Mathematics in “The Curious Incident of the Dog at Nighttime”

I recently read the novel “The Curious Incident of the Dog at Nighttime”, a wonderful and deep novel, that also contained some sophisticated Mathematics! I have tried to elaborate on some of that Mathematics below:

Monty Hall Problem

The Monty Hall Problem is a famous problem in Mathematics. Although I have known about the problem for a long time, I could never fully understand it. I recently read about it in the book “The Curious Incident of the Dog in the Nighttime”, and thought I finally had some understanding of it. I will try to write down my thoughts on it.

There are three doors- we shall call them $A, B$ and $C$. There is a car behind one of those doors, and nothing behind the other doors. You are asked to choose a door. Let us suppose you choose $A$. The host will now open one of the remaining doors to show that the car is not behind it. Let us suppose that he opens $C$. Should you now stick to your previous choice of doors, or should you change your choice of doors to $B$?

The best way to understand this problem is to generalize it; perhaps by increasing the number of “doors”. Let us suppose that there are $1000000$ cups (instead of doors), labeled $1$ to $1000000$. There is a ball in one of those cups, and we have to choose the cup that we think contains the ball. Clearly, the probability of the ball being in cup $1$ is $\frac{1}{1000000}$, and the probability of the ball not being in cup $1$ is $\frac{999999}{1000000}$. As we can see, the probability of the ball **not** being in cup $1$ is substantially higher; in other words, we can be almost certain that the ball is not in cup $1$. Let us now suppose that there is a host, who asks you to choose a cup which you think contains the ball. Let us say you choose $1$. Now out of the remaining $999999$ cups, he opens $999998$ cups which do not contain the ball. So there are only two cups remaining. We shall call the remaining cup $C$. Should you switch to $C$?

Remember that we can be almost sure the ball was never in cup $1$ (the probability of it being in cup $1$ was $\frac{1}{1000000}$). Hence, it almost certainly had to have been in some other cup. Now all cups except for $C$ have been opened. Hence, because the probability of the ball being in cup $1$ is almost $0$, and all other cups except for $C$ have been opened, the ball is almost certainly in cup $C$. Hence you should switch to $C$!!

The same thing happens in the Monty Hall problem with $3$ doors. The probability of the car being behind $A$ is $\frac{1}{3}$, and the probability of the car not being behind $A$ (and hence being behind $B$ or $C$) is $\frac{2}{3}$. Now that the host has opened $C$ to show that there is nothing behind it, its probability of $\frac{1}{3}$ gets transferred to $B$. Hence, $B$ has a $\frac{2}{3}$ probability of having the car behind it, and you should switch to it!

I shall soon be updating this blog post with other mathematical gems from the book.