It's Time For Algebraic Probability!

$f_4$ is minimised when $\max \frac{x_i}{x_j} < 2$ ; in this case $f_4=24$ .

To see we can't do better than this, consider two cases:

Case $x_i=x_j$ : we have $\left \lceil \frac{x_i}{x_j} \right \rceil + \left \lfloor \frac{x_i}{x_j} \right \rfloor + \left \lceil \frac{x_j}{x_i} \right \rceil + \left \lfloor \frac{x_j}{x_i} \right \rfloor = 1+1+1+1=4$

Case $x_i<x_j$ : $\left \lceil \frac{x_i}{x_j} \right \rceil + \left \lfloor \frac{x_i}{x_j} \right \rfloor + \left \lceil \frac{x_j}{x_i} \right \rceil + \left \lfloor \frac{x_j}{x_i} \right \rfloor = 0+1+\left \lceil \frac{x_j}{x_i} \right \rceil + \left \lfloor \frac{x_j}{x_i} \right \rfloor$

Now, since $x_i<x_j$ , $\left \lceil \frac{x_j}{x_i} \right \rceil \ge 2$ and $\left \lfloor \frac{x_j}{x_i} \right \rfloor \ge 1$

so that $\left \lceil \frac{x_i}{x_j} \right \rceil + \left \lfloor \frac{x_i}{x_j} \right \rfloor + \left \lceil \frac{x_j}{x_i} \right \rceil + \left \lfloor \frac{x_j}{x_i} \right \rfloor \ge 4$

with equality iff $\frac{x_j}{x_i} < 2$ .

Since the sum is made up of six terms, it follows the least possible total is $24$ as claimed.

This condition is easily achieved; to count up the possibilities, categorise by the smallest element of $\{x_1,x_2,x_3,x_4\}$ ; call this element $x_{min}$ :

If $x_{min}=1$ , none of the $x_i$ can be greater than $1$ ; this gives the single possibility $\{1,1,1,1\}$ .

If $x_{min}=2$ , none of the $x_i$ can be greater than $3$ ; this gives the $15$ possibilities $\{2,2,2,2\},\{2,2,2,3\},\cdots \{3,3,3,2\}$ .

If $x_{min}=3$ , none of the $x_i$ can be greater than $4$ ; this gives the $15$ possibilities $\{3,3,3,3\},\{3,3,3,4\},\cdots \{4,4,4,3\}$ .

If $x_{min}=4$ , we again have a single possibility, $\{4,4,4,4\}$ .

Hence $32$ sets minimise $f_4$ .

There are $4^4=256$ possible sets; so the probability that $f_4$ is minimised is $\frac{32}{256}=\boxed{\frac18}$ .