# Inherent negativity of seemingly symmetric finite integer sets

Hi everyone.
My first post on this great forum, keep up all the good ideas.
Apologies if this is in the wrong section and for any lack of appropriate jargon in my post. I am not a mathematician.

I have a theory / lemma which I would like your feedback on:

- Take a finite set S of integers which has an equal number (n) of positive and negative elements.
The trivial example would be n= 1 and S = [-1,1], or n = 2 and S = [-2,-1,1,2] and so on to something less trivial like n = 3 and S = [-35,-3,-2, 1,1,100]

- Take two elements from the set without replacement. ie. do not choose the same element twice. for the trivial case n=1 we get [-1,1].
When n = 2 we have possibilities [-2, -1] , [-2, 1] , [-2, 2] , [-1, 1] , [-1, 2] , [1, 2]

Note that 4 of these possibilities (at n = 2) have one element of each sign, and only 2 are of the same sign.

- In the general case this holds (as far as I can tell). That is using non replacement choosing of two elements on such a set will result in more possibilities with one element from each sign than two elements of the same sign.
As n increases the probability approaches 50% but is always slightly higher favoring the opposite signs.
n = 1 = 100%
n = 2 = 66.%
etc.

- Perform a multiplication or division on your chosen elements. Because there is a higher probability for the elements to have opposite signs, there is therefore a higher probability that the result of your operation will be negative.

That is my understanding of the situation. If I have gotten something wrong please correct me.

And here is my question. Why in a seemingly symmetrical set is there a tendency towards a negative result?
And can any one think of any application of this to any field or real world scenario?
Does any one know if this type of set is used anywhere?

Thank you all for your time.
I look forward to your responses.

Djordje

pwsnafu