# Find the probability that $[x+y+z]=[x]+[y]+[z]+2$

Find the probability that the equation $$[x+y+z]=[x]+[y]+[z]+2$$ is true, where $$x,y,z \in R$$. [.] Represents the greatest integer function.

I got two different answers by two different methods.

1st method: $$x=[x]+\{x \}$$ etc in LHS, where {} is the fractional part function.

So $$[\{x \}+\{y \}+ \{z \}]=2$$

So $$\{x \}+\{y \}+ \{z \}$$ is between $$[2,3)$$.

All $$\{x \},\{y \},\{z \}$$ are between $$[0,1)$$ and are uniformly distributed in this interval.

So if we consider the "expectation" instead of actual probability. The expectation that $$\{x \},\{y \},\{z \}$$ is between $$[2,3)$$.

= Expectation that $$3\{x \}$$ is between $$[2,3)$$.

= Expectation that $${x}$$ is between $$[2/3,1)$$

$$= (1-\frac{2}{3})/1 = \frac{1}{3}$$

(Due to uniform distribution of {x}).

Can we call this the final required probability?

Method 2: consider a unit cube with vertices $$(0,0,0),(1,0,0),(0,1,0),(0,0,1),(1,1,0),(1,0,1),(0,1,1),(1,1,1)$$ and the plane $$x+y+z=2$$.

The required probability (of $$\{x \}+\{y \}+ \{z \}$$ is between $$[2,3)$$). is the volume of the cube cut out of the plane(not including the origin) /volume of cube = volume of tetrahedron with vertices $$(1,1,1),(1,1,0),(1,0,1),(0,1,1)$$/1 = $$\frac{1}{6}$$.

Which method is correct (if at all any) and is there any other way to solve this question?

• Please exercise more care with tags. From the algebraic-geometry tag description: " This tag should not be used for elementary problems which involve both algebra and geometry." Sep 10, 2020 at 5:32
• Oh sorry @KReiser. Any other tag which would be apt here? Sep 10, 2020 at 5:48

Linearity of expectation says that, if $$X$$ and $$Y$$ are two random variables (if this isn't a term you've come across, don't worry; it means pretty much what you'd think it should), then $$\mathbb E[X+Y]=\mathbb E[X]+\mathbb E[Y].$$ So, one thing you could say is that, since $$\{x\}$$, $$\{y\}$$, and $$\{z\}$$ are all random variables, $$\mathbb E[\{x\}+\{y\}+\{z\}]=\mathbb E[\{x\}]+\mathbb E[\{y\}]+\mathbb E[\{z\}],$$ and then calculate each of the things on the right individually. What you can't do is conflate expectation and probability -- they are different things. Given an event $$A$$, you can define a random variable $$X=\begin{cases}1&\text{if }A\text{ occurs} \\ 0&\text{if }A\text{ doesn't occur,}\end{cases}$$ and you will get that $$\operatorname{Pr}(A)=\mathbb E[X]$$, but $$X$$ and whatever $$A$$ is measuring are different objects. In particular, if $$A$$ is the event $$2\leq \{x\}+\{y\}+\{z\}<3,$$ the random variable $$X$$ you get with the above process is sometimes $$0$$ and sometimes $$1$$, but it isn't the sum of anything useful depending on what $$\{x\}$$, $$\{y\}$$, and $$\{z\}$$ are. The random variable $$\{x\}+\{y\}+\{z\}$$ is the sum of useful things, but you care about $$\mathbb E[X]$$, not $$\mathbb E[\{x\}+\{y\}+\{z\}]$$.