Probability of 4 disjoint randomly selected segments Select 4 random real numbers $x_1,x_2,x_3,x_4$ from $[0, 9]$. What is the probability that the 4 segments $[x_1,x_1+1],[x_2,x_2+1],[x_3,x_3+1],[x_4,x_4+1]$ are disjoint?
I have considered using linear programming. However in this question there are four dimensions and I think it is not the best way. I have no clue on how to solve this problem so can someone help?
Sorry for not putting any further information/attempts - as I don't have any.
Thank you. 
 A: Notice that if we select four values, each drawn uniformly and independently from $[0, 6]$, and call these $y_1, y_2, y_3, y_4$ with $y_1 < y_2 < y_3 < y_4$, then there is a bijection
$$
x_1 = y_1 \\
x_2 = y_2+1 \\
x_3 = y_3+2 \\
x_4 = y_4+3
$$
that provides all satisfactory combinations and only those combinations.  Thus the desired probability is $\frac{6^4}{9^4} = \frac{2^4}{3^4} = 16/81 \approx 0.19753$.
ETA: Python code to simulate the question, just in case I've misunderstood it.
#!/usr/bin/python

import random

random.seed()

n = 1000000
t = 0
for i in range(n):
    x1 = random.random()*9
    x2 = random.random()*9
    x3 = random.random()*9
    x4 = random.random()*9
    if abs(x1-x2) > 1 and abs(x1-x3) > 1 and abs(x1-x4) > 1 \
    and abs(x2-x3) > 1 and abs(x2-x4) > 1 and abs(x3-x4) > 1:
        t += 1
print(float(t)/n)


Here's a breakdown of the two-dimensional analogue of this problem.  Suppose you choose two numbers $x_1$ and $x_2$ from the interval $[0, 9]$, and you want the probability that the intervals $[x_1, x_1+1]$ and $[x_2, x_2+1]$ are disjoint.  That's the same as requiring that $|x_1-x_2| > 1$.
We can represent this situation graphically on the Cartesian plane:

The red triangles represent the portion of the $9$-by-$9$ space over which the condition is met.  Since the area of that $9$-by-$9$ square is $9^2 = 81$, and the combined area of the triangles is $64$, the desired probability is $64/81$.
Analogous to my answer above, we can alternatively generate the answer by choosing two numbers $y_1$ and $y_2$ from the square $[0, 8] \times [0, 8]$, and then adding $1$ to whichever one is larger.  This forces the two numbers to be separated by at least $1$.  Conversely, given any two numbers that satisfy the conditions, we can get a point in the $8$-by-$8$ square by subtracting $1$ from the larger number (without changing which number is larger).
Graphically, this represents splitting the $8$-by-$8$ square into two triangles and sliding the upper-left one up by one unit, and sliding the lower-right one right by one unit:

Something similar happens in higher dimensions, except that there are different numbers of identical parts being slid together to make a square/cube/hypercube.  In two dimensions, there are $2$ triangles that come together to make an $8$-by-$8$ square, because there are $2! = 2$ different ways to order two variables.  In three dimensions, there are $6$ pyramids that come together to make a $7$-by-$7$-by-$7$ cube, because there are $3! = 6$ different ways to order three variables.  In four dimensions, there are $24$ hyper-pyramids that come together to make a $6$-by-$6$-by-$6$-by-$6$ hypercube, because there are $4! = 24$ different ways to order four variables.  And so on.
A: I assume the $x_i$ are real numbers.  If they are integers, this is a simple counting exercise.
Let $D$ be the event that the $4$ intervals are disjoint.  Let $E$ be the event that $x_1 < x_2 < x_3 < x_4$.


*

*$P(D \cap E)$ can be evaluated with a quadruple integral:


$${1 \over 9^4} \int_0^6 d x_1 \int_{1+x_1}^7 d x_2 \int_{1+x_2}^8 d x_3 \int_{1+x_3}^9 d x_4 = {2 \over 243}$$


*

*$P(E) = 1/4! = 1/24$ via a combinatorial argument: any permutation is equally likely.

*From above you get $P(D \mid E) = P(D \cap E) / P(E) = {48 \over 243} = {16 \over 81}$.  

*Finally you argue that $D$ and $E$ are independent and so $P(D \mid E) = P(D)$.
