I was watching the film Crazy Rich Asians the other day, as there’s not a lot to do at the moment besides watching Netflix and watching more Netflix. I thoroughly enjoyed the film and would highly recommend it. However, there was something that happened in the first few minutes which really got me thinking and inspired the subject of this Chalkdust article.

The main character in the film is an economics professor (the American kind where you can achieve the title of professor while still being in your 20s and without having to claw your way over a pile of your peers to the top of your research field). Within the opening scenes of the film we see her delivering a lecture, in which she is playing poker with a student, while also making remarks about how to win using ‘mathematical logic’. The bell rings seemingly halfway through the lecture the way it always does in American films and TV shows, and our professor calls out to her students “…and don’t forget your essays on conditional probability are due next week!” Now, I am not going to delve into the question of what type of economics course she is teaching that involves playing poker and mathematical logic, but it got me thinking—what exactly would an ‘essay on conditional probability’ entail?

What is conditional probability?

Conditional probability is defined as a measure of the probability of an event occurring, given that another event (by assumption, presumption, assertion, or evidence) has already occurred. For all intents and purposes here, for two events $A$ and $B$, we’ll write the conditional probability of $A$ given $B$ as $P(A \mid B)$, and define it as \[P(A \mid B) = \frac{P(A\;\text{and}\;B)}{P(B)}.\] The little bar $ \mid $ can just be thought of as meaning ‘given’.

Now that we’ve got some technicalities out of the way, let’s look at some examples of conditional probability. Imagine you are dealt exactly two playing cards from a well-shuffled standard 52–card deck. The standard deck contains exactly four kings. What is the probability that both of your cards are kings? We might, naively, say it must be simply $(4 / 52)^2 \approx 0.59$, but we would be gravely mistaken. There are four chances that the first card dealt to you (out of a deck of 52) is a king. Conditional on the first card being a king, there remains three chances (out of a deck of 51) that the second card is also a king. Conditional probability then dictates that: \begin{align*} P(\textrm{both are kings}) &= P(\textrm{second is a king} \mid \textrm{first is a king}) \times P(\textrm{first is a king}) \\ &= \frac{3}{51}\times \frac{4}{52}\approx 0.45\%.\ \end{align*} The events here are dependent upon each other, as opposed to independent. In the realm of probability, dependency of events is very important. For example, coin tosses are always independent events. When tossing a fair coin, the probability of it landing on heads, given that it previously landed on heads 10 times in a row, is still $1/2$. Even if it lands on heads 1000 times, the chance of it landing on heads on the 1001st toss is still 50%.

Bayes’ theorem

Any essay on conditional probability would be simply incomplete without a mention of Bayes’ theorem. Bayes’ theorem describes the probability of an event, based on prior knowledge of conditions that might be related to the event. It is stated mathematically as:

Bayes’ theorem

$P(A\mid B) = \frac{P(B \mid A)P(A)}{P(B)}.$

We can derive Bayes’ theorem from the definition of conditional probability above by considering $P(A \mid B)$ and $P(B \mid A)$, and using that $P(A\;\text{and}\;B)$ equals $P(B\;\text{and}\;A)$.

A fun (and topical!) example of Bayes’ theorem arises in a medical test/screening scenario. Suppose a test for whether or not someone has a particular infection (say scorpionitis) is 90% sensitive, or equivalently, the true positive rate is 90%. This means that the probability of the test being positive, given that someone has the infection is 0.9, or $P(\textrm{positive}\mid\textrm{infected}) = 0.9$. Now suppose that this is a somewhat prevalent infection, and 6% of the population at any given time are infected, ie $P(\textrm{infected}) = 0.06$. Finally, suppose that the test has a false positive rate of 5% (or equivalently, has 95% specificity), meaning that 5% of the time, if a person is not infected, the test will return a positive result, ie $P(\textrm{positive}\mid\textrm{not infected}) = 0.05$.

Now imagine you take this test and it comes up positive. We can ask, what is the probability that you actually have this infection, given that your test result was positive? Well, \[P(\textrm{infected} \mid \textrm{positive}) = \frac{P(\textrm{positive} \mid \textrm{infected}) P(\textrm{infected})}{P(\textrm{positive})}.\] We can directly input the probabilities in the numerator based on the information provided in the previous paragraph. For the $P(\textrm{positive})$ term in the denominator, this probability has two parts to it: the probability that the test is positive and you are infected (true positive), and the probability that the test is positive and you are not infected (false positive). We need to scale these two parts according to the group of people that they apply to—either the proportion of the population that are infected, or the proportion that are not infected. Another way of thinking about this is considering the fact that \[P(\textrm{positive}) = P(\textrm{positive and infected}) + P(\textrm{positive and not infected}).\] Thus, we have \[P(\textrm{positive}) = P(\textrm{positive} \mid \textrm{infected})P(\textrm{infected}) + P(\textrm{positive} \mid \textrm{not infected})P(\textrm{not infected}).\]

And we can infer all the probabilities in this expression from the information that’s been given. Thus, we can work out that\[P(\mathrm{infected}|\mathrm{positive}) = \frac{0.9\times 0.06}{0.9\times 0.06 + 0.05\times 0.94}\approx 0.5347.\]

In a population with 6% infected by scorpionitis, this test will come back positive (purple shaded background) in 10.7% of cases.

Unpacking this result, this means that if you test positive for an infection, and if 1 in 17 people in the population (approximately 6%) are infected at any given time, there is an almost 50% chance that you are not actually infected, despite the test having a true positive rate of 90%, and a false positive rate of 5% (compare to the proportion of the shaded area in the diagram filled by infected people). That seems pretty high. Here are some takeaways from this example: the probability that you have an infection, given that you test positive for said infection, not only depends on the accuracy of the test, but it also depends on the prevalence of the disease within the population.

Unprecedented applicability

If 3% are infected with Covid-19 (purple person), a lateral flow test will come back positive (purple shaded background) 2.7% of the time.

Of course, in a real-world scenario, it’s a lot more complicated than this. For something like (and, apologies in advance for bringing it up) Covid-19, the prevalence of infection (our $P(\textrm{infected})$ value) changes with time. For example, according to government statistics, the average number of daily new cases in July 2020 was approximately 667, whereas in January 2021 it was 38,600. Furthermore, $P(\textrm{infected})$ depends on a vast number of factors including age, geographical location, and physical symptoms to name only a few. Still, it would be nice to get a sense of how Bayes’ theorem can be applied to these uNpReCeDeNtEd times.

An article from the UK Covid-19 lateral flow oversight team (catchy name, I know) released on 26 January 2021 reported that lateral flow tests (which provide results in a very short amount of time but are less accurate than the ‘gold standard’ PCR tests) achieved 78.8% sensitivity and 99.68% specificity in detecting Covid-19 infections. In the context of probabilities, this means that \begin{align*} &P(\textrm{positive} \mid \textrm{infected}) = 0.788 \textrm{ and}\\ &P(\textrm{positive} \mid \textrm{not infected}) = 0.0032. \end{align*} On 26 January 2021, there were 1,927,100 active cases of Covid-19 in the UK. Out of a population of 66 million, this is gives us a prevalence of approximately 3%, or $P(\textrm{infected}) = 0.03$.

Taking all these probabilities into account, we have \[P(\textrm{infected} \mid \textrm{positive}) = \frac{0.788 \times 0.03}{0.788 \times 0.03 + 0.0032 \times 0.97} \approx 0.8839,\] which means that the chances of you actually having Covid-19, given that you get a positive result from a lateral flow test, is about 88%. This seems pretty good, but can we make this any better?

If the prevalence increases to 80%, you can be much more certain of a positive result, but there are also more false negatives.

Instead of just taking the number of active cases as a percentage of the total population of the UK to give us our prevalence, we can alternatively consider $P(\textrm{infected})$ for a particular individual. For someone who has a cough, a fever, or who recently interacted with someone who was then diagnosed with Covid-19, we could say that their $P(\textrm{infected})$ is substantially higher than the overall prevalence in the country. The article Interpreting a Covid-19 test result in the BMJ suggests a reasonable value for such an individual would be $P(\textrm{infected}) = 0.8$. It’s worth mentioning that this article has a fun interactive tool where you can play around with sensitivity and specificity values to see how this affects true and false positivity and negativity rates. Taking this new value of prevalence, $P(\textrm{infected})$, into account, then \[P(\text{infected} \mid \text{positive}) = \frac{0.788 \times 0.8}{0.788 \times 0.8 + 0.0032 \times 0.2} \approx 0.9990,\vphantom{\frac{a}{\frac{c}{b}}}\] giving us a 99.9% chance of infection given a positive test result, which is way closer to certainty than the previous value of 88%.

Can we do any better than this? Well, compared with the lateral flow Covid-19 tests, it has been found that PCR tests (which use a different kind of technology to detect infection) have substantially higher sensitivity and specificity. Another recent article in the BMJ published in September 2020 reported that the PCR Covid-19 test has 94% sensitivity and very close to 100% specificity. In a survey conducted by the Office for National Statistics in the same month, they measured how many people across England and Wales tested positive for Covid-19 infection at a given point in time, regardless of whether they reported experiencing symptoms. In the survey, even if all positive results were false, specificity would still be 99.92%. For the sensitivity and specificity reported in the BMJ article, this is equivalent to having a false negative rate of 6% and a false positive rate of 0%. If we plug these numbers in, regardless of what the prevalence is taken to be, we have: \[P(\textrm{infected} \mid \textrm{positive}) = \frac{0.94 \times P(\textrm{infected})}{0.94 \times P(\textrm{infected}) + 0 \times P(\textrm{not infected})} = 1.\] So when a test has a false positive rate of almost 0%, if you achieve a positive test result, there is essentially a 100% chance that you do in fact have Covid-19.

So what can we take away from this? Well, we have seen that if a test has higher rates of sensitivity and specificity, then the probability of the result being a true positive is also higher. However, prevalence and the value of the probability of infection also play a big role in this scenario. This could be used as an argument for targeted testing only, for example if only people with symptoms were tested then this would increase the probability of the result being a true positive. Unfortunately, it is the case that a large number of Covid-19 infections are actually asymptomatic—in one study it was found that over 40% of cases in one community in Italy had no symptoms. So, if only people with symptoms were tested, a lot of infections would be missed.

Bae’s theorem

$P(\text{Netflix} \mid \text{chill}) =$

$\frac{P(\text{chill} \mid \text{Netflix})P(\text{Netflix})}{P(\text{chill})}$

In conclusion, I’m no epidemiologist, just your average mathematician, and I don’t really have any answers. Only that conditional probability is actually pretty interesting, and it turns out you can write a whole essay on it. The ending of Crazy Rich Asians was much better than the ending to this article. Go watch it, if you haven’t already.

It was 8pm on a wintry Saturday and I was pleading for my life. “I would never betray you. I promise.” I searched desperately for someone, anyone, to back me up. Of course, they were right. I had been responsible for the murder of many of their friends but I wasn’t about to admit to that. My co-conspirators had gone quiet, well aware that to support me was to put themselves into the firing line. At 8.55pm a vote was held. By 9pm I had been executed.

OK, so that first paragraph may have been a bit misleading. Thankfully I was not actually put to death, and I haven’t killed anyone in real life. This all happened whilst I was playing Mafia. For those unfamiliar, Mafia is strategy game in which players are (secretly) assigned to be either citizens or mafia. The game is split up into day and night phases (when playing in person, night is simulated by everybody closing their eyes). During the night phase, the Mafia are able to communicate with each other and can vote to kill one person. During the day phase, all the residents (both citizens and mafia) discover who died, and then vote to execute one resident. The aim for each group is to eliminate the other.

Some of you may be reading this thinking “Hmmmm this is sus. It sounds very much like Among Us” and you’d be right. The popular online game was inspired by Mafia and is one of many adaptations of the game. The particular version I was playing—when I was outed as a Mafia member—was Harry Potter themed (if there’s one thing you’ll learn about me during this article, it’s that I’m incredibly cool). People found themselves either on Team Hogwarts or Team Death Eater, and there were some special Potter themed rules, which led to an interesting situation mathematically. Continue reading →

My quest to answer this simple question began for the noblest of reasons—to win an argument with my wife. We are both football fans and have been following England all our lives (the phrase ‘long-suffering’ has never been more apt). Our house is usually an oasis of calm and tranquillity, but one thing is guaranteed to get things kicking off: was Sven-Göran Eriksson a good England manager?

This year, we have had more time than usual at home together and the discussion has become heated. I believe that Sven took a golden generation of England players and led them to disappointing performances in three major tournaments and she points to the team reaching the last 8 at consecutive World Cups under his stewardship. I’ve been a maths teacher for over 25 years, so surely, I can prove I am correct using maths, right? (At this point, it is definitely not worth mentioning that in the 15 years we’ve been having this row, I never thought of applying any maths to the problem until my wife suggested it.)

How do I prove that I am right? Well, the first, and most obvious, thing to do is to look at the playing record of Sven-Göran Eriksson. He was manager of England from January 2001 until July 2006. In that time the team played 67 games and won 40 of them—a win percentage of 59.7%.

On its own that is a bit meaningless, so we need something to compare it to. It’s time for a spreadsheet.

Walter Winterbottom

I am going to tidy up the data a little though. Firstly, up to 1946, there was no England coach. Even under Walter Winterbottom, the players were selected by committee, so I am going to exclude them. Secondly, caretaker managers like Stuart Pearce or Joe Mercer (or Sam Allardyce who was sacked after one game for ‘reasons’) didn’t have enough games and so I’ll drop them from consideration. That gives us the trimmed list below, ranked by winning percentage.

This shows that Sven had a pretty average record, only just reaching the top half of the table. I was feeling suitably pleased with myself for having come up with such a convincing statistic, only to be shot down with, “Yeah, but a lot of those games were meaningless friendlies.” I mean, you could argue that playing for England in any game is the pinnacle of a footballer’s career, and that international friendlies are always important games, but I decided to look at this as it seemed interesting (and I was confident that it would support my point even more).

Since we were looking at competitive internationals, I decided to look at the overall results record rather than using just the win percentage. After all, there are three possible outcomes in football and a draw has value (although this value varies with the opponent—a draw against Brazil is generally seen as a fairly decent result, whereas as a draw against Greece is not).

To calculate this, I used 3 points for a win and 1 for a draw. This has been the standard across football since the 1980s as it rewards positive play. This may disadvantage the managers from before it was introduced because playing for a draw would have been more profitable in group games and qualifiers, but I feel it is the best of the options available (and I’m trying to prove that Sven was a negative manager and I think this will help me)…

This did not go well and there was a significant amount of smugness, which I felt was inappropriate and irritating.

To be honest, this is a compelling result and I needed to come back strong if I was to maintain any credibility in this argument. I felt a little disappointed that I had done all that work to prove this important point and it wouldn’t be any use. I was starting to get concerned that manipulating statistics to get the result I wanted was not working, when a thought occurred to me—I might be able to use the Fifa ranking data to demonstrate that Sven-Göran Eriksson’s England team was only able to beat lesser teams and often struggled against higher ranking sides. In short, I chose to take a leaf from the Trump playbook—when you’re in trouble, smear the opposition.

OK, so it’s not classy but, in this case, I think it is a valid point to explore. Were most of Sven’s competitive games against weaker opposition? This is a possibility because qualifiers and group games are seeded, and so England would be facing so-called lesser teams. For example, let’s consider the 2006 World Cup qualifying group.

Only Poland finished ranked in the world’s top 50 international teams, which supports my contention that England were flat-track bullies under Sven. But this raised two interesting (in my opinion) questions:

1. How are the Fifa rankings calculated?
2. How can I use them to win this argument?

The Fifa ranking system

The Fifa ranking system was introduced in December 1992, and initially awarded teams points for every win or draw, like a traditional league table. However, Fifa quickly (five years later) realised that there were many other factors affecting the outcome of a football match and, over time (over twenty years) moved to a system based on the work of Hungarian–American mathematician Árpád Élő,more on him in a moment (I mean, why use an established and respected system when you can faff about making your own useless one? To be fair, the women’s rankings have used a version of the Elo system since their inception, which may make Fifa’s unwillingness to use it for the men even stranger).

The Fifa rankings are not helpful to me because they don’t cover all the managers I’m considering and because their accuracy, reliability and the many methods used to generate them were always questioned. Luckily, football fans have had these arguments before and there is an Elo ranking for all men’s international teams, which has been calculated back to the first international between England and Scotland in 1872 (a disappointing goalless draw).

Competitors in an esports event

The Elo rating system compares the relative performance of the competitors in two-player games. Although it was initially developed for rating chess players, variations of the system are used to rate players in sports such as table tennis, esports and even Scrabble. Strictly speaking, we should be saying an Elo system, rather than the Elo system as each sport has modified the formula to suit their own needs.

So how does an Elo system calculate a ranking? Well, at the most basic level, each team has a certain number of points and at the end of each game, one team gives some points to the other. The number of points depends on the result and the rankings of the two teams. When the favourite wins, only a few rating points will be traded, or even zero if there is a big enough difference in the rankings (eg in September 2015, England beat San Marino 6–0, but no Elo points were exchanged). However, if the underdog manages a surprise win, lots of rating points will be transferred (for example, when Iceland beat England at Euro 2016, they took 40 points from England). If the ratings difference is large enough, a team could even gain or lose points if they draw. So teams whose ratings are too low or too high should gain or lose rating points until the ratings reflect their true position relative to the rest of the group.

But how do you know how many points to add or take away after each game? Elo produced a formula for this, but there is a bit of maths—brace yourself.

Firstly, Elo assumed that a team would play at around the same standard, on average, from one game to next. However, sometimes they would play better or worse but with those performances grouped towards the average. This is known as a normal distribution (Elo uses a logistic distribution rather than the normal, but the differences are small—I mean, what’s a couple of percent between friends?) or bell curve, where outstanding results are possible but rare. In the graph below, the $x$-axis would represent the level of performance, and the $y$-axis shows the probability of that happening. So, we can see that the chance of an exceptional performance is smaller than that of an unremarkable one and the bulk of games will have a middling level of skill shown.

This means that if both teams perform to their standard, we can predict an expected score, which Elo defined as their probability of winning plus half their probability of drawing. Because we do not know the relative strengths of both teams, this expected score is calculated using their current ratings and the formulas \begin{align*} E_A&=\frac1{1+10^{(R_B-R_A)/400}} & &\text{and} & E_B&=\frac1{1+10^{(R_A-R_B)/400}}. \end{align*}

The expected score for a range of differences in team ratings.

In these formulas, $E_A$ and $E_B$ are the expected results for the teams, and $R_A$ and $R_B$ are their ratings. If you plot a graph of the $E$ values for different values of $R_A-R_B$ you get the graph shown to the left.

It’s interesting (again, interesting to me) to note the shape of this graph, which is a sigmoid, a shape that anyone who has drawn a cumulative frequency graph for their GCSE maths will recognise. It is an expression of the area under the distribution (ie the cumulative distribution function). The graph shows that if the difference between ratings is zero, the expected result is 0.5. The system uses values of 1 for a win, 0.5 for a draw and 0 for a loss, so this suggests a draw is the most likely outcome. And if the difference is 380 in your favour, the expected score is 0.9, which suggests you are likely to win (an $E_A$ of 0.9 doesn’t necessarily mean you’ll win 90% of the games and lose the rest as other combinations also give an expected score of 0.9. For example, winning 80%, and drawing the rest or winning 85%, drawing 10% and losing 5% gives the same value). The system then compares the actual result to the expected outcome and uses a relatively simple calculation (honestly, it’s easier than it looks) to calculate the number of points exchanged: $$R_A’=R_A+K(S_A-E_A).$$ In this equation, $R_A’$ is the new rating for team A, $S_A$ is the actual result of the game, and $K$ is a scaling factor. We’ll come back to $K$ in a moment. Recently, England (rating 1969) played Belgium (rating 2087) at the King Power stadium in Leuven, Belgium. It is generally thought that the home team is at an advantage and to reflect this, the home team gets a bonus 100 points to their rating which means there is a 218-point difference between the teams. England are clear underdogs, and we can calculate the expected result as follows: $$E_A=\frac1{1+10^{(2187-1969)/400}}\approx0.22$$ This shows that this will be a tricky game for England, and a draw would be a good result. Unfortunately, England lost the game 2–0, an $S_A$ of 0 (still using 1 for a win, etc). Therefore we can calculate the rating change using the formula: $$R_A’=1969+K(0-0.22)$$ Now we need to understand the $K$ value. In simple terms, the bigger the $K$ value we use, the more the rating will change with each result. We need to choose a suitable value so that it isn’t too sensitive, which would lead to wild swings, but also allows for teams to change position when they start to improve. $$R_A’=1969+60(0-0.22)\approx1956$$ The world football Elo rankings adjust the $K$ value depending on the score and the competition. In our example, which was a Nations League game (a new competition between European teams with similar Fifa rankings), the base value for $K$ is 40. This is multiplied by 1.5 for a win by 2 clear goals giving a $K$ value of 60.

This is a change of $-13$ points, and so Belgium would change by $+13$ points to a new rating of 2100.

Although I have focused on the world football Elo rankings, the Fifa rankings now use a system which is basically similar, with slight variations in the weightings and allowances. This brings me to the second, and more important part, of the question: can I use this to prove that I’m right?

Unfortunately, this explanation shows that you can only use this type of ranking, whether it’s the Elo or the Fifa system, to compare with teams that were playing at that time. This means that trying to use it to look back over time is pointless. You can’t compare the performance of Alf Ramsey’s England with that of Steve McClaren using the Elo rankings, because it is not designed to do that.

What can I do?

I can, however, use a similar idea—looking at England’s performance against differently rated teams—to judge Sven.

To achieve this, I’ve collated all of England’s results in competitive games under Sven and used some spreadsheet magic to create the tables shown to the right. (Do not ask how long this took.)

This is conclusive (it is—just trust me on this). Under Sven-Göran Eriksson, England were brilliant—if the team they were playing were outside the top twenty. Against good teams, England were awful. For comparison, in the 2020–21 season, Manchester United have a win percentage of 63.2% and a points percentage of 70.2%. On the other hand, Chelsea had a win percentage of 42.1% and a points percentage of 50.9% (based on results up to 27 January 2021), and they sacked the manager.

I can finally conclude that I was right. Sven was a rubbish manager who was worse than Frank Lampard.

Appendix – Sven-Goran Eriksson’s competitive match record

High-speed internet and digital storage get cheaper, but the challenge of sharing large files is one that anybody who spends their time working on computers faces. Digital images in particular can be a pain if they lead to long loading times and high server costs. If you have ever seen an image on the internet, then you have certainly encountered the JPEG format because it has been the web standard for almost 30 years. I am sure, however, that you have never heard of its potential successor, JPEG2000, even though it recently celebrated its 20th anniversary. If so, then that is unfortunate because it produces much better results than its predecessor. Continue reading →