Is this mathematically/statistically possible?

Let’s say there are 4 groups of people, with an unknown number of people in each group. Within each group, every person is either a 1 or a 0. The percentages below describe the percentage of people who are 1s.

Group A: 50%
Group B: 51%
Group C: 50%
Group D: 50%

Is it possible for Groups A, B, C, D aggregated to be less than 50%? Is it still true if Group B is 50% instead of 51%?

I think it depends on whether or not you allow

  1. A group to have 0 people.
  2. If you define a group with an odd number of people as meaningful for A,B,C; and exactly how you handle it if you can have an odd number.

NO (assuming your percentages are exact and not 49.8% rounded to 50%). Yes, it’s still true.

Doesn’t matter. Your aggregate has to be a convex combination of 50%, 50%, 51% and 50%, so it can’t be any smaller than 50%.

No, it’s not possible. At least half the people in each group are 1s. Adding this up, we get that at least half the total population are 1s.

Indeed, if Group B is exactly 50%, we have that exactly half the total population are 1s.

How would a group have 0 people and still have 50% or 51% 1s?

When dealing with averages of various sized groups like this, and then trying to think about the total aggregate average, you are dealing with weighted averages. Perhaps the best-known situation where we all (most of us) have dealt with weighted averages is in computing our school grade-point average.

Here’s a helpful idea for thinking about this: Every individual contributor to the total average has the effect of pulling the aggregate average toward itself. Whatever grade you get in a 5-unit class, and whatever your GPA was before that, and however many total units you had before that, the new 5-unit grade will pull the overall average toward itself. It will pull the average more toward itself that a 4-unit grade would do, which will also pull the average toward itself.

If you only have a few units already, the new grade may affect the average a large amount. If you already have hundreds of units, the new grade may be just a drop in the bucket. If your average is a C, then either an A or a B will pull you up, but the A will pull you up more. But in ALL cases, the new grade pulls the average toward itself.

ETA: If you allow a group of 0 people, it would have a weight of 0 and not affect the average at all. So even if you had an empty group and (somehow) computed its average to be 51% ( :dubious: ) that still could not draw the aggregate average down below 50%.
You have the same sort of thing in the OP’s example. Each group contributes to the aggregate average according to the (unspecified) size of the group. The size of the group is its weight. A big group affects the average more than a little group. If the average is 50%, then adding a 75% group makes a bigger difference than adding a 51% group, and the bigger the group is, the more difference it makes.

Here, you have three groups with 50% average, for an aggregate average (of those three groups) of 50%. Adding the 51% group can only pull the aggregate average up toward (but not beyond) 51%. Even if the three 50% groups are small and the 51% group has six billion people in it, it can only pull the average up toward 51% and not above, and not below 50% either.

If the groups are large, then dealing with odd-number groups and halves of people can only make a minuscule and insignificant difference. If the groups are small, then fractional people might matter, and the result would depend on how you want to count them. But there’s no way the 51% group, added to the other 50% groups, could draw the average down below 50%.

0 * .5 = 0?

Yeah, the more pertinent question is, how would a group have an odd number of people and still have exactly 50% (or, for that matter, exactly 51%)?

But, at any rate, the answer is unconditionally “No”, as everyone else has said.

bizerta is right. Basically, this is because if you have two non-empty groups, the average of the aggregated group must be between the averages of the two separate groups. (And an empty group can’t have an average, because you can’t divide by zero.)

Here’s the proof:
Let group A have n members, and an average of m/n.
Let group B have q members, and an average of p/q.
Suppose that m/n < p/q. (The other case where m/n > p/q has a similar proof, and if m/n = p/q the aggregated average is the same number.)
Then, since n and q are positive integers, their product nq is positive, and you can multiply by nq to get mq < np.
Add pq to each side to give (p + m)q < (n + q)p
Now q/(n + q) is positive (since n and q are positive), so multiply each side by that:
(p + m)/(n + q) < p/q
So the average of the aggregate [(p + m)/(n + q)] must be less that the larger average.

You can construct a similar argument to prove that the average of the aggregate must be larger than the smaller average.

Maybe it’s a math thing, but I’m pretty sure there’s no 1s in a group with 0 people.

It depends on how you define the problem.

You say "a group of people of an arbitrary number of people are split into four groups under the following conditions:

A such that half are assigned a 1, and half are assigned a 0.
B such that just over half (51%) are assigned a 1, else 0.
C as A
D as A"

Say three people enter the room, how do you handle the problem? The problem is not worded to exclude odd numbers or people, nor does it place a lower bound, so therefore the following scenario is possible

A has one member
B has zero members
C has one member
D has one member

So for the person in A,C,D, you must assign a one or zero to the person. Since there’s only three people, you must either have the case (unordered tuples) (0,0,1), (0,1,1), (0,0,0), or (1,1,1).

I don’t understand what I’m missing here, the problem is not defined in a sufficiently exclusive way, so corner cases have to be handled. Unless the problem specifically defined that either you can’t have less than 4 people, or else specifically predefines a behavior in such cases, the question allows potentially all possible configurations of corner cases to be valid, meaning that there must exist cases where there isn’t an even split.

Okay, then you assume that if B only has one member, it must be a 1.

Then you have the configurations

(1,0,0,0), (1,1,0,0), (1,1,1,0), (1,1,1,1). I just don’t get why corner cases don’t have to be predefined.

In a group with 0 people, it is simultaneously true that the number of 1s is 0% of everyone, 100% of everyone, 50% of everyone, 51% of everyone, and every which other thing.

As I and everyone else read it, the problem is: Is there any way to have four groups of people, such that in each group 50% of the group are 1s and the rest are 0s (or this is the case in three of the four groups, and in the fourth, 51% are 1s), and overall, less than 50% of the total population is 1s?

The answer to that question is no. There’s no way to have that. There’s no ambiguity that needs to be resolved; it’s simply impossible to have that.

It’s also true, while we’re at it, that the answer to the question “Is there any way to have four groups of people, such that in each group exactly 50% of the group are 1s and the rest are 0s (or this is the case in three of the four groups, and in the fourth, exactly 51% are 1s), and also, at least one of the groups has an odd number of people?” is also no, but that’s irrelevant to the first question.

I don’t understand, the OP made me think we were doing assignment, meaning n people walk into a room and then we assign them groups, and then assign group members numbers due to some algorithm, in which case don’t these cases need to be accounted for in case of strange input?

I don’t know why you thought the OP had anything to do with assignment or an algorithm. They straight up ask “Is this mathematically/statistically possible?” and then describe a situation. The situation they describe has nothing to do with assignment or an algorithm. It also happens to be a situation which is not mathematically/statistically possible. Accordingly, the answer is “No”.

It’s analogous to this:

"Is the following situation mathematically/statistically possible?

There is a group, with some number of men and some number of women.

There are more men than women.

There are twice as many women as men."

The answer to that question is “No”, not “What happens if the number of women is odd?”.

See, this is why I leave math to mathematicians, and cats to physicists.

It’s no crazier than the fact that 0 = 0.00 * 0 = 1.00 * 0 = 0.50 * 0 = 0.51 * 0, etc., which I am surely you will readily agree to (anything times zero is zero). Indeed, it’s the exact same fact.

But, yes, this is why people are generally and reasonably loath to discuss proportions or average values among an empty population.

I guess I just thought that multiplication and division was defined for every number in the set of natural numbers (save division by zero), so that meant there was a definition for n/2 if n is odd, even though there is no INVERSE operation (i.e. is not isomorphic). This means that defining a problem such that 2n = x is not the same as defining a problem such that n/2 = x if the domain is the set of natural numbers (which is the only set that really makes sense with people). Which would mean that your example with “twice as many” isn’t analogous to the OP’s “half of them” problem. I don’t know, I just find this really, really confusing that it uses the set of natural numbers, but is somehow in a different space with different operations defined.

Um, what? Perhaps I should rephrase, then.

"Is the following situation mathematically/statistically possible?

There is a group, with some number of men and some number of women.

There are less women than men.

There are half as many men as there are women."

The answer is “No, that’s not possible”, not “What happens if there are an odd number of women?”.

Or, put another way,

"Is the following situation mathematically/statistically possible?

There is a group of people, each of which is one of American, British, or Canadian.

One third of the group is Americans.

There are as many Americans as Brits and Canadians combined".

The answer is “No, that’s not possible”, not “What if the number of people in the group is not divisible by three?”.