Engaging students: Expressing probability as a fraction and as a percentage

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission again comes from my former student Jenna Sieling. Her topic, from probability: expressing a probability as a fraction and as a percentage.

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How could you as a teacher create an activity or project that involves your topic?

 

This topic is something that can really be applied in many places. Especially in sports, weather, and economics, probabilities as fractions and percentages are used daily. This can become very relatable to high school students no matter what they are interested in or plan to study in college. An activity that can be used in the classroom is starting a fake fantasy football league. Although I have never played in a fantasy football league, I know that to win in your group you need to look at the statistics of each player doing well. Given a class of hopefully around 30 students, we can start a week long activity of our own fantasy football league in the classroom and the students can be given different statistics each day to calculate the probability of their players being a good advantage for their team. This is just one activity that could catch the interest of students who may not usually be interested in probabilities.

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How can this topic be used in your students’ future courses in mathematics or science?

 

            One of the most popular majors for young students to fall into is business and probabilities become an important concept to understand if you plan to work in the business world. By making this point to a class, I feel the students will take the importance of this subject to heart. Business is not the only future path that would be using probabilities in the form of fractions or percentages. Fields like meteorology, economics, and even education majors would use the concept of probabilities to help teach elementary school students the basics to help them further on. If a student goes on to study history, at one point he or she will have to look at the economic history and understand the probability of these events happening and the probability of them happening again. The student would need to know how to multiply integers by fractions or percentages to gain conceptual knowledge of probability and its use.

 

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How can technology (YouTube, Khan Academy [khanacademy.org], Vi Hart, Geometers Sketchpad, graphing calculators, etc.) be used to effectively engage students with this topic?

 

I googled different online games to use for probability games and the most useful games, I found from Mathwire.com. Most games on this website were dice-based probability games but I think these are fun, easy games that could be assigned as homework. One game on the website was a game named SKUNK. The aim of the game is to guess the probability that a pair a dice will give you the highest amount of points. Each letter in the name SKUNK counts as one round and at the end of all the rounds, the person with the highest amount of points wins. Each player has to roll the dice once within one round and calculate the probability of getting the highest amount on each round. After looking at this game and others on this website, I realized that I could also explain the probability you need to understand to play poker if it was a popular game between friends and family. I could easily find a website to create a mock poker game and show students the idea of probability within poker.

 

 

Paranormal distribution

Source: https://www.tumblr.com/search/paranormal%20distribution

Engaging students: Independent and dependent events

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Danielle Pope. Her topic, from Probability: independent and dependent events.

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What interesting (i.e., uncontrived) word problems using this topic can your students do now?

Students use the idea of independent and dependent events in their lives without even realizing it. Many of the word problems used to introduce probability are basic concepts that students can understand. The basic definition of an independent event is “the probability that one event occurs in no way affects the probability of the other event occurring”. Word problems can be used to demonstrate this. Asking if the probability of flipping a coin changes if you were to roll a die as well is a prime example. These two acts are something that can be easily implemented in the classroom and the technical definition can be taught. Students can then help come up with more scenarios and teach themselves the terms. Similarly this idea can be used for dependent variables with a few changes. If the “probability of one event occurring influences the likelihood of the other event” then the event is defined as dependent. Word problems could be “if you were to draw two cards from a deck of 52 cards and if on your first draw you had an ace and you put that aside, would the probability of drawing another ace change? This card questions could be more challenging by taking out more cards each time.

 

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How can this topic be used in your students’ future courses in mathematics or science?

The topic of independent and dependent events can later be translated into variables when used with functions in Algebra class. Knowing and understanding the difference will help students know how to classify an event and use the correct variable and axis if asked to sketch a list of data. Just in a probability course students will learn about conditional probability, which will use the idea of dependence. Other terminology like with replacement and without replacement will be used to define a dependent event in probability. This topic can even be translated into a physics classroom when talking about time, position, velocity, acceleration etc. For example, when calculating the velocity students will either find or be given the displacement and change in time. Not knowing that the dependent event divides the independent variable or specifically with velocity, displacement divides time, If those numbers are not plugged in correctly then that will lead to the wrong answer.

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How has this topic appeared in the news?

In the news, independent and dependent events show up everyday. The most common example is weather. One of the longest debates that we have been having is if global warming and climate change has influenced the world. According to an article in the Smithsonian Magazine, scientists “couldn’t prove that global warming had “caused” the heat wave of 2003, (they) did assert that warming from human emissions had doubled the risk of extreme weather events.” This observation can then be taken to a student’s science class and they can research the risk of continuing this pattern of damage to Earth. Natural disasters can also have a say in many events around the world. For example just recently “Gas prices spiked in the Baltimore-area — and nationwide — in recent days and are expected to continue to rise after a major pipeline that runs from Texas to the East Coast had to be shut down following Hurricane Harvey”. This is a prime example of a dependent event. This shortage in gas specifically in Texas then led to many people rushing to fill up their cars, resulting in gas stations running out of gas, which is just another example of dependent events.

 

 

References:

https://www.wyzant.com/resources/lessons/math/statistics_and_probability/probability/further_concepts_in_probability

http://www.smithsonianmag.com/science-nature/does-climate-change-cause-extreme-weather-events-180964506/

http://www.baltimoresun.com/business/bs-bz-gas-prices-20170901-story.html

Facebook Birthday Problem: Part 5

Recently, I devised the following problem:

Suppose that you have n friends, and you always say “Happy Birthday” to each friend on his/her birthday. On how many days of the year will you not say “Happy Birthday” to one of your friends?

Until somebody tells me otherwise, I’m calling this the Facebook birthday problem in honor of Facebook’s daily alerts to say “Happy Birthday” to friends.

Here’s how I solved this problem. Let I_k be an indicator random variable for “no friend has a birthday on day k, where k = 366 stands for February 29 and k = 1, \dots, 365 stand for the “usual” 365 days of the year. Therefore, the quantity N, representing the number of days of the year on which no friend has a birthday, can be written as

N = I_1 + \dots + I_{365} + I_{366}

In yesterday’s post, I began the calculation of the standard deviation of N by first computing its variance. This calculation is complicated by the fact that I_1, \dots, I_{366} are dependent. Yesterday, I showed that

\hbox{Var}(N) = 365\displaystyle \left( \frac{1457}{1461} \right)^n \left[ 1 - \left( \frac{1457}{1461} \right)^n \right] + \displaystyle \left( \frac{1460}{1461} \right)^n \left[ 1 - \left( \frac{1460}{1461} \right)^n \right]

+ \displaystyle (365)(364) \left[ \displaystyle \left( \frac{1453}{1461} \right)^n - \displaystyle \left( \frac{1457}{1461} \right)^{2n} \right] + 2 \sum_{k=1}^{365}\hbox{Cov}(I_k,I_{366}).

To complete this calculation, I’ll now find \hbox{Cov}(I_k,I_{366}), where 1 \le k \le 365. I’ll use the usual computation formula for a covariance,

\hbox{Cov}(I_k,I_{366}) = E(I_k I_{366}) - E(I_k) E(I_{366}).

We have calculated E(I_k) earlier in this series. In any four-year span, there are 4 \times 365 + 1 = 1461 days, of which only one is February 29. Assuming the birthday’s are evenly distributed (which actually doesn’t happen in real life), the chance that someone’s birthday is not on day k is

\displaystyle 1 - \frac{4}{1461} = \displaystyle \frac{1457}{1461},

so that the probability that no friend has a birthday on day k is

\displaystyle \left( \frac{1457}{1461} \right)^n.

Therefore, since the expected value of an indicator random variable is the probability that the event happens, we have

E(I_k) = \displaystyle \left( \frac{1457}{1461} \right)^n

for k = 1, \dots, 365. Similarly,

E(I_{366}) = \displaystyle \left( \frac{1460}{1461} \right)^n,

so that

\hbox{Cov}(I_k,I_{366}) = E(I_k I_{366}) - \displaystyle \left( \frac{1457}{1461} \right)^n \left( \frac{1460}{1461} \right)^n.

To find E(I_k I_{366}), we note that since I_k is equal to either 0 or 1 and I_{366} is equal to either 0 or 1, the product I_k I_{366} can only equal 0 and 1 as well. Therefore, I_k I_{366} is itself an indicator random variable. Furthermore, I_k I_{366} = 1 if and only if I_k = 1 and I_{366} = 1, which means that no friends has a birthday on either day k or day 366 (that is, February 29). The chance that someone doesn’t have a birthday on day k or February 29 is

\displaystyle 1 - \frac{4}{1461} - \frac{1}{1461} = \displaystyle \frac{1456}{1461},

so that the probability that no friend has a birthday on day k or February 29 is

\displaystyle \left( \frac{1456}{1461} \right)^n.

Therefore, as before,

E(I_k I_{366}) = \displaystyle \left( \frac{1456}{1461} \right)^n,

so that

\hbox{Cov}(I_k,I_{366}) = \displaystyle \left( \frac{1456}{1461} \right)^n - \displaystyle \left( \frac{1457}{1461} \right)^n \left( \frac{1460}{1461} \right)^n.

Therefore,

\hbox{Var}(N) = 365\displaystyle \left( \frac{1457}{1461} \right)^n \left[ 1 - \left( \frac{1457}{1461} \right)^n \right] + \displaystyle \left( \frac{1460}{1461} \right)^n \left[ 1 - \left( \frac{1460}{1461} \right)^n \right]

+ \displaystyle (365)(364) \left[ \displaystyle \left( \frac{1453}{1461} \right)^n - \displaystyle \left( \frac{1457}{1461} \right)^{2n} \right] + 2(365) \left[ \left( \frac{1456}{1461} \right)^n - \displaystyle \left( \frac{1457}{1461} \right)^n \left( \frac{1460}{1461} \right)^n \right],

and we find the standard deviation of N using

\hbox{SD}(N) = \sqrt{\hbox{Var}(N)}.

The graph below shows the expected value of N, which was shown earlier to be

E(N) = 365 \displaystyle \left( \frac{1457}{1461} \right)^n + \left( \frac{1460}{1461} \right)^n,

along with error bars representing two standard deviations.

Interestingly, the standard deviation of N changes for different values of n; a direct calculation shows that the \hbox{SD}(N) is maximized at n = 459 with maximum value of approximately 6.1. Accordingly, for n = 450 and n = 500, the error bars in the above figure have a total width of approximately 24 days (two standard deviations both above and below the expected value).

Facebook Birthday Problem: Part 4

Recently, I devised the following problem:

Suppose that you have n friends, and you always say “Happy Birthday” to each friend on his/her birthday. On how many days of the year will you not say “Happy Birthday” to one of your friends?

Until somebody tells me otherwise, I’m calling this the Facebook birthday problem in honor of Facebook’s daily alerts to say “Happy Birthday” to friends.

Here’s how I solved this problem. Let I_k be an indicator random variable for “no friend has a birthday on day k, where k = 366 stands for February 29 and k = 1, \dots, 365 stand for the “usual” 365 days of the year. Therefore, the quantity N, representing the number of days of the year on which no friend has a birthday, can be written as

N = I_1 + \dots + I_{365} + I_{366}

In yesterday’s post, I began the calculation of the standard deviation of N by first computing its variance. This calculation is complicated by the fact that I_1, \dots, I_{366} are dependent. Yesterday, I showed that

\hbox{Var}(N) = 365\displaystyle \left( \frac{1457}{1461} \right)^n \left[ 1 - \left( \frac{1457}{1461} \right)^n \right] + \displaystyle \left( \frac{1460}{1461} \right)^n \left[ 1 - \left( \frac{1460}{1461} \right)^n \right]

+ \displaystyle 2 \!\!\!\!\! \sum_{1 \le j < k \le 365} \!\!\!\!\! \hbox{Cov}(I_j,I_k) + 2 \sum_{k=1}^{365} \hbox{Cov}(I_k,I_{366})

To complete this calculation, I’ll now find the covariances. I’ll begin with \hbox{Cov}(I_j,I_k) if 1 \le j < k \le 365; that is, if j and k are days other than February 29. I’ll use the usual computation formula for a covariance,

\hbox{Cov}(I_j,I_k) = E(I_j I_k) - E(I_j) E(I_k).

We have calculated E(I_k) earlier in this series. In any four-year span, there are 4 \times 365 + 1 = 1461 days, of which only one is February 29. Assuming the birthday’s are evenly distributed (which actually doesn’t happen in real life), the chance that someone’s birthday is not on day k is

\displaystyle 1 - \frac{4}{1461} = \displaystyle \frac{1457}{1461},

so that the probability that no friend has a birthday on day k is

\displaystyle \left( \frac{1457}{1461} \right)^n.

Therefore, since the expected value of an indicator random variable is the probability that the event happens, we have

E(I_k) = \displaystyle \left( \frac{1457}{1461} \right)^n

for k = 1, \dots, 365. Therefore,

\hbox{Cov}(I_j,I_k) = E(I_j I_k) - \displaystyle \left( \frac{1457}{1461} \right)^n \left( \frac{1457}{1461} \right)^n = E(I_j I_k) - \displaystyle \left( \frac{1457}{1461} \right)^{2n}.

To find E(I_j I_k), we note that since I_j is equal to either 0 or 1 and I_k is equal to either 0 or 1, the product I_j I_k can only equal 0 and 1 as well. Therefore, I_j I_k is itself an indicator random variable, which I’ll call I_{jk}. Furthermore, I_{jk} if and only if I_j = 1 and I_k = 1, which means that no friends has a birthday on either day j or day k. The chance that someone doesn’t have a birthday on day j or day k is

\displaystyle 1 - \frac{4}{1461} - \frac{4}{1461} = \displaystyle \frac{1453}{1461},

so that the probability that no friend has a birthday on day j or k is

\displaystyle \left( \frac{1453}{1461} \right)^n.

Therefore, as before,

E(I_j I_k) = \displaystyle \left( \frac{1453}{1461} \right)^n,

so that

\hbox{Cov}(I_j,I_k) = \displaystyle \left( \frac{1453}{1461} \right)^n - \displaystyle \left( \frac{1457}{1461} \right)^{2n}.

Since there are \displaystyle {365 \choose 2} = \displaystyle \frac{365\times 364}{2} pairs (j,k) so that 1 \le j < k \le 365, we have

\hbox{Var}(N) = 365\displaystyle \left( \frac{1457}{1461} \right)^n \left[ 1 - \left( \frac{1457}{1461} \right)^n \right] + \displaystyle \left( \frac{1460}{1461} \right)^n \left[ 1 - \left( \frac{1460}{1461} \right)^n \right]

+ \displaystyle 2 \times \displaystyle \frac{365\times 364}{2} \left[ \displaystyle \left( \frac{1453}{1461} \right)^n - \displaystyle \left( \frac{1457}{1461} \right)^{2n} \right] + 2 \sum_{k=1}^{365}\hbox{Cov}(I_k,I_{366}),

or

\hbox{Var}(N) = 365\displaystyle \left( \frac{1457}{1461} \right)^n \left[ 1 - \left( \frac{1457}{1461} \right)^n \right] + \displaystyle \left( \frac{1460}{1461} \right)^n \left[ 1 - \left( \frac{1460}{1461} \right)^n \right]

+ \displaystyle (365)(364) \left[ \displaystyle \left( \frac{1453}{1461} \right)^n - \displaystyle \left( \frac{1457}{1461} \right)^{2n} \right] + 2 \sum_{k=1}^{365}\hbox{Cov}(I_k,I_{366}).

The calculation of \hbox{Cov}(I_k,I_{366}) is similar to the above calculation; I’ll write this up in tomorrow’s post.

Facebook Birthday Problem: Part 3

Recently, I devised the following problem:

Suppose that you have n friends, and you always say “Happy Birthday” to each friend on his/her birthday. On how many days of the year will you not say “Happy Birthday” to one of your friends?

Until somebody tells me otherwise, I’m calling this the Facebook birthday problem in honor of Facebook’s daily alerts to say “Happy Birthday” to friends.

Here’s how I solved this problem. Let I_k be an indicator random variable for “no friend has a birthday on day k, where k = 366 stands for February 29 and k = 1, \dots, 365 stand for the “usual” 365 days of the year. Therefore, the quantity N, representing the number of days of the year on which no friend has a birthday, can be written as

N = I_1 + \dots + I_{365} + I_{366}

In yesterday’s post, I showed that

E(N) = E(I_1) + \dots + E(I_{365}) + E(I_{366}) = 365 \displaystyle \left( \frac{1457}{1461} \right)^n + \left( \frac{1460}{1461} \right)^n.

The calculation of the standard deviation of N is considerably more complicated, however, since the I_1, \dots, I_{366} are dependent. So we will begin by computing the variance of N:

\hbox{Var}(N) = \displaystyle \sum_{k=1}^{366} \hbox{Var}(I_k) + 2 \!\!\!\!\! \sum_{1 \le j < k \le 366} \!\!\!\!\! \hbox{Cov}(I_j,I_k),

or

\hbox{Var}(N) = \displaystyle \sum_{k=1}^{365} \hbox{Var}(I_k) + \hbox{Var}(I_{366}) + 2 \!\!\!\!\! \sum_{1 \le j < k \le 365} \!\!\!\!\! \hbox{Cov}(I_j,I_k) + 2 \sum_{k=1}^{365} \hbox{Cov}(I_k,I_{366})

For the first term, we recognize that, in any four-year span, there are 4 \times 365 + 1 = 1461 days, of which only one is February 29. Assuming the birthday’s are evenly distributed (which actually doesn’t happen in real life), the chance that someone’s birthday is not on day k is

\displaystyle 1 - \frac{4}{1461} = \displaystyle \frac{1457}{1461}.

Therefore, the chance that all n friends don’t have a birthday on day k is

\displaystyle \left( \frac{1457}{1461} \right)^n.

Using the formula \hbox{Var}(I) = p(1-p) for the variance of an indicator random variable, we see that

\hbox{Var}(I_k) = \displaystyle \left( \frac{1457}{1461} \right)^n \left[ 1 - \left( \frac{1457}{1461} \right)^n \right]

for k = 1, \dots, 365. Similarly, for the second term,

\hbox{Var}(I_{366}) = \displaystyle \left( \frac{1460}{1461} \right)^n \left[ 1 - \left( \frac{1460}{1461} \right)^n \right]

Therefore, so far we have shown that

\hbox{Var}(N) = 365\displaystyle \left( \frac{1457}{1461} \right)^n \left[ 1 - \left( \frac{1457}{1461} \right)^n \right] + \displaystyle \left( \frac{1460}{1461} \right)^n \left[ 1 - \left( \frac{1460}{1461} \right)^n \right]

+ \displaystyle 2 \!\!\!\!\! \sum_{1 \le j < k \le 365} \!\!\!\!\! \hbox{Cov}(I_j,I_k) + 2 \sum_{k=1}^{365} \hbox{Cov}(I_k,I_{366})

In tomorrow’s post, I’ll complete this calculation by finding the covariances.

Facebook Birthday Problem: Part 2

Recently, I devised the following problem:

Suppose that you have n friends, and you always say “Happy Birthday” to each friend on his/her birthday. On how many days of the year will you not say “Happy Birthday” to one of your friends?

Until somebody tells me otherwise, I’m calling this the Facebook birthday problem in honor of Facebook’s daily alerts to say “Happy Birthday” to friends.

Here’s how I solved this problem. Let I_k be an indicator random variable for “no friend has a birthday on day k, where k = 366 stands for February 29 and k = 1, \dots, 365 stand for the “usual” 365 days of the year. Therefore, the quantity N, representing the number of days of the year on which no friend has a birthday, can be written as

N = I_1 + \dots + I_{365} + I_{366}

Let’s start with any of the “usual” days. In any four-year span, there are 4 \times 365 + 1 = 1461 days, of which only one is February 29. Assuming the birthday’s are evenly distributed (which actually doesn’t happen in real life), the chance that someone’s birthday is not on day k is

\displaystyle 1 - \frac{4}{1461} = \displaystyle \frac{1457}{1461}.

Therefore, the chance that all n friends don’t have a birthday on day k is

\displaystyle \left( \frac{1457}{1461} \right)^n.

Since the expected value of an indicator random variable is the probability of the event, we see that

E(I_k) = \displaystyle \left( \frac{1457}{1461} \right)^n

for k = 1, \dots, 365. Similarly, the expected value for the indicator for February 29 is

E(I_{366}) = \displaystyle \left( \frac{1460}{1461} \right)^n.

Since E(X+Y) = E(X) + E(Y) even if X and Y are dependent, we therefore conclude that

E(N) = E(I_1) + \dots + E(I_{365}) + E(I_{366}) = 365 \displaystyle \left( \frac{1457}{1461} \right)^n + \left( \frac{1460}{1461} \right)^n.

This function is represented by the red dots on the graph below.

In tomorrow’s post, I’ll calculate of the standard deviation of N.

Facebook Birthday Problem: Part 1

The “birthday problem” is one of the classic problems in elementary probability because of its counter-intuitive solution. From Wikipedia:

In probability theory, the birthday problem or birthday paradox concerns the probability that, in a set of n randomly chosen people, some pair of them will have the same birthday. By the pigeonhole principle, the probability reaches 100% when the number of people reaches 367 (since there are only 366 possible birthdays, including February 29). However, 99.9% probability is reached with just 70 people, and 50% probability with 23 people. These conclusions are based on the assumption that each day of the year (except February 29) is equally probable for a birthday.

Recently, I devised the following different birthday problem:

Suppose that you have n friends, and you always say “Happy Birthday” to each friend on his/her birthday. On how many days of the year will you not say “Happy Birthday” to one of your friends?

Until somebody tells me otherwise, I’m calling this the Facebook birthday problem in honor of Facebook’s daily alerts to say “Happy Birthday” to friends.

In this series, I will solve this problem. While this may ruin the suspense, here’s a graph of the solution for 100 \le n \le 1000 along with error bars indicating two standard deviations.

Before deriving this solution, I’ll start with a thought bubble if you’d like to take some time to think about how to do this.

What is a Random Walk?

From PBS Infinite Series:

A Long-Sought Proof, Found and Almost Lost

I enjoyed this article from Quanta Magazine, both for its mathematical content as well as the human interest story.

A Long-Sought Proof, Found and Almost Lost

From the opening paragraphs:

Known as the Gaussian correlation inequality (GCI), the conjecture originated in the 1950s, was posed in its most elegant form in 1972 and has held mathematicians in its thrall ever since. “I know of people who worked on it for 40 years,” said Donald Richards, a statistician at Pennsylvania State University. “I myself worked on it for 30 years.”

[Thomas] Royen hadn’t given the Gaussian correlation inequality much thought before the “raw idea” for how to prove it came to him over the bathroom sink… In July 2014, still at work on his formulas as a 67-year-old retiree, Royen found that the GCI could be extended into a statement about statistical distributions he had long specialized in. On the morning of the 17th, he saw how to calculate a key derivative for this extended GCI that unlocked the proof. “The evening of this day, my first draft of the proof was written,” he said.

Not knowing LaTeX, the word processer of choice in mathematics, he typed up his calculations in Microsoft Word, and the following month he posted his paper to the academic preprint site arxiv.org. He also sent it to Richards, who had briefly circulated his own failed attempt at a proof of the GCI a year and a half earlier. “I got this article by email from him,” Richards said. “And when I looked at it I knew instantly that it was solved” …

Proofs of obscure provenance are sometimes overlooked at first, but usually not for long: A major paper like Royen’s would normally get submitted and published somewhere like the Annals of Statistics, experts said, and then everybody would hear about it. But Royen, not having a career to advance, chose to skip the slow and often demanding peer-review process typical of top journals. He opted instead for quick publication in the Far East Journal of Theoretical Statistics, a periodical based in Allahabad, India, that was largely unknown to experts and which, on its website, rather suspiciously listed Royen as an editor. (He had agreed to join the editorial board the year before.)

With this red flag emblazoned on it, the proof continued to be ignored… No one is quite sure how, in the 21st century, news of Royen’s proof managed to travel so slowly. “It was clearly a lack of communication in an age where it’s very easy to communicate,” Klartag said.