Nationstates Mathyness

[vh]

I haz done it

I was wondering about the probability of a younger nation catching up to a older nation, population-wise.

So after some mathyness,

After 'n' days, a nation, initially starting out with 'd' less population than another, larger nation, will have this much chance of being ahead:



This formula is approximate, but it works well enough, and gets more and more accurate as 'n' approaches infinity.

('n' for days and 'd' for difference sounds strange, but d is supposed to represent difference).

And link to wolfram alpha page where you can replace 'd' with the difference in population and 'n' in the number of days

The probability of a smaller nation overtaking a larger nation will never be over 50%, of course, because a smaller nation has a disadvantage. Even if two nations started out with the same population, they wouldn't have more than 50% of growing larger than each other.

For example, Kyupaa, the second largest nation, has 15 million less people than Korinekia, and in a year, there's a 25% probability it'll be larger than Korinekia. In 10 years, a 41% probability, and in 100 years, a 47% chance.

Explanation:
The formula calculates the amount of area of the normal distribution that is from x=0 to x=infinty. (Normal distribution because of the Central Limit theorem). The normal distribution has an mean of -d. If the two nation's populations are larger, then the curve will be shifted down and less of it will be right of x=0. The variance of the distribution is sqrt(4n/3), because the variation of a single nation's variance is sqrt(2n/3) and the variation of two independent variables is the sum of the individual variables.

The square of a single nation's population can be calculated as 2n/3 because
'After n steps, there are 3^n equally likely distinct possible paths (made up of sequences of -1, 0, and +1). A path p has final position f(p) (note that many different paths can have the same final position, because the sequence -1 0 +1 ends at the same place as +1 -1 0, and so on).
Let S(n) be the sum of all the f(p)^2 after n steps. Then after n+1 steps, each element x^2 (where x is one of the f(p)s) in this sum is replaced by (x-1)^2+x^2+(x+1)^2, because each path can be continued by subtracting 1, staying the same, or adding 1. This is 3x^2+2, and since it's true for all paths, the sum after n+1 steps is 3*S(n)+2*3^n (since there's a 2 added for each of the 3^n paths included in S(n).) If S(n) = 2n*3^(n-1), then
S(n+1)=3(2n3^(n-1))+2(3^n)=2n3^n+2(3^n)=2(n+1)3^n . Because it's clearly true for n=0 that S(0) = 0 = 2*0*3^(0-1), it is true by induction that the sum of the squares of the endpoints of all possible paths after n steps is 2n3^(n-1). . Then the expected value is this divided by 3^n, to get 2n/3.'-gmalivuk

[matty406]

I understand nothing.

[FiahOwl]

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[vh]

Well pretty much all you need to understand is that 'd' stands for the difference in population and 'n' after the amount of days. Just click on the link and replace the values in wolfram alpha.

For example, Starrie has 19 million more citizens than Aeridani. I plugin 19 for 'd' and 365 for 'n' which will give me the amount of chance that in one year from now, Aeridani will have a higher population than Starrie: 19.5%