Inherent negativity of seemingly symmetric finite integer sets

[Evic]

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 anyone think of any application of this to any field or real world scenario?
Does anyone know if this type of set is used anywhere?

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

Djordje

 

[pwsnafu]

It's very simple actually. Instead of integers, suppose we have a bag which contains two white balls and two black balls. You draw one, but don't replace it. So there is 3 balls left, two of which will be the opposite colour to the one you just drew. The second draw will always be biased towards the opposite colour. In your case you are enumerating all possible draws, so obviously you should get more "opposites" than "sames".

Oh, and as n increases it converges to 50/50 as you said. Why? Because the effect of removing one ball is proportionally smaller as n increases.