When the sky reaches the ground

Relevant wiki: Geometric Probability

$\large \lfloor x+0.7 \rfloor = \lceil x - 0.7 \rceil$

The range of values allowing the equation to be true is the same between every consecutive integer. Thus, regardless of the values of $a$ and $b$ , the odds will be the same.

For number $x$ , we denote $[x]$ as its whole part and $\{x\}$ as its fractional part.

Case 1: $x=x$

When does $\lfloor x+0.7 \rfloor = x$ ?

That is when $\{x\} + 0.7 <1$ , or $\{x\} < 0.3$ .

That would give us $\lfloor [x] + w \rfloor$ with $w = \{x\} + 0.7$ , so $w<1$ . This means that $\lfloor [x] + w = x$ .

So if we have $\{x\} <0.3$ , The RHS, on the other hand, would result to $\lceil [x-1] + z \rceil$ with $z<1$ . So ultimately that will give us $[x-1] +1 = x$ . This now validates the case.

Subsequently, if $\{x\} >0.3$ , that would result to $\lfloor [x] + y\rfloor$ with $y >1$ , and that will give us $[x] + 1$ . This will mean that the LHS is $x+1$ , while the RHS may be $x$ , or also $x+1$ , depending on $\{x\}$ .

Case 2: $x+1 = x+1$

We already know that $\lfloor x+0.7\rfloor = x+1$ when $\{x\} >0.3$ .

So when does $\lceil x - 0.7 \rceil = x+1$ ?

That is when $\{x\} - 0.7 > 0$ , or $\{x\} >0.7$ . In this case, that would give us $\lceil [x] + v \rceil$ , with $v = \{x\} -0.7$ . That would provide a fractional part to the expression resulting it to $[x] + 1$ . Thus, this case is validated.

So, for any two consecutive integers $a$ and $a+1$ , the mentioned equation holds when $\{x\} <0.3$ or $\{x\} >0.7$ . So the range of permissible values is $[a, a+0.3) \cup (a+0.7, a+1]$ .

So the probability for any interval $(a, b)$ would be

$\frac{0.6 (b-a)}{b-a} = 0.6$

The functions $y=x+0.7$ and $y=x-0.7$ are in grey.

The function $\lceil x-0.7\rceil$ is in blue and function $\lfloor x+0.7\rfloor$ is in red. They overlap when the fractional part of x is either in the interval [0, 0.3] or in (0.7, 1). Together these intervals make up a fraction $\boxed{0.6}$ of the interval [0,1).