IB Maths Resources from Intermathematics

On this site you will find IB Maths and IGCSE Maths Resources for IB Maths explorations and investigations. I’ve tried to build connections with real life maths, Theory of Knowledge (ToK) and ideas for maths careers. There are also maths videos, puzzles and lesson resources.

Be sure to download a completely free 63 page Exploration Guide to help you achieve an excellent score on your coursework.  You can download this for free here.

Welcome to the site.  There are hundreds of pages of maths ideas to explore.  Scroll down to see.

Many thanks!

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If you are a teacher then please also visit my new site:  intermathematics.com.

My new site has been designed specifically for teachers of mathematics at international schools.  The content now includes over 2500 pages of pdf content for the entire SL/HL Analysis and SL/HL Applications syllabus Some of the content includes:

  1. Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics.  These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
  2. Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
  3. Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
  4. A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.

There is also a lot more.  I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!

You can see some of examples of the content in the Teacher resources section of this site – which has a lot of worksheets and ideas for IB maths teachers.

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IB Maths Revision

There’s a really fantastic website been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams.  I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL and SL students for both Analysis and Applications.  Some of my favourite resources include:

1. Questionbank

Screen Shot 2018-03-19 at 4.42.05 PM.pngThere is a beautifully designed and comprehensive Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and then provides a large bank of graded questions.  What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial.  Really useful!

2. Practice Exams

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The Practice Exams section takes you to a large number of ready made quizzes, exams and predicted papers.   These all have worked solutions and allow you to focus on specific topics or start general revision.  This also has some excellent challenging questions for those students aiming for 6s and 7s.

3.  IB Maths Key Concepts Videos

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Each course has a dedicated video tutorial section which provides 5-15 minute tutorial videos on every single syllabus part – handily sorted into topic categories.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.

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Proving Pythagoras Like Einstein?

There are many ways to prove Pythagoras’ theorem – Einstein reputedly used the sketch above to prove this using similar triangles.  To keep in the spirit of discovery I also just took this diagram as a starting point and tried to prove this myself, (though Einstein’s version turns out to be a bit more elegant)!

Step 1: Finding some links between triangles

We can see that our large right angled triangle has sides a,b,c with angles alpha and beta.  Hopefully it should also be clear that the two smaller right angled triangles will also have angles alpha and beta.  Therefore our triangles will all be similar.  It should also be clear that the area of the 2 small triangles will be the same as the area of the large triangle.

Step 2: Drawing a sketch to make things clearer:

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It always helps to clarify the situation with some diagrams.  So, let’s do that first.

Step 3:  Making some equations

As the area of the 2 small triangles will be the same as the area of the large triangle this gives the following equation:

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We also can make the following equation by considering that triangles 2 and 3 are similar

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We can now substitute our previous result for x into this new equation (remember our goal is to have an equation just in terms of a,b,c so we want to eliminate x and y from our equations).

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We can also make the following equation by considering that triangles 1 and 2 are similar:

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And as before, our goal is to remove everything except a,b,c from these equations, so let’s make the substitution for y using our previous result:

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And if by magic, Pythagoras’ theorem appears!  Remember that the original a,b,c  related to any right angled triangle with hypotenuse c, so we have proved that this equation must always be true for right angled triangles.

You can explore some other ways of proving Pythagoras here.  Which is the most elegant?

I’ve just made a big update to both the teacher and student resources sections:

Student resources

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These now have some great free resources for students to help them with the IB maths course – including full course notes, formula books, Paper 3s, an Exploration guides and a great mind-map.  Make sure to check these all out to get some excellent support for the IB maths course.

Teacher resources

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These now have over 25 worksheets, investigations, paper 3s, treasure hunts and more resources – both with question pdfs and markscheme pdfs.  I’ve added a lot of enriching activities that would support explorations and paper 3 style problems and also put a selection of some excellent other resources from IB teachers too.

So be sure to check these both out!

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Finding the average distance in a polygon

Over the previous couple of posts I’ve looked at the average distance in squares, rectangles and equilateral triangles.  The logical extension to this is to consider a regular polygon with sides 1.   Above is pictured a regular pentagon with sides 1 enclosed in a 2 by 2 square.  The points N and O represent 2 randomly chosen points which we find the distance between.  On average what is the distance between these randomly chosen points N and O?

Starting with a hexagon

It’s a little easier to start with a hexagon as we get some nicer coordinate points.  So, our first challenge is to find the coordinates of a regular hexagon with sides 1.  Luckily we can use the complex roots of unity to do this.  We start by finding the 6th roots of unity and then converting these to coordinates in an Argand diagram:

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This then allows us to plot the following:

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We can then work out the inequalities which define the inside of the hexagon when we generate points within the 2×2 square centred at (0,0).  This gives:

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We can then run the following code to find the average distance:

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This gives the following result:

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We can check this result as the exact value is:

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which is 0.8262589495.  So we can see we are accurate here to 3 sf.


For the pentagon we can find the coordinates by finding the 5th roots of unity:

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We then need to scale all coordinate points by a factor, because in a pentagon the distance from the centre to the points is not 1 (as is the case in roots of unity).  We can find the distance from the centre to the edge of a pentagon by the following trigonometry:

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So, when we scale all coordinate points by this factor we get:

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And we can then do the same method as before and run the following Python code:

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This gives:

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n-sided polygon

We can now consider an n-sided polygon with sides 1.  Let’s start with the values we’ve found for an equilateral triangle (0.364), a square (0.522), a pentagon (0.697) and a hexagon (0.826.

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When we plot these they appear to follow a linear relationship:

average distance = 0.14n

We can check that this is correct by considering the fact that an n sided polygon will approximate a circle when n gets large.  So an n sided polygon with sides length 1 can be approximated by a circle with circumference n.  This allows us to work out the radius.

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We can then substitute this into the equation for the average distance of 2 points in a circle.

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So we would expect the average distance between 2 points in a regular polygon of sides 1 to approach the equation (as n gets large):

average distance = 0.144101239n

And we’ve finished!  Everything cross-checks and works nicely.  We’ve been able to use a mixture of complex numbers, geometry, coding and trigonometry to achieve this result.

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Finding the average distance in an equilateral triangle

In the previous post I looked at the average distance between 2 points in a rectangle.  In this post I will investigate the average distance between 2 randomly chosen points in an equilateral triangle.

Drawing a sketch.

The first step is to start with an equilateral triangle with sides 1.  This is shown above.  I sketched this using Geogebra – and used some basic Pythagoras to work out the coordinates of point C.

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I can then draw a square of sides 1 around this triangle as shown above.  I’m then going to run a Python program to randomly select points and then work out the distance between them – but I need to make sure that the 2 points chosen are both inside this triangle.  For this I need to work out the equation of the line AC and CB.

Using basic coordinate geometry we can see that the line AC has equation y = √3x.  We want the inequality y < √3x so that we are on the correct side of this line.

The line BC has equation y = -√3x + √3.  Therefore the triangle must also satisfy the inequality y < -√3x + √3.

I can then run the following code on Python, with finds the average distance between points (a,c) and (b,d) both within the unit square but also subject to the 2 inequality constraints above.

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When this is run it performs 999,999 trials and then finds the average distance.  This returns the following value:

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So we can see that the average distance is just over a third of a unit.

Finding the average distance of an equilateral triangle of length n.

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We can then draw the sketch above to find the equation of lines AC and CB for an equilateral triangle with lengths n.  This leads to the following inequalities:

 y < √3x

 y < -√3x + √3n

So we can then modify the code as follows:

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This then returns the average distances for equilateral triangles of sizes 1 to 10.

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And when we plot this on Desmos we can see that there is a linear relationship:

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The regression line has gradient 0.36 (2sf) so we can hypothesise that for an equilateral triangle of size n, the average distance between 2 points is approximately 0.36n.

Checking the maths

I then checked the actual equation for the average distance between 2 points in an equilateral triangle of sides n:

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This gives us:

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So we can see that we were accurate to 2 significant figures.  So this is a nice mixture of geometry, graphing and computational power to provide a result which would be otherwise extremely difficult to calculate.

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What is the average distance between 2 points in a rectangle?

Say we have a rectangle, and choose any 2 random points within it.  We then could calculate the distance between the 2 points.  If we do this a large number of times, what would the average distance between the 2 points be?

Monte Carlo method

The Monte Carlo method is perfect for this – we can run the following code on Python:

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This code will find the average distance between 2 points in a 10 by 10 square.  It does this by generating 2 random coordinates, finding the distance between them and then repeating this process 999,999 times.  It then works out the average value.  If we do this it returns:

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This means that on average, the distance between 2 random points in a 10 by 10 square is about 5.21.

Generalising to rectangles

I can now see what happens when I fix one side of the rectangle and vary the other side.  The code below fixes one side of the rectangle at 1 unit, and then varies the other side in integer increments.  For each rectangle it then calculates the average distance.

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This then returns the first few values as:

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This shows that for a 1 by 1 square the average distance between two points is around 0.52 and for a 1 by 10 rectangle the average distance is around 3.36.

Plotting some Desmos graphs

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Because I have included the comma in the Python code I can now copy and paste this straight into Desmos.  The dotted green points show how the average distance of a 1 by x rectangle changes as x increases.  I then ran the same code to work out the average distance of a 10 by x rectangle (red), 20 by x rectangle (black), 30 by x rectangle (purple) and 100 by x rectangle (yellow).

We can see if we continue these points further that they all appear to approach the line y = 1/3 x (dotted green).  This is a little surprising – i.e when x gets large, then for any n by x rectangle (with n fixed), an increase in x by one will tend towards an increase in the average distance by 1/3.

Heavy duty maths!

There is actually an equation that fits these curves – and will give the average distance, a(X) between any 2 points in a rectangle with sides a and b (a≥b).  Here it is:

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I added this equation into Desmos, by changing the a to x, and then adding a slider for b.  So, when I set b=1 this generated the case when the side of a rectangle is fixed as 1 and the other side is increased:

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Plotting these equations on Desmos then gives the following:

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Pleasingly we can see that the points created by our Monte Carlo method fit pretty much exactly on the lines generated by these equations.  By looking at these lines at a larger scale we can see that they do all indeed appear to be approaching the line y = 1/3 x.

General equation for a square

We can now consider the special case when a=b (i.e a square).  This gives:

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Which we can simplify to give:

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We can see therefore that a square of side 1 (a=1) will have an average distance of 0.52 (2dp) and a square of side 10 (a=10) will have an average distance of 5.21  – which both agree with our earlier results.

Plotting Pi and Searching for Mona Lisa

This is a very nice video from Numberphile – where they use a string of numbers (pi) to write a quick Python Turtle code to create some nice graphical representations of pi.  I thought I’d quickly go through the steps required for people to do this by themselves.

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Firstly you can run the Turtle code on trinket.io.  If you type the above code this will take the decimal digits of pi one at a time and for each one move forward 10 steps and then turn by 36 degrees times by that digit.  So for example the 1 will lead to a right turn of 36 degrees and the 4 will lead to a right turn of 36 x 4 = 144 degrees.

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Next it would be nice to have more digits of pi to paste in rather than type.  So we can go to the onlinenumbertools website and generate as many digits of pi as we want.  Select them to be comma separated and also to not include the first digit 3.  You can then copy and paste this string in place of the 1,4,1 in the code above.

1000 digits of pi

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If we run this program after pasting the first 1000 digits of pi we get (after waiting a while!) the above picture. There are a number of questions that they then raise in the video – if this program was ran infinitely long would the whole screen eventually be black?  Would this create every possible image that can be created by 36 degree turns?  Would it be possible to send this picture (say to an alien civilization) and for the recipient to be able to reverse engineer the digits of pi?

2000 digits of pi

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If you increase the digits of pi to around 2000 you get the above picture.  The graph spends a large time in the central region before finally “escaping” to the left.  It then left my screen at the top.

3000 digits of pi

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We can see that the turtle “returned” from off the top of the screen and then joined back up with the central region.  This starts to look like a coastline – maybe the south of the UK!

Different bases: Base 3

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We can consider the digits of pi in base three – which means that they are all equivalent to 0,1,2.  This means that we can use these to specify either 0 degree, 120 degree or 240 degree turns.  We can change the code as shown above to achieve this.  Note the i%3 gives i mod 3.  For example if the digit is 8, then 8 mod 3 is 2 (the remainder when 8 is divided by 3) and so this would turn 120 x 2 = 240 degrees.

This then creates a pattern which looks similar to the Sierpinski triangle fractal design:

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Base 4

Using a similar method, we can create the following using a base 4 design:

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This creates what looks like a map layout.

Base 5:

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In base 5 the turtle quickly departed from my screen!  With turns of 72 we don’t see the tessellating shapes that we do with base 3 and 4.

Base 6:

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With a 60 degree turn we can now see a collection of equilateral triangles and hexagons.

You can explore with different numbers and different bases to see what patterns you can create!

Witness Numbers: Finding Primes

The Numberphile video above is an excellent introduction to primality tests – where we conduct a test to determine if a number is prime or not.  Finding and understanding about prime numbers is an integral part of number theory.  I’m going to go through some examples when we take the number 2 as our witness number.  We have a couple of tests that we conduct with 2 – and for all numbers less than 2047 if a number passes either test then we can guarantee that it is a prime number.

Miller-Rabin test using 2 as a witness number:

We choose an odd number, n >2. First we need to write it in the form:

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Then we have to conduct a maximum of 2 different tests:

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If either of the above are true then we have a prime number.

Testing whether n = 23 is prime.

First we need to write 23 in the following form:

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Next we need to check if the following is true:

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Remember that mod 23 simply means we look at the remainder when we divide by 23.  We can do this using Wolfram Alpha – but in this case let’s see how we could do this without a calculator:

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Therefore this passes the test – and we can say that it is prime.

Testing whether 1997 is prime

For 1997 we have:

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So we need to first test if the following is true:

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However using Wolfram Alpha we get:

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So this fails the first part of the test.

Trying the second part of the test, we need:

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We have already tested the case when r=0 (this gives the earlier result), so just need to look at what happens when r=1.  Again we use Wolfram Alpha to get:

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This passes the 2nd part of the test and so confirms that 1997 is prime.

What happens with 2047?

2047 is not prime as we can write it as 2 x 3 x 11 x 31.  However it is the first number for which the witness 2 gives a false positive (i.e we get a positive result even though it is not prime).  We write 2047 as:

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But we do indeed get:

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So we can call 2047 a pseudoprime – it passes this prime number test but is not actually prime.

Larger primes

For numbers larger than 2047 you can combine witnesses – for example if you use both 2 and 3 as your witness numbers (and regard a positive result as at least one of them returning a positive result) then this will find all primes for n < 1,373,653.

More interestingly for extremely large numbers you can use this test to provide a probability estimate for the likelihood that a number is prime.  Lots to explore here!

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Maths Games and Markov Chains

This post carries on from the previous one on Markov chains – be sure to read that first if this is a new topic.  The image above is of the Russian mathematician Andrey Markov [public domain picture from here] who was the first mathematician to work in this field (in the late 1800s).

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Above I have created a simple maths game – based on the simplified rules of Monopoly.  All players start at Go.  All players roll 2 dice and add the scores and move that number of squares forward (after square 9 you move to square 1 etc).  If you roll a double you get sent to Jail (square 9) – but are released on your next turn.  If you land on square 5 you immediately are transported to Go (and end your turn).  

The question is, if we play this game over the long run which famous mathematician will receive the most visits?  (In order we have Newton (2), Einstein (3), Euler (4), Gauss (6), Euclid (7), Riemann (8)).

Creating a Markov Chain

The first task is to work out the probabilities of landing on each square for someone who is starting on any square.  Using a sample space diagram you can work out the following:

p(move 2) = 0.  p(move 3) = 2/36.  p(move 4) = 2/36.  p(move 5) = 4/36.  p(move 6) = 4/36.  p(move 7) = 6/36.  p(move 8) = 4/36.  p(move 9) = 4/36.  p(move 10) = 2/36.  p(move 11) = 2/36. p(move 12) = 0.  p(double) = 6/36.

We can see that the only variation from the standard sample space is that we have separated the doubles – because they will move us to jail.  

Matrix representation

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I now need to create a 9×9 matrix which represents all the states of the game.  The first subscript denotes where a player starts and the second subscript denotes where a player finishes after 1 turn.  So for example m_13 denotes that a player will start on square 1 and finish on square 3, and p(m_13) is the probability that it will happen.  If we do some (rather tedious!) calculations we get the following matrix:

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We can note that the probability of ending up on square 5 is 0 because if you do land on it you are transported to Go.  Equally you can’t start on square 5 – so the probability of starting there and arriving somewhere else is also 0.  Also note that each row represents all possible states – so always adds up to 1.

Now all I need to do is raise this matrix to a large power.  If I raise this to the power 100 this will give me the long term probabilities of which square people will land on.  This will also give me a 9×9 matrix but I can focus on the first row which will tell me the long term probabilities of where people end up when starting at Go.  This gives:

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So I can see that the long term probabilities (to 3sf) show that Jail is going to be visited around 40% of the time, followed by Go (around 26.7% of the time).  And the mathematician that receives the most visits will be Euclid (around 16.8%).  We can logically see why this is true – if 40% of the time people are in Jail, then the next roll they are most likely to get a 7 which then takes them to this square.


Clearly we can then refine our game – we could in theory use this to model the real game of Monopoly (though this would be quite complicated!)  The benefit of Markov chains is that it is able to reduce complex rules and systems into a simple long term probability – which is hugely useful for making long term predictions.  

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New Paper 3s for Applications!

I’ve just finished making six Paper 3 practice papers for HL students sitting the Applications examination.   The Paper 3 pack is 41 pages and includes over 180 marks of questions and full typed up markscheme.  I’ve paid close attention to the IB’s provided examples for the course to make sure these look and feel very similar to how I would expect the papers to be in the summer. 

The six questions are:

Investigating Body Mass Index [30 marks]

The students carry out a statistical investigation to inform a decision on how to decrease BMI. The mathematics includes: regression lines, z tests, paired t-tests, 2 sample t-tests and Chi squared test for independence.

Life’s a Beach [32 marks]

The students carry out an investigation using Markov chains to investigate time spent on the beach. The mathematics includes: transition matrices, matrix operations, eigenvalues and eigenvectors and steady state calculations.

Avoiding a Magical barrier [33 marks]

The students explore a scenario where they must minimise their journey. The mathematics includes: forming equations, finding averages, differentiation using the chain rule, optimization, and graphing.

Who killed Mr. Potato? [32 marks]

Students explore a scenario where they must decide the time that a potato was first removed from an oven using Newton’s Law of Cooling. The mathematics includes: solving differential equations by separating variables, curve fitting, log regression, differentiation of exponentials, graphing to solve equations.

Hare vs. Lynx [32 marks]

Students explore a predator prey system to see what happens when population parameters are changed. The mathematics includes: understanding equilibrium and saddle points, finding eigenvalues, Euler’s method to approximate differential equations, sketching phase portraits, curve fitting a trigonometric model and graphical skills.

Rolling dice [32 marks]

Students conduct a dice rolling investigation which mimics radioactive decay. The mathematics includes: equations of linear regression, exponential regression, Chi squared goodness of fit, Euler’s method to approximate differential equations, solving differential equations by separating variables, percentage error.

You can purchase this Paper 3 exam pack below (you do not need a PayPal account, just click on the relevant credit card picture).

Paper 3 exam pack (Applications HL)

Six paper 3 investigations to help prepare for the summer exams. Please note this is not an automatic download but will be sent the same day.


Life on the Beach with Markov Chains

Markov chains are exceptionally useful tools for calculating probabilities – and are used in fields such as economics, biology, gambling, computing (such as Google’s search algorithm), marketing and many more.  They can be used when we have the probability of a future event dependent on a current event.  The image above is of the Russian mathematician Andrey Markov [public domain picture from here] who was the first mathematician to work in this field (in the late 1800s).

Beach life

The picture above is an example of a situation which can be modeled using a Markov chain.  We have two states:  Beach (state 1) and Home (state 2).  In our happy life we spend all the hours of the day in either one of these two states.  If we are on the beach then the probability we remain on the beach in one hour is 0.6 and the probability we go home in one hour is 0.4.  If we are at home the probability we remain at home in one hour is 0.8 and the probability we go to the beach is 0.2.  Hopefully you can see how this information is represented above.

Using matrices

First we need to represent our information in a matrix form.  A general 2×2 matrix is written as:

Where the subscript tells you the row and column (e.g. a_12 tells you it is in the first row and 2nd column).

For our Markov chain we define the following matrix:

Here m_11 is the situation of starting in state 1 and moving to state 1.  m_12 is the situation of starting in state 1 and moving to state 2.  Therefore p(m_12) is the probability of starting in state 1 and moving to state 2.  So for our beach  existence we have:

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The 0.6 shows that if we are already on the beach we have a 0.6 chance of still being on the beach in one hour.

Where will we be in the future?

The benefit of Markov chains is that they allow us to utilise computer power to now calculate where someone will be in the future – simply by taking the power of the matrix.  To find the probabilities after 2 hours I can square my matrix:

Using the rules of matrix multiplication this then gives:

Which we can the calculate as:

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I’ve used arrow notation here to represent the start and end state so p(m_1 arrow 1) means starting at 1 and ending at 1 after 2 hours.  We can see that for someone who started in the beach, the chance of them being on the beach in 2 hours is 0.44.  Equally the probability of someone who started in the house being on the beach in 2 hours is 0.28.

I can then carry on with matrix multiplication to work out where someone will likely be for any given number of hours in the future.  

p(m_1 arrow 1) means starting at 1 and ending at 1 after n hours.  So for example if I want to see where someone will be in 24 hours I simply do:

Screen Shot 2021-11-15 at 7.12.42 AM

We can see that now it doesn’t actually matter (to 3sf anyway) where someone started – if they started on the beach there is a 0.333 chance they are on the beach in 24 hours, if they started in the house there is also a 0.333 chance they are on the beach in 24 hours.  So I can conclude that as the number of hours increase towards infinity that the person in this scenario would spend 1/3 of their time on the beach and 2/3 of their time at home – not a bad life!

A more demanding beach life

We can see that things get much more complicated, even by adding an extra state.  Now we have 3 possible states, Beach (state 1), Home (state 2) and SCUBA (state 3).  This time we need a 3×3 matrix:

and as before we define our probability matrix as:

So from our diagram we have the following:

For example the 0.8 in row 2 column 2 shows that there is a 0.8 chance of starting in state 2 (Home) and ending in state 2 (Home) in one hour.

Then using our same notation we have:

Which shows that after 2 hours there is (for example) a 0.19 chance that someone who started in state 2 (Home) is now found in state 1 (Beach).

After 24 hours we have the following matrix:

So we notice the same situation as last time – as the number of hours increase it gets less important where we started from – we can see that to 3sf it doesn’t now matter where we started – the probability after 24 hours of being found on the beach is 0.407, the probability of being found at home is 0.333 and the probability of being found diving is 0.260.

Hopefully this is a quick example to demonstrate the power of Markov chains in working with probabilities.  There is a lot more to explore (maybe in another post!)

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All content on this site has been written by Andrew Chambers (MSc. Mathematics, IB Mathematics Examiner).

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