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.


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Graduate student

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