Help Me Understand Averages

I need some math help. Tell me why two averages do not equal the average of all sets.

Why does (a+b)/2 + (c+d)/2 not equal (a+b+c+d)/4?

I know these equations are not equal but logically I don’t understand why they are not equal. Explain it to me as if I am five years old. Thank you.

*I know these equations are not equal *

Maybe you’d like to try to evaluate that equation yourself either longhand or in a spreadsheet, and then come back to us with your findings? Might save unnecessary irritation of that 5yo.

Maybe if you rewrote it one notch down it would click.

(a+b)/1 + (c+d)/1 = (a+b+c+d)/2 [not!]
(a+b) + (c+d) = (a+b+c+d)/2 [nope!]
(a+b+c+d) = (a+b+c+d)/2 [still no! but maybe clearer]

or rethink as what you are saying in the first one is add two averages - and the second is make one average. Even if all numbers were equal a=b=c=d (let’s say 25) you are saying…

On the left side - give me the average of two numbers that are equal - now double that.
On the right side you are saying give me the average of four numbers that are equal

A question that may help you think about what you’re trying to think about: Why should they be equal?

What your example is about is the sum of two averages… you should be talking about the average of two averages…

e.g. (((a + b) / 2) + ((c + d) / 2)) / 2

if a = 1, b = 2, c = 3, d =4…

= ((3 / 2) + (7 / 2)) / 2 = (10/2) / 2) = 2.5 which is the same…

Obviously it would work if the denominators were the same.
So if it helps, just read out the equations like this:

A half of a, plus a half of b, plus a half of c, plus a half of d
A quarter of a, plus a quarter of b, plus a quarter of c, plus a quarter of d

…not the same

(Noting of course that (a+b)/2 = a/2 + b/2)

If a=1, b=2, c=3 and d=4, then 3/2 + 7/2 = 1.5+3.5=5 and 10/4=2.5. They are different. Tell me why.

Because the first one is ten divided by two, and the second one is ten divided by four.

Averages aren’t your problem here if you don’t understand how adding fractions works.

Okay. This sort of helps. In the equations I am working on, the second set might just be one number, not averaged. So I can see that the equation is not equal, but I don’t understand it logically. It feels like it should be equal but I know it is not.

The first involves adding two averages (NOT the average of two averages) and the second is the average of all of the numbers.

What about this:

Say there was

(1+2)/2 + (3+4)/2 + (5+6)/2… the sum of those averages is 1.5+3.5+5.5 = 10.5

The average is 3.5…

What you need to do is find the AVERAGE of the averages, not the sum.

e.g.
(1.5+3.5+5.5)/3
= 10.5 / 3
= 3.5

Okay. So obviously 10 does not equal 10/2. So an average of two numbers plus an average of two other numbers do not equal an average of all four numbers. Because they are two totally different sets that have no relation to each other. C and D do not have anything to do with A and B. Math is so logical to me normally. Maybe I am just getting old!

You’ve got to divide by two still. And in my previous post you’ve got to divide by 3. (I had 6 numbers - 3 sets of averages)

It’s like having 100 averages then adding them all up - you’re meant to divide the sum by 100 to get the overall average.

I think the easiest way to think about it is to forget about the origin of the first two averages. Let’s say (a+b)/2 = 1 and (c+d)/2 = 1000.

Now we just have 1 and 1000. Your question is why the average of 1 and 1000 (500.5) is not equal to the sum of 1 and 1000 (1001). I don’t think you’d have asked the question if you had started with those numbers, right? It’s only because you’re using averages to get them that it seems weird.

Okay. I think this makes the most sense. The average of two averages does not equal the average of all numbers included in two different averages. Does that make sense?

You’re not quite there. The average of a and b, plus the average of c and d, doesn’t equal the average of all four numbers. The average of all four is the average of the two separate averages.

A real world example, suitably simplified. Suppose there are as many female Dopers as male in a short survey you take. Suppose the average height of all the female Dopers is 5’ 6". Suppose the average height of all the male Dopers is 6’. It therefore follows that the average height of all the Dopers in your sample is (pick one):

Eleven feet six inches, or
Five feet nine inches.

The correct way to work out the average of two sets is

(1’s average * 1’s size + 2’s average * 2’s size) / (1’s size + 2’s size).

But if both sets are of equal size, there is a short cut ( which you can easily see in the above formula) is

(1’s average + 2’s average) /2

Actually in my examples the average of two (or three) averages is equal to the average of all of the numbers combined… that’s assuming that each average has the same number of items…

But what if the items in each average are of different sizes…

e.g. the average of (1,2) and (3,4,5,6)… well the overall average is 3.5 and the average of the first group is 1.5 and the average of the second group is 4.5… the average of the two averages is 3… that’s because it gave equal weight to the first group (1,2) even though it only had 33% of the numbers.

Ignoring that these are averages, think of it in terms of simple fractions, since these simple examples are simple fractions:

1/2 + 1/2 = 2/2, not 2/4

In one room you have Mark Zuckerberg and Bill Gates.

In the next room you have a million random Americans.
The average income in room 1 is $100,000,000.
The average income in room 2 is $30,000
What’s the average income of everyone in both rooms together?

$100,000,000 + $30,000 = $100,030,000. Well, THAT doesn’t help, that’s just one average ADDED to another average, ending up with a number that is larger than either one, and not the average of anything.

($100,000,000 + $30,000) / 2 is $50,015,000. That’s more interesting. If for each room you say “ok, I’m going to take the average-income-of-occupants, and call that the AIO… so we have one room with an AIO of 100,000,000 and one room with an AIO of 30,000. So, what’s the average AIO across all rooms?”, it’s $50,015,000. That’s what you get when you take two averages, add them together, and divide by two. The average of the averages.
But what you MOST likely want is the average income of everyone in both rooms.

So the two averages are:

(1) (gates + zuckerberg) / 2
(2) (guy1 + guy2 + … + guy1000000) / 1000000

And the TOTAL average is:

(gates + zuckerberg + guy1 + guy2 + … + guy1000000) / 1000002

And that is neither of the previous numbers we’ve calculated. Instead it’s somewhere around $30200, if I did my math right.

Not sure if that helps or not :slight_smile: