Unequally Floored Probability

Since $x\in(0,1]$ ,

$\lfloor x\rfloor=0$

$\lfloor \cfrac{\lfloor x\rfloor+1}{x}\rfloor=\lfloor \cfrac{1}{x}\rfloor\leq3$

$x>0$

$\cfrac{1}{x}>0$

By definition of floor function,

$\lfloor \cfrac{1}{x} \rfloor \leq \cfrac{1}{x}< \lfloor \cfrac{1}{x} \rfloor +1\leq 3+1=4$

$\therefore \cfrac{1}{x}<4$

$x>\cfrac{1}{4}$

$x\in(\cfrac{1}{4},1)$

The required probability $=1-\cfrac{1}{4}=\cfrac{3}{4}$

• First I tried substituting in $x=1$ to the inequality, and $x=1$ makes the statement true since $2\leq3$ .
• If $\left \lfloor \frac{\left \lfloor x \right \rfloor+1}{x} \right \rfloor\leq 3$ then $\frac{\left \lfloor x \right \rfloor+1}{x} < 4.$
• Multiplying both sides by $x$ yields $\left \lfloor x \right \rfloor+1< 4x$ .
• Lastly, adding 1 to both sides yields $\left \lfloor x \right \rfloor< 4x-1$ .
• If $0< x< 1$ , then $\left \lfloor x \right \rfloor=0$ and $x> \frac{1}{4}$ .
• On the interval $(0,1]$ , $x$ is allowed to be $(\frac{1}{4},1]$ . The solutions for $x$ make up $\frac{3}{4}$ of the given interval.

The fact that x is distributed uniformly, and the fact that the floor function returns an integer, both make this problem relatively simple. Note that because x is continuously distributed,
P( floor(x) = 0 ) = 1.
This simplifies the function -- let's call it f(x):
f(x) = floor(1/x)
where x ~ U(0,1).
Now we solve for the probability:
P ( f(x) <= 3 )
= P ( f(x) < 4 )
= P ( x >= 1/4)
= 3/4

We are choosing $x$ from $(0,1]$ .
$\begin{aligned} \lfloor x \rfloor &= 0 \quad (\text{since we can ignore } x=1) \\ \left \lfloor \frac {\lfloor x \rfloor+1} x \right\rfloor = \left \lfloor \frac 1 x \right\rfloor &\le 3 \\ \frac1x &< 4 \\ x&\in \left(\frac14, 1\right) \end{aligned}$
Therefore the probability is $\frac34$ .