foundations of probability

Let our stick be the interval [0,4], and let X and Y be two points randomly placed on it. X is the first point we place, and Y is the second.

In order to not lose this game immediately, X must land on the interval (1,3). If it did not, then X would be within a distance of 1 from either 0 or 4. Since this interval is half the size of the larger interval, the probability of X landing here is 1/2.

Let's assume X lands in (1,3). Y must also land on this interval for the same reasons as X, but also Y must be more than a distance of 1 from X. This means the closer X is to 2, the less space Y will have on the interval.

Y's interval can be written as either (X+1,3) or (1,X-1), depending on if X is on the left or right side of 2. But both have a length of |X-2|. And just like for X, the probability of Y landing on either of these intervals is $\frac{|X-2|}{4}$.

To consider all possible cases of X, we can integrate $\frac{|X-2|}{4}$ over the interval (1,3):

$\int_1^3 {\frac{|X-2|}{4}dX}$

The absolute value makes the integral symmetric, so we can rewrite it as:

$2\int_2^3 {\frac{X-2}{4}\space dX} =\int_2^3 {\frac{X-2}{2}\space dX}$

Letting U = X-2:

$\int_0^1 {\frac{U}{2}\space dU} = 1/4$

The placement of the two points are independent of each other, so our final answer is then 1/2 * 1/4 = 1/8