On the probability of forming a triangle, when $(0,1)$ is divided into three segments , where dividing points are i.i.d. Uniform $(0,1)$

Question

Divide $(0,1)$ into three line segments, let $X,Y$ be the dividing points. Assume $X,Y$ are independent and follows Uniform $(0,1)$. What is the probability that the three line segments can form a triangle ?

My try: The lengths of the three parts are $\min (X,Y), |Y-X|=\max(X,Y)-\min(X,Y)$ and $1-\max(X,Y)$. In order to form a triangle, we must have

$1-\max (X,Y)\le \min(X,Y)+|Y-X|=\max(X,Y)$, so $1/2\le \max(X,Y)$;

and $\max(X,Y)-\min(X,Y)=|Y-X|\le \min(X,Y)+1-\max(X,Y)$ i.e. $\max(X,Y)-\min(X,Y)\le 1/2$ ;

and $\min(X,Y) \le 1-\max(X,Y)+\max(X,Y)-\min(X,Y)$ so $\min(X,Y)\le 1/2$

So we need to find $P(1/2 \le \max (X,Y) ; \max(X,Y)-\min(X,Y)\le 1/2; \min(X,Y)\le 1/2)$.

But then I'm lost ... I don't know how to find this probability ...

Please help.

Answer 1

By symmetry, we can assume $x\le y$.

Thus, $0\le x \le y \le 1$.

Also, we can assume the inequalities are strict, since the probability that some two consecutive terms of $0,x,y,1$ are equal has probability zero.

Thus, our sample space is the triangular subregion $T$ of the standard unit square such that $0 < x < y < 1$, which has area ${\small{\frac{1}{2}}}$.

Note that a triangle is possible if and only if the largest part has length less than ${\small{\frac{1}{2}}}$.

Equivalently, a triangle is not possible if the largest part has length at least ${\small{\frac{1}{2}}}$.

We proceed to find the probability that a triangle is not possible . . .

Ignoring the case of equality (which has probability zero), consider $3$ cases . . .

Case $(1)$:$\;x > {\small{\frac{1}{2}}}$.

The probability that case $(1)$ occurs is the area of the triangular subregion of the standard unit square satisfying the inequalities \begin{align*} x &< y\\[4pt] x &> {\small{\frac{1}{2}}}\\[4pt] \end{align*} divided by the area of $T$, hence the probability for case $(1)$ is $$ \frac {\left({\large{\frac{1}{8}}}\right)} {\left({\large{\frac{1}{2}}}\right)} = \frac{1}{4} $$ Case $(2)$:$\;y-x > {\small{\frac{1}{2}}}$.

The probability that case $(2)$ occurs is the area of the triangular subregion of the standard unit square satisfying the inequalities \begin{align*} x &< y\\[4pt] y-x &> {\small{\frac{1}{2}}}\\[4pt] \end{align*} divided by the area of $T$, hence the probability for case $(2)$ is again $$ \frac {\left({\large{\frac{1}{8}}}\right)} {\left({\large{\frac{1}{2}}}\right)} = \frac{1}{4} $$ Case $(3)$:$\;1-x > {\small{\frac{1}{2}}}$.

The probability that case $(3)$ occurs is the area of the triangular subregion of the standard unit square satisfying the inequalities \begin{align*} x &< y\\[4pt] 1-x&> {\small{\frac{1}{2}}}\\[4pt] \end{align*} divided by the area of $T$, hence the probability for case $(3)$ is yet again $$ \frac {\left({\large{\frac{1}{8}}}\right)} {\left({\large{\frac{1}{2}}}\right)} = \frac{1}{4} $$ Since the cases are mutually exclusive, the probability that a triangle is not possible is ${\large{\frac{3}{4}}}$.

Hence, the probability that a triangle is possible is $1-{\large{\frac{3}{4}}}={\large{\frac{1}{4}}}$.

Answer 2

We break our line segment at $x_1,x_2$

WLOG we will insist that $x_1 \le x_2$

From there we can map any $(x_1,x_2)$ pair to the equilateral triangle. The mapping is made finding the equilateral triangle with base vertexes at $(x_1,0),(x_2,0)$

This mapping is 1-1 and surjective, and uniform.

If the mapped point is in the white area, no segment is longer than $\frac 12.$ If you are in the blue area, one segment is too long.

This should then suggest and analytical approach to the problem.

$P(x_1 < \frac 12, (1-x_2)<\frac 12, \text{ and }x_2 - x_1 < \frac 12) = \dfrac {\int_\limits{\frac 12}^{1}\int_\limits{x_2-\frac 12}^{\frac 12}\ dx_1\ dx_2}{{\int_\limits{0}^{1}\int_\limits{0}^{x_2}}\ dx_1\ dx_2}$