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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).
Creating a maths game
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
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:
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:
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.
Extensions
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.
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:
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:
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:
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!)
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
The Barnsley Fern: Mathematical Art
This pattern of a fern pictured above was generated by a simple iterative program designed by mathematician Michael Barnsely. I downloaded the Python code from the excellent Tutorialspoint and then modified it slightly to run on repl.it. What we are seeing is the result of 40,000 individual points – each plotted according to a simple algorithm. The algorithm is as follows:
Transformation 1: (0.85 probability of occurrence)
xi+1 = 0.85xi +0.04yi
yi+1= -0.04xi+0.85yi+1.6
Transformation 2: (0.07 probability of occurrence)
xi+1 = 0.2xi -0.26yi
yi+1= 0.23xi+0.22yi+1.6
Transformation 3: (0.07 probability of occurrence)
xi+1 = -0.15xi -0.28yi
yi+1= 0.26xi+0.24yi+0.44
Transformation 4: (0.01 probability of occurrence)
xi+1 = 0
yi+1= 0.16yi
So, I start with (0,0) and then use a random number generator to decide which transformation to use. I can run a generator from 1-100 and assign 1-85 for transformation 1, 86-92 to transformation 2, 93-99 for transformation 3 and 100 for transformation 4. Say I generate the number 36 – therefore I will apply transformation 1.
xi+1 = 0.85(0)+0.04(0)
yi+1= -0.04(0)+0.85(0)+1.6
and my new coordinate is (0,1.6). I mark this on my graph.
I then repeat this process – say this time I generate the number 90. This tells me to do transformation 2. So:
xi+1 = 0.2(0) -0.26(1.6)
yi+1= 0.23(0)+0.22(1.6)+1.6
and my new coordinate is (-0.416, 1.952). I mark this on my graph and carry on again. The graph above was generated with 40,000 iterations – let’s see how it develops over time:
1000 iterations:
10,000 iterations:
100,000 iterations:
500,000 iterations:
If we want to understand what is happening here we can think of each transformation as responsible for a different part of our fern. Transformation 1 is most likely and therefore this fills in the smaller leaflets. Transformations 2 and 3 fill in the bottom left and right leaflet (respectively) and transformation 4 fills in the stem.
It’s quite amazing to think that a simple computer program can create what looks like art – or indeed that is can replicate what we see in nature so well. This fern is an example of a self-similar pattern – i.e one which will look the same at different scales. You could zoom into a detailed picture and see the same patterns repeating. You might want to explore the idea of fractals in delving into this topic in more detail.
Changing the iterations
We can explore what happens when we change the iterations very slightly.
Christmas tree
Crazy spiral
Modern art
You can modify the code to run this here. Have a go!
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- 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.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- 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!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Martingale II and Currency Trading
We can use computer coding to explore game strategies and also to help understand the underlying probability distribution functions. Let’s start with a simple game where we toss a coin 4 times, stake 1 counter each toss and always call heads. This would give us a binomial distribution with 4 trials and the probability of success fixed as 1/2.
Tossing a coin 4 time [simple strategy]
For example the only way of losing 4 counters is a 4 coin streak of T,T,T,T. The probability of this happening is 1/16. We can see from this distribution that the most likely outcome is 0 (i.e no profit and no loss). If we work out the expected value, E(X) by multiplying profit/loss by frequencies and summing the result we get E(X) = 0. Therefore this is a fair game (we expect to neither make a profit nor a loss).
Tossing a coin 4 time [Martingale strategy]
This is a more complicated strategy which goes as follows:
1) You stake 1 counter on heads.
b) if you lose you stake 2 counters on heads
c) if you lose you stake 4 counters on heads
d) if you lose you stake 8 counters on heads.
If you win, the your next stake is always to go back to staking 1 counter.
For example for the sequence: H,H,T,T
First you bet 1 counter on heads. You win 1 counter
Next you bet 1 counter on heads. You win 1 counter
Next you bet 1 counters on heads. You lose 1 counter
Next you bet 2 counters on heads. You lose 2 counters
[overall loss is 1 counter]
For example for the sequence: T,T,T,H
First you bet 1 counter on heads. You lose 1 counter
Next you bet 2 counter on heads. You lose 2 counters
Next you bet 4 counters on heads. You lose 4 counter
Next you bet 8 counters on heads. You win 8 counters
[overall profit is 1 counter]
This leads to the following probabilities:
Once again we will have E(X) = 0, but a very different distribution to the simple 4 coin toss. We can see we have an 11/16 chance of making a profit after 4 coins – but the small chance of catastrophic loss (15 counters) means that the overall expectation is still zero.
Iterated Martingale:
Here we can do a computer simulation. This is the scenario this time:
We start with 100 counters, we toss a coin for a maximum of 3 times. We then define a completed round as when we get to a shaded box. We then repeat this process through 999 rounds, and model what happens. Here I used a Python program to simulate a player using this strategy.
We can see that we have periods of linear growth followed by steep falls – which is a very familiar pattern across many investment types. We can see that the initial starting 100 counters was built up to around 120 at the peak, but was closer to just 40 when we finished the simulation.
Let’s do another simulation to see what happens this time:
Here we can see that the 2nd player was actually performing significantly worse after around 600 rounds, but this time ended up with a finishing total of around 130 counters.
Changing the multiplier
We can also see what happens when rather than doubling stakes on losses we follow some other multiple. For example we might choose to multiply our stake by 5. This leads to much greater volatility as we can see below:
Multiplier x5
Here we have 2 very different outcomes for 2 players using the same model. Player 1 (in blue) may believe they have found a sure-fire method of making huge profits, but player 2 (green) went bankrupt after around 600 rounds.
Multiplier x1.11
Here we can see that if the multiplier is close to 1 we have much less volatility (as you would expect because your maximum losses per round are much smaller).
We can run the simulation across 5000 rounds – and here we can see that we have big winning and losing streaks, but that over the long run the account value oscillates around the starting value of 100 counters.
Forex charts
We can see similar graphs when we look at forex (currency exchange) charts. For example:
In this graph (from here) we plot the exchange between US dollar and Thai Baht. We can see the same sort of graph movements – with run of gains and losses leading to a similar jagged shape. This is not surprising as forex trades can also be thought of in terms of 2 binary outcomes like tossing a coin, and indeed huge amounts of forex trading is done through computer programs, some of which do use the Martingale system as a basis.
The effect of commission on the model
So, to finish off we can modify our system slightly so that we try to replicate forex trading. We will follow the same model as before, but this time we have to pay a very small commission for every trade we make. This now gives us:
E(X) = -0.000175. (0.0001 counters commission per trade)
E(X) = -0.00035. (0.0002 counters commission per trade)
Even though E(X) is very slightly negative, it means that in the long run we would expect to lose money. With the 0.0002 counters commission we would expect to lose around 20 counters over 50,000 rounds. The simulation graph above was run with 0.0002 counters commission – and in this case it led to bankruptcy before 3000 rounds.
Computer code
The Python code above can be used to generate data which can then be copied into Desmos. The above code simulates 1 player playing 999 rounds, starting with 100 counters, with a multiplier of 5. If you know a little bit about coding you can try and play with this yourselves!
I’ve also just added a version of this code onto repl. You can run this code – and also generate the graph direct (click on the graph png after running). It creates some beautiful images like that shown above.
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- 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.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- 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!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
The Martingale system
The Martingale system was first used in France in 1700s gambling halls and remains used today in some trading strategies. I’ll look at some of the mathematical ideas behind this and why it has remained popular over several centuries despite having a long term expected return of zero.
The scenario
You go to a fair ground and play a simple heads-or-tails game. The probability of heads is 1/2 and tails is also 1/2. You place a stake of counters on heads. If you guess correctly you win that number of counters. If you lose, you double your stake of counters and then the coin is tossed again. Every time you lose you double up your stake of counters and stop when you finally win.
Infinitely deep pockets model:
You can see that in the example above we always have a 0.5 chance of getting heads on the first go, which gives a profit of 1 counter. But we also have a 0.5 chance of a profit of 1 counter as long as we keep doubling up our stake, and as long as we do indeed eventually throw heads. In the example here you can see that the string of losing throws don’t matter [when we win is arbitrary, we could win on the 2nd, 3rd, 4th etc throw]. By doubling up, when you do finally win you wipe out your cumulative losses and end up with a 1 counter profit.
This leads to something of a paradoxical situation, despite only having a 1/2 chance of guessing heads we end up with an expected value of 1 counter profit for every 1 counter that we initially stake in this system.
So what’s happening? This will always work but it requires that you have access to infinitely deep pockets (to keep your infinite number of counters) and also the assumption that if you keep throwing long enough you will indeed finally get a head (i.e you don’t throw an infinite number of tails!)
Finite pockets model:
Real life intrudes on the infinite pockets model – because in reality there will be a limit to how many counters you have which means you will need to bail out after a given number of tosses. Even if the probability of this string of tails is very small, the losses if it does occur will be catastrophic – and so the expected value for this system is still 0.
Finite pockets model capped at 4 tosses:
In the example above we only have a 1/16 chance of losing – but when we do we lose 15 counters. This gives an expected value of:
Finite pockets model capped at n tosses:
If we start with a 1 counter stake then we can represent the pattern we can see above for E(X) as follows:
Here we use the fact that the losses from n throws are the sum of the first (n-1) powers of 2. We can then notice that both of these are geometric series, and use the relevant formula to give:
Therefore the expected value for the finite pockets model is indeed always still 0.
So why does this system remain popular?
So, given that the real world version of this has an expected value of 0, why has it retained popularity over the past few centuries? Well, the system will on average return constant linear growth – up until a catastrophic loss. Let’s say you have 100,000 counters and stake 1 counter initially. You can afford a total of 16 consecutive losses. The probability of this is only:
but when you do lose, you’ll lose a total of:
So, the system creates a model that mimics linear growth, but really the small risk of catastrophic loss means that the system still has E(X) = 0. In the short term you would expect to see the following very simple linear relationship for profit:
With 100,000 counters and a base trading stake of 1 counter, if you made 1000 initial 1 counter trades a day you would expect a return of 1000 counters a day (i.e 1% return on your total counters per day). However the longer you continue this strategy the more likely you are to see a run of 16 tails – and see all your counters wiped out.
Computer model
I wrote a short Python code to give an idea as to what is happening. Here I started 9 people off with 1000 counters each. They have a loss limit of 10 consecutive losses. They made starting stakes of 1 counter each time, and then I recorded how long before they made a loss of 10 tosses in a row.
For anyone interested in the code here it is:
The program returned the following results. The first number is the number of starting trades until they tossed 10 tails in a row. The second number was their new account value (given that they had started with 1000 counters, every previous trade had increased their account by 1 counter and that they had then just lost 1023 counters).
1338, 1315
1159, 1136
243, 220
1676, 1653
432, 409
1023, 1000
976, 953
990, 967
60, 37
This was then plotted on Desmos. The red line is the trajectory their accounts were following before their loss. The horizontal dotted line is at y = 1000 which represents the initial account value. As you can see 6 people are now on or below their initial starting account value. You can also see that all these new account values are themselves on a line parallel to the red line but translated vertically down.
From this very simple simulation, we can see that on average a person was left with 884 counters following hitting 10 tails. i.e below initial starting account. Running this again with 99 players gave an average of 869.
999 players
I ran this again with 999 players – counting what their account value would be after their first loss. All players started with 1000 counters. The results were:
31 players bankrupt: 3%
385 players left with less than half their account value (less than 500): 39%
600 players with less than their original account value (less than 1000): 60%
51 players at least tripled their account (more than 3000): 5%
The top player ended up with 6903 counters after their first loss.
The average account this time was above starting value (1044.68). You can see clearly that the median is below 1000 – but that a small number of very lucky players at the top end skewed the mean above 1000.
Second iteration
I then ran the simulation again – with players continuing with their current stake. This would have been slightly off because my model allowed players who were bankrupt from the first round to carry on [in effect being loaned 1 counter to start again]. Nevertheless it now gave:
264 players bankrupt: 26%
453 players left with less than half their account value (less than 500): 45%
573 players with less than their original account value (less than 1000): 57%
95 players at least tripled their account (more than 3000): 10%
The top player ended up with 9583 counters after their second loss.
We can see a dramatic rise in bankruptcies – now over a quarter of all players. This would suggest the long term trend is towards a majority of players being bankrupted, though the lucky few at the top end may be able to escape this fate.
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- 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.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- 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!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
IB Maths Resources from Intermathematics 
