind of... But if the group in question is left so vaguely defined there is no justification to putting any specific number values on such non-specific entities and also nothing to stop someone from including, for example, the 10 year old versions of A, B, and C to drag up the average. such would not be prohibited due to not specifically defining the trait measurements (in this case age in years) and assigning them to specific entities (4 year old Abe as opposed to just 'Abe', etc.)

Any other example con may come up with runs into the same problem. Different versions of a person (not just different ages but any trait one may measure) are either separate entities part of the group or they are not. If they are  then some entity will be below the average. If they are not part of the group as separate entities they cannot artificially drag the average down. If this is left vaguely defined the whole thing kinda just crumbles.

Con
#6
When is it possible? When is the data taken? When is the average calculated? Where are each of these pieces of information located to obtain and then use?

When we say it's impossible for all men to be sexoer than the later-calculated average, do you see the issue? When we measure something and then later hold that data against the current average, what happens when a group of one age all are older now than when we took their results then?

Thank you.