Skip to main content

Why am I frequently meeting my crush?

Gourav Banerjee, a 21MS student, goes to the main canteen of IISER Kolkata for dinner at some arbitrarily scheduled time between 8 and 9 pm. He frequently meets an anonymous, beautiful girl in the mess and begins to wonder whether the girl is stalking him or if their meeting is just a coincidence. So he tries to compute the probability of meeting that girl in the mess during dinner time given the following constraints:

  1. Both Gourav and the girl go to mess for having dinner at some random time between 8 - 9 pm.
  2. Because of the Queue at the mess, both stay in the mess for minimum of 30 min.

What do you think?

Solution

Let $x$ denote the time when Gourav enters the mess and let y denote the time when girl enters the mess. Here we take origin to be the 8 pm mark and a distance of 1 unit represents 1 hour on both $x$ and $y$ axis so all possible coordinates within the unit square $ABCD$ represents an event where Gourav and the girl both visit the canteen. Now the favourable coordinates which denote the event that Gourav meets the girl must be satisfied by the constraint $$ |x-y| \leq \frac{30}{60}, $$ as Gourav should arrive within 30 minutes of her arrival or vice-versa.


If Gourav arrives first then  $$ y-x \leq \frac{1}{2}, $$ i.e. the region below the orange line; if the girl arrives first then $$ x-y \leq \frac{1}{2}, $$ i.e. the region above the magenta line.

So, in the figure given above, the square region represents all possible outcomes and the red region denotes favourable outcomes, since a proper measure of total number of coordinate points enclosed in the 2D region is given by the area enclosed by the region. Therefore, since each of the outcomes represented by the coordinate points enclosed within the square are equally likely, the probability of meeting the girl in the mess is denoted by $$\begin{align*}
P(M) &=\frac{\mbox{area of red region}}{\mbox{area of whole square}} \\
&= \frac{1 - \mbox{area of blue region}}{\mbox{area of whole square}} \\
&= \frac{1 - 2\times \mbox{area of triangular region}}{\mbox{area of whole square}} \\
&= 1 -  \frac{1}{2^2} \\
&= \frac{3}{4}.
\end{align*}$$

The probability of meeting her at random is 75%!

Conclusion

It is highly disappointing that their frequent meetings are just a random phenomenon, and the girl is not interested in Gourav. This result applies for all of Gourav's other friends who go to have dinner in the mess at some arbitrary time in between 8 and 9 pm, but it is just that Gourav was not searching for their faces, so he was not meeting them as frequently; but he was specifically searching for that girl's face, so he was meeting her regularly.




Comments

Popular posts from this blog

The height of probabilistic interpretation

Girls only love men as tall as 6' and above. Socrates, ca. 2023 It is undeniable that heights strongly influence our daily lives. Be it our heights, or the height of a mountain we scale, or the height of all problems - humans. Mathematics too hasn't been able to escape its clutches, with height functions being useful in several fields, including but not limited to - Diophantine Geometry, Automorphic forms and the Weil-Mordell theorem - something you should have heard before if you attend my talks. If you have attended school (or maybe you are a climate activist) - then try recalling the elementary school days when fractions were introduced. Albeit unknowingly, but we had as children classified fractions into proper and improper - based on whether the denominator was larger than the numerator or vice versa. Well, it seems mathematicians have stuck with this classification - giving us the crux of todays discussion - height of a rational number. Given a rational number $x=\frac mn...

Monotonic functions and the first derivative

A couple of days ago, Rohan Didmishe shared this problem with us: show that the function defined by \[ f\colon \mathbb{R} \to \mathbb{R}, \qquad f(x) = \begin{cases} x + x^2\sin(1 / x), &\text{ if }x \neq 0, \\ 0, &\text{ if } x = 0. \end{cases} \] is not monotonic (increasing or decreasing) in any interval $(-\delta, \delta)$ around zero. Graphing this function (say, using Desmos ) shows that it oscllates rapidly, curving up and down with increasing frequency the closer its gets to zero. This is due to the $x^2\sin(1 / x)$ term; the $x$ added in front 'tilts' the curve upwards. The first thing to look at is the derivative of $f$. Using $\lim_{x \to 0} x\sin(1 / x) = 0$ and the chain rule, we can compute \[ f'(x) = \begin{cases} 1 + 2x\sin(1 / x) - \cos(1 / x), &\text{ if }x \neq 0, \\ 1, &\text{ if } x = 0. \end{cases} \] Specifcally, $f'(0) = 1$ which seems to tell us that $f$ is increasing at $0$ ... or doe...