By Sean Whitear

In 1918 a sickly Indian mathematician was lying in a hospital bed in Putney, when fellow researcher G.H. Hardy paid him a visit. Knowing this unwell mathematician’s fascination for numbers, Hardy brought up the number of the cab he took on the way to the hospital, 1729, and claimed that it wasn’t particularly interesting. The man immediately piped up with conjecture, stating that the number WAS interesting, being the smallest number that can be expressed by the sum of two cubes in two different ways, being 123+13 and 93+103. While not a particular moment of realised genius, with the man making great strides in number theory, infinite series and continued fractions despite no formal training, it is an excellent example of the lifelong genius of Srinivasa Ramanujan.

Ramanujan was born on the 22nd of December 1887 in Erode, Madras Presidency to K. Srinivasa, a clerk, and Komalatammal, a housewife. At the age of two, Ramanujan contracted smallpox, beginning an infinite series of illnesses that were to plague him for his entire life. In this period Ramanujan spent large periods of time with his mother, who was heavily involved in the local Brahmin culture, sowing the seeds for Ramanujan’s deep spirituality as he grew into adulthood. Ramanujan enrolled in a local Primary school in 1892 and graduated in 1897, with high scores in mathematics.

In 1898 Ramanujan attended Town Higher Secondary School, where a true fascination in mathematics first began to form. By the age of 13 he was studying advanced trigonometry, having received material from two local college borders. In 1902 he independently found methods of solving cubic and quartic equations, and spent a good deal of time attempting to solve the quintic. In 1903 Ramanujan received what was to be his cornerstone text: “A Synopsis of Elementary Results in Pure and Applied Mathematics” by G.S. Carr. The book was vital in igniting Ramanujan’s consistent self-discovery, and formed the style in which Ramanujan published his theorems. Next year the budding mathematician graduated, receiving the R. Ranganatha Rao prize and received a scholarship to study at a nearby college.

Unfortunately, college life did not suit Ramanujan. While he excelled in mathematics in his first year, all his other courses were neglected, and consequently he soon failed and dropped out. The same thing happened the next year at another college. Determined to continue in his pursuit of mathematical discovery, he dropped out again in order to study independently. This was a local minimum for Ramanujan, as during this period he lived in extreme poverty. Despite these hazardous conditions, at this point Ramanujan began recording his theorems in what would eventually amount to four notebooks containing over 5000 theorems.

Returning home in 1910, Ramanujan began searching for clerical work, but by the end of the year he was ill again and required surgery. He recovered in 1911 and, despite relapsing, continued looking for work. In this endeavour, he demonstrated some of his theorems to a potential employer. These theorems were quickly sent to R. Ramachandra Rao of the Indian Mathematical Society. The papers were initially thought to be fraud, due to the astounding nature of the theorems, which were rarely provided with full proofs. However, after Rao met Ramanujan in person, the now clerk received funding to conduct research and began publishing problems in Mathematical Journals.

In early 1913 Ramanujan began writing to English mathematicians in order to share his theorems and research. While ignored by some, G.H. Hardy immediately noticed some extraordinary insight in the 100 theorems he was sent. While some had already been discovered and there were some errors, many theorems had never before been identified. While initially thought to be fraudulent again, Hardy was soon won over and requested Ramanujan travel to England in order to continue research. Ramanujan initially refused, with his travel being in conflict with his Brahmin beliefs and against his parents’ wishes, however by the end of the year, he resolved to go in the pursuit of enlightenment and with his parents’ wishes, travelled to England in March 1914.

Ramanujan arrived in April in 1914 and immediately began mathematical research with Hardy. Upon Ramanujan’s arrival, Hardy was again astonished upon gaining full access to Ramanujan’s notebooks. While both were accomplished mathematicians, the two had very different working methods. While Hardy’s methods focused heavily on logic and proofs, Ramanujan relied mainly on intuition and spontaneity. Ramanujan’s method of thought was very heavily spiritually influenced, believing that fully formed theorems were given to him divinely. However, despite these differences, the two managed to work efficiently and productively throughout Ramanujan’s stay in Europe, producing 21 papers in the four years. During these last few years, Ramanujan’s genius finally began to be realised, with the mathematician being awarded the BA Degree for Research from Cambridge University in 1916 and a Fellow of the Royal Society in February 1918.

Unfortunately, England itself was not so accommodating. The weather and lack of a sustainable vegetarian diet due to World War One rationing had a startling impact on the Indian’s health. He soon contracted tuberculosis and spent much of his remaining stay in and out of hospitals, consequently being the biggest obstacle in Ramanujan’s endeavours. In 1918 he began to recover and the next year moved back to India for his health. Unfortunately, he fell ill again and passed away in Madras on 26 April 1920 at the age of 32.

This could have been the end of Ramanujan’s story. However, due to the continued work of Hardy and his London associates, Ramanujan’s theorems continued to be investigated and proven, and with 600 theorems he wrote in the last year of his life, the genius even gifted the world of mathematics posthumously. Even now, ‘Ramanujan’s name is currently at the cutting edge of mathematics, which is pretty impressive when you consider that he’s been dead for 95 years’ (Heyer, Marie, 2015).

Sean is in a double degree of civil and structural engineering and maths.

Flogging a dead horse, my two cents about the Monty Hall problem

By Adam Hamilton

Almost all of you reading this blog post will have heard of the Monty Hall problem. Those of you who’ve taken Puzzle Based Learning will probably have been given this problem to ponder, possibly as an assignment or as extra credit lecture question. Even if you hadn’t encountered this probability puzzle as part of your degree you may have heard this problem in the movie 21, or a mention in the tv show Numb3rs, if you’re a true 90’s kid then you might have read this in the “Ask Marilyn” column of The Parade which flung this problem into the spotlight. Or, like me, some smartarse gives you this problem so they can watch you get the more intuitive but “wrong” answer so they can feel smarter than you (I hate those people). For those of you who don’t know or would like to refresh their memories the Monty Hall problem goes something like this.

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

Figure 1: the obligatory XKCD comic

Most people upon hearing this problem (myself included) give answer which goes something along these lines “If I’ve picked door No.1 and the host has revealed that there is a goat behind door No.3, then that means that the car is really behind door No.1 or No.2. This means that there is a 50/50 chance that the car is behind either door and switching will not change my odds.” This is what people consider to be the “wrong” answer. I’ll get back as to why I’ve put the word “wrong” in quotation marks later in this article.

The answer that is typically regarded as “right” is that you should always switch your doors and in fact this will even double your chances of winning the car. The logic behind this answer goes something like this. “When you picked door No.1 there is a 2/3 chance that you will not choose the door that the car is behind. This means that there is a 2/3 chance that the car is behind one of the doors that you didn’t pick i.e there is a 2/3 probability that the car is behind door No.2 or it is behind door No.3. We now know that door No.3 doesn’t contain the car but we still know that our initial guess had only a 1/3 chance of resulting in winning a car. This means that there is still a 2/3 probability that the car is behind doors No.2 or No.3 but since we know that it can’t be behind door No.3 it must, with 2/3 probability be behind door No.2. Q.E.D”

figure 2: A tree diagram of all possible decisions you could make.

A more intuitive way to see this is to just write out every single possible outcome and just count the ones that result in you winning a car. Since this problem is small enough it is perfectly reasonable to write out every combination but for larger mathematical problems this approach is not recommended. Because I love Boolean algebra, in the above diagram all possible outcomes that are the result of switching are marked with a 1 and all outcomes that come from not switching are marked with a 0. In the three outcomes marked with 1 (which are the result of switching) two of these outcomes result in you winning the car, but in the three outcomes marked with a 0(which are the result of sticking with your initial choice) only one outcome results in winning the car. It is plain to see that if you swap your choice of door then you will double your odds of winning the car.

The reason I put the word “wrong” in quotation marks in the start is that both answers are technically correct. However the reason that both answers could be correct regard a few subtle assumptions in how we define the problem. Like most of these mathematical questions phrased with words it requires you to make some assumptions as to how the problem translates to mathematics. Sometimes these assumptions might be a little abstruse it might be difficult to justify them. A lot of people that I’ve asked about their assumptions with regards to this problem give the same set of answers; usually “your goal is to win the car”, “there is an equal probability that the car is behind of each of the three doors”, “there is no behind the scenes switching of goats and cars so you always lose”, and “Adam, don’t you think you’re a little bit obsessive about this problem”.

figure 3: The first assumption might be a little hard to justify. Who could say no to that face.

The implicit assumption that rarely gets mentioned is that the host will always reveal that a goat is behind one of the three doors. It may be the case that after selecting a door the host merely asks if you want to switch doors without revealing whether a goat is behind any of them, or he could just tell you whether you had guessed correctly and that would be the end of the game. Imagine putting yourself in the shoes of the game show host where a contestant has picked the door that the car is behind. It is well within your interests to stop the contestant from winning the car. At some point you may find yourself thinking something like “Shit! This guy picked the car. What am I gonna do? I know. There’s this thing called the Monty Hall problem that says that they should switch doors, and it was in that movie with Kevin Spacey so they’ve probably heard of it. I’ll show him that a goat is behind one of these doors so they’ll inclined to change their answer and better yet they’ll think it was their idea.”

So under this new assumption that the host will actively try and sabotage you rather than just open doors like some goat revealing robot a very interesting question is raised. What should I do now?. How should you take the behaviour of the host into account to maximise your chances of winning the car? This is where the Monty Hall Problem stops being a puzzle of probability and starts being a Game Theory problem. If you knew anything about the psychology of the host and whether he reveals goats in order to throw you off your game, as some sort of double bluff, or just because he doesn’t know any better then you might be able to assign probabilities of success to any strategy that you could employ. If not then your best bet is really just to switch doors only half of the time. It is under these different set of assumptions that can be used to justify the initial, naïve answer of 50/50 probabilities.

Who knows what’s going on in that head of his?

The main point to take away from this blog post is that in any applied mathematics problem you are trying to translate the problems of the real world into mathematics. This can’t always be done neatly and might require you to make some assumptions which may or may not accurately reflect what’s going on in reality. Each of these assumptions can lead to different mathematical models which will in turn lead to different solutions. Whilst making mathematical models you should always keep track of which assumptions you make and why you made them and if that’s not your thing then you should do pure maths and leave the probability puzzles alone.

Adam is in his third year of a BMaSc. He likes reading, painting, writing lists of things he enjoys, and self referential sentences.