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?

  • $\begingroup$ 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." $\endgroup$
    – KReiser
    Sep 10, 2020 at 5:32
  • $\begingroup$ Oh sorry @KReiser. Any other tag which would be apt here? $\endgroup$
    – user600016
    Sep 10, 2020 at 5:48

1 Answer 1


The second method is correct. The first fails because you've improperly applied linearity of expectation.

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 $\mathbb E[\{x\}+\{y\}+\{z\}]$.

  • $\begingroup$ Thanks a lot for the very clear and elaborate answer! $\endgroup$
    – user600016
    Sep 10, 2020 at 7:11

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