# An ethical question with a concrete answer

Ethics questions often give a scenario (often unrealistic and missing key details) where you have to choose between a set of difficult options. For example, it could be “an old lady who has no family or a young man with a promising career and loved by a lot of people are both admitted to the ER. Who would you prioritize on treating if their probability of survival is the same?”

However, sometimes the answer is quite a bit more straightforward. The question I saw is the following: Suppose that an intelligent machine has the ability to predict, with 99.99% accuracy, whether someone will commit murder. Would it be permissible to arrest people based on the predictions of the machine?

There seems to be no ‘correct’ way to answer this question, but that’s because the person who asked the question doesn’t understand statistics. There is a phenomenon in statistics called the false positive paradox, where even a very accurate test will produce many more false positives than actual positives. This is relevant when the actual probability of a true positive is very low.

Here is an example. You are a unique person, one of say 7 billion. Suppose that there is a machine that can identify someone with 99.9999% accuracy. A person gets scanned by the machine, and the machine says it’s you. What is the probability that the machine is right?

There are two possibilities for the machine to give that reading. Either the person scanned by the machine actually is you and the machine is accurate, or the person scanned by the machine is not you and the machine malfunctioned. The probability that the machine is accurate is 99.9999%, and the probability that the person scanned is you is one in 7 billion. The other possibility is that the machine is wrong and the person being scanned is not you. The probability that the machine is wrong is 0.0001%, and the probability that the person is not you is 6,999,999,999 out of 7 billion. Thus, the probability that the person actually is you, given that the machine says it scanned you, is given by the equation

$\dfrac{\frac{999999}{1000000} \times \frac{1}{7000000000}}{\frac{999999}{7000000000} \times \frac{1}{7000000000} + \frac{1}{1000000} \times \frac{6999999999}{7000000000}}.$

Evaluating, the probability that the machine is right is less than 0.015%. The paradox is caused by the fact that the machine’s accuracy is very poor compared to the astronomically unlikely phenomenon that you would be picked out of 7 billion people.

Now back to the question. Whether it is ethical or not (that is, whether the machine produces a desirable result an acceptable proportion of the time) will depend on the actual murder rate of a place. The global average is currently 6.2 people out of every 100,000, or 31 out of every half a million. Assuming the machine has accuracy 99.99%, the probability that a person identified by the machine as a murderer is actually a murderer is given by

$\dfrac{\frac{9999}{10000} \times \frac{31}{500000}}{\frac{9999}{10000} \times \frac{31}{500000} + \frac{1}{10000} \times \frac{499969}{500000}}$

or roughly 38.3%. Therefore, the machine is right way less than half the time, far below what can be considered reasonable. Therefore this question has a concrete answer and is not really debatable.

# An elementary statistics problem from Hearthstone

In Hearthstone, a popular online card game, there is a minion called C’Thun which deals damage equal to its attack value, one damage at a time. This prompted a popular streamer named Trump (not the insufferable presidential candidate) to ask the following question once on his channel: when playing a C’Thun with 12 attack (hence, it will deal 12 damage one at a time), what is the probability that it will kill a 6 health minion when played, with no other minions on the board?

To those who don’t know how Hearthstone works, one can think of the question as follows: suppose you flip a coin 12 times in a row. Each time you flip the coin, you record down whether you got heads or tails. As long as you have fewer than 6 heads, the coin is fair and you have equal probability of getting heads or tails on your next flip. However, as soon as you hit 6 heads, then the coin becomes unfair and you can only get tails after. The question that Trump (real name Jeffrey Shih) asks is then equivalent to “what is the probability that you get 6 heads”.

To calculate this probability, we can ask the reverse question: what is the probability that we don’t get six heads? If we don’t get 6 heads, then the probability distribution is the same as if we always had a fair coin, and so the probability is the same. The probability of getting at most five heads when flipping a fair coin 12 times is given by

$\displaystyle 2^{-12} \sum_{k=0}^5 \binom{12}{k}.$

The well-known binomial theorem tells us that

$\displaystyle 2^{-12} \sum_{k=0}^{12} \binom{12}{k} = 1.$

This shows that the probability of getting at most 5 heads is equal to

$\displaystyle \frac{1}{2} - 2^{-13} \binom{12}{6} = \frac{793}{2048}.$

Therefore, the probability of getting 6 heads in the original game is equal to

$\displaystyle 1 - \frac{793}{2048} = \frac{1255}{2048},$

or roughly $61.28\%$. Thus it is far more likely that C’Thun will kill a 6 health minion with only 12 damage than too many shots going to face.

# Some thoughts after reading an article on Peter Scholze

Recently the following article appeared on my Facebook feed: https://www.quantamagazine.org/20160628-peter-scholze-arithmetic-geometry-profile/Aside from the usual bland mix of singing praises of an archetypal ‘genius’, the article does contain some genuine insights. The most striking of which is Scholze’s description of him learning the proof of Fermat’s Last Theorem by Sir Andrew Wiles: that he “worked backward, figuring out what he needed to learn to make sense of the proof”. Later he also said things like “I never really learned the basic things like linear algebra, actually – I only assimilated it through learning some other stuff.”

If you have experiences learning mathematics at the senior undergraduate level or post-graduate level, you will likely find that these experiences are orthogonal to your own. We spend an inordinate amount of time learning the ‘basics’ for various things, which very much includes linear algebra for example, in order to do research… or so we are told. If you have passed the part of your career where you do more courses than self-learning, then you have likely reached the epiphany that usually it’s not efficient to learn everything there is to know on a subject before actually doing work on the subject.

Of course, I am but a pebble to the avalanche that is Peter Scholze, so my advice may not be worth much. Nevertheless, I feel like I should say this to all prospective and current graduate students: be bold, and give every difficult paper in your field a read. Don’t be intimidated by them. If you don’t understand something, google it until you find what you need to learn the language of the subject. Don’t feel like you need to understand all of Harthshorne before you can read any research papers related to algebraic geometry. Your future self will thank you for this.

# A question for students you dislike

At the recent CNTA conference, Professor Joe Silverman gave an explicit homomorphism of $\text{GL}_3(\mathbb{R})$ into $\text{GL}_6(\mathbb{R})$ which he jokingly called a great question to ask undergraduates to work out explicitly… if you don’t like them very much. In a similar vein, here is a question that one might ask undergraduates they don’t particularly like:

Let $\mathbf{a} = (\alpha, \beta, \gamma)$ be a triple of co-prime integers which are not all zero and such that $\alpha \gamma > 1$. Prove that the ternary quadratic form given by:

$K_{\mathbf{a}} (F) = \dfrac{1}{8\alpha^3} \bigg( 72 \beta^2 \gamma A^2 + 9 \alpha(\beta^2 + 4 \alpha \gamma) B^2 + 8 \alpha^3 C^2 - 18\beta (\beta^2 + 4 \alpha \gamma)AB$
$+ 12 \alpha (3 \beta^2 - 4 \alpha \gamma)AC - 24 \alpha^2 \beta BC \bigg)$

takes on integer values whenever the triplet $(A,B,C)$ lies in the lattice defined by the congruence conditions

$4 \beta \gamma x - (2 \beta^2 + \alpha \gamma) y + 2 \alpha \beta z \equiv 0 \pmod{\alpha^2}$

and

$\gamma(2 \beta^2 + \alpha \gamma)x - \beta(\beta^2 + \alpha \gamma)y + \alpha \beta^2 z \equiv 0 \pmod{\alpha^3}.$

I will reveal the solution in due time (and it does not involve explicit computation of congruences), but if come up with a solution let me know!

# Manjul Bhargava’s advice to mathematics graduate students

Last night, I had the honour of attending a panel discussion featuring eminent mathematician Manjul Bhargava. During the panel, the moderator, Professor Kumar Murty asked the very productive Fields Medalist to give some advice to graduate students in the audience who may be struggling with their research. Professor Bhargava’s response, paraphrased, is essentially the following:  always work on several problems at a time (at least three), of varying difficulty. There should be a problem which is quite difficult and if you make any progress on it it will be a major breakthrough; there should be one of moderate difficulty, and there should be an ‘easy’ problem that you know you can make progress on eventually. Further, never think too much about a problem at a time and instead rotate between the problems to change your mindset. Sometimes when you approach a problem with a fresh perspective you will gain some insight that would’ve been impossible if you stared at the same problem continuously, since you are subconsciously trying to apply the same techniques.

I thought Professor Bhargava’s advice was very helpful to the graduate students in the audience. It is something I started doing a few years ago, but it wasn’t something that I was aware of consciously. Hopefully heeding this advice will be helpful to your work.

# Pure Mathematics Analysis Exam – 2016 edition

Hi all,

I have included the solutions to the 2016 analysis exam here. Enjoy!

# A grade school question from China

My wife showed me a homework problem of a friend’s son in China that he could not solve, and his mother also could not solve. I thought it was quite an interesting problem, so I thought I would share.

I do not have an acceptable diagram (which admittedly would make the problem much easier to digest), so I will be as precise as possible in describing the problem. Suppose that you have a triangle $ABC$ with a right angle at vertex $B$. The length of the side $AC$ is equal to 14. Construct squares $AGFB$ (read counter-clockwise) on side $AB$ and $BEDC$ on side $BC$. Draw the line from vertex $A$ to the vertex $D$ of square $BCDE$, and let $H$ be the intersection of the line segment $AD$ and $BC$. Suppose further that the line segment $GH$  is parallel to $AC$. What is the area of the quadrilateral $BHDE$?

We let $x$ denote the side length of $AB$, $y$ denote the side length of $BC$, so that $x^2 + y^2 = 14^2 = 196$. Put $z$ for the length of $BH$. Since $AC$ and $GH$ are parallel, it follows that $z = y - x$. Therefore the desired area is given by $[AED] - [ABH]$ (the square brackets denote the area of the polygon with the given vertices), or

$\frac{y(x+y)}{2} - \frac{x(y-x)}{2} = \frac{x^2 + y^2}{2} = 98.$

This problem is nice because it requires adroit use of many different geometric facts, and once the proper principles are applied, the solution is beautiful and elegant. Hard to imagine an 11 year old being able to do this regularly though!

Edit: in the original post there was an error where I forgot to divide by 2 in the penultimate step. Silly mistake!