P and Q are uniformly distributed in a square of side AB. What is the probability that segments AP and BQ intersect?

enter image description here


3 Answers 3


Inspired by Lee David Chung Lin's answer.

Consider the four events $$AP \cap BQ \neq \varnothing, \\ AP \cap DQ \neq \varnothing, \\ CP \cap BQ \neq \varnothing, \\ CP \cap DQ \neq \varnothing.$$

Ignoring degenerate configurations, the events are mutually disjoint, equiprobable, and cover the whole space. $p = 1/4$.

  • 1
    $\begingroup$ This is really great. The reason I felt it was necessary to awkwardly make rigorous the $1/4$ (argument in the last part of my post), was exactly due to the lack of symmetry on the $AB$-and-$PQ$ configuration with respect to the square. I think this is THE canonical treatment of it as a geometric probability problem. $\endgroup$ May 2, 2018 at 8:14

Suppose we start with $Q = (x, y)$. The admissible positions for $P$ will be in the triangle $BQR$ if $y < x$ or in the quadrilateral $BQRC$ if $y > x$. The areas are calculated from the sides and the heights, and $$p = \int_0^1 \int_0^x \frac {y (1 - x)} {2 x} dy dx + \int_0^1 \int_x^1 \left( 1 - \frac x {2 y} - \frac y 2 \right) dy dx = \frac 1 4.$$


Here's a calculation free "cut-and-paste" argument. Feel free to skip the words and directly examine the figures.

Consider the point $Q$ only in a quarter of the square $\square ABCD$, as shown in the left plot below. We will combine the "region of $P$ that creates a crossing" associated with $Q$ and its 4-fold symmetric mirror images.

As the right plot below shows, point $Q'$ is reflected with respect to (WRT) the main diagonal $\overline{AC}$, point $q$ is reflected WRT the off-diagonal $\overline{BD}$, and point $q'$ is doubly-reflected (equivalent to rotated $180^{\circ}$). enter image description here The region that is admissible (borrowing the great term from @Maxim) associated with $Q$ is highlighted below on the left (in magenta), and the region for $Q'$ is highlighted on the right plot (in blue).

The labels for corner points $A,B,C,D$ will be omitted in the plots from now on, because they are unnecessary most of the time and a bit distracting. enter image description here We can cut-and-paste the triangular blue "$Q'$ region" in the manner demonstrated below, thanks to the mirror symmetry by construct of $Q'$.

Note that the resultant quadrilateral consists of an obtuse triangle ($\triangle DQB$) plus THE isosceles right triangle, which is half of $\square ABCD$. enter image description here Similarly, the admissible regions associated with $q$ and $q'$ can be cut-and-paste into a quadrilateral that is half of $\square ABCD$ minus an obtuse triangle ($\triangle DqB$).

I sincerely apologize for the poor choice of colors if the readers find the figures unclear.

enter image description here

Due to the symmetry WRT the off-diagonal, we can cut-and-paste (flip) the combined-$(q,q')$-region and have the exact full square.

In short, since we effectively quadrupled the admissible region to arrive at $1$ (unity), the actual probability is $\displaystyle \frac14$.

enter image description here

If one is unsatisfied with the brief ending remark just above, below is the more detailed explanation.

Formally, instead of scanning point $Q$ over the entire square, we scan it only in a quadrant and reallocate the probability mass from the 3 mirrored positions $(Q',q,q')$. That is, we redefine the mass associated with $(Q',q,q')$ as the contribution associated with $Q$.

By reallocation, it means that the (conditional) probability of crossing is to be defined as zero when $Q$ is outside of the quadrant.

After the reallocation, when $Q$ is in the quadrant, the (conditional) probability of making an intersection is $1$, because $\square ABCD$ (result of cut-and-paste) divided by $\square ABCD$ (the domain of point $P$) is one.

In other words, the re-distributed conditional probability is $1$ over one-fourth of the domain (of $Q$) and $0$ elsewhere. Thus the overall probability is one-fourth.

  • 2
    $\begingroup$ A really nice answer. I suppose the last part can also be explained by saying that when computing the integral sums $S(x, y) \Delta x \Delta y$ over a symmetric grid, we can combine the four related terms, and the result will be the same as summing $\Delta x \Delta y / 4$. I've added another answer, exploiting the symmetry over the vertices of the square instead of the symmetries over the positions of $P$ and $Q$. $\endgroup$
    – Maxim
    May 2, 2018 at 8:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.