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distance 'd' will be between 0 < d < 1

1/3

I can write a simple program to get this answer, but it's hard for me to mathematically prove it.

Let me try to prove it mathematically:

let's assume the position of the first point is y (0<=y<=1). So the probability dense function of the distance to the second point is:
f(x) = 2, o<=x<=y
f(x) = 1, a<x<=1-y
therefore, we can get the expectation as: definite integral (x*f(x))dx from 0 to 1, and get E(x) = y*y-y+1/2 (here y is a parameter).
Now we need to compute the expectation of E(x) given all the possible values of y.
Then according to the Central limit theorem, we can calculate the expectation as: definite integral (y*y-y+1/2)dy from 0 to 1 (here y is a variable), and get result as 1/3.

Not sure if this is a correct proof.

Assume a uniform distribution between 0 and 1. Let f(x) = 1, for x in [0, 1].
Then the joint density of two point x, y is f(x,y)=f(x)f(y)=1, for x,y in [0,1].

Compute the integral (in Tex script)
\int_0^1 \int_0^1 |x-y| f(x,y) dxdy = \int_0^1 \int_0^1 |x-y| dxdy = ... = 1/3

E[x] = Integral[x*(1-x^2)^1/2] from limits 0 to 1.. = 1/3
since f(x,y) = x^2 + y^2 - 1

Let a and b be random, uniformly distributed variables such that 0<=a<=1 and 0<=b<=1. These are our two points on the line. We can say that the expected distance between the two points is the expected probability that a third, randomly distributed point, c, 0<=c<=1 lies between a and b. c is no different from a and b so the odds that c lies between a and b is 1/3.

@Anonymous nice but I couldn't get.