You are currently browsing the category archive for the ‘IB Maths’ category.
IB Maths and GCSE Maths Resources from British International School Phuket. Theory of Knowledge (ToK). Maths explorations and investigations. Real life maths. Maths careers. Maths videos. Maths puzzles and Maths lesson resources.
British International School Phuket
Welcome to the British International School Phuket’s maths website. I am currently working at BISP and so I am running my site as the school’s maths resources website for both our students and students around the world.
We are a British international school located on the tropical island of Phuket in Southern Thailand. We offer a number of scholarships each year, catering for a number of national and international standard sports stars as well as for academic excellence. You can find out more about our school here.
Thailand Maths Challenge
British International School, Phuket (BISP) and Rangsit University’s College of Information and Communication Technology are proud to announce the winner of the 2015-16 Thailand Maths Challenge. Yu Qing Wu from Bangkok Patana school turned in a incredibly impressive performance to claim first place. Her achievement is all the more impressive given that she is still only in Year 11 at school and was competing with students up to two years older than herself.
The finalists round was held at Rangsit University’s Sky Lounge on February 19th. Students had to complete 3 rounds of problem solving questions with a focus on a branch of mathematics called Number Theory – which requires rigorous proof and mathematical logic. Assitant Professor Wongsakorn Charoenpanitseri who was one of the coordinators for the World Mathematics Olympiad held last year in Thailand helped with the judging process.
Second place was also claimed by a Bangkok Patana student, Benjada Karprasertsri, whilst third place went to Santkorn Gorsagun from Mahidol University Demonstration School. There were also good results from Anglo Singapore and Trinity International School. The top three students were all offered full and partial scholarships to study at Rangsit University at the College of ICT.
Over 200 of Thailand’s top schools were invited to participate: international schools, bilingual schools, Thai private schools, and state schools in order to find the best young mathematicians in the country. Well done to all who took part!
One of the main benefits of flipping the classroom is allowing IB maths students to self-teach IB content. There are currently a good number of videos on youtube which allow students to self teach syllabus content, but no real opportunity to watch videos going through IB Higher Level past paper questions. So, I’ve started to put some of these together:
Playlist, Worked Exam Solutions:
The videos above are all around 10 minutes long and consist of talking through the solutions to 2-3 IB HL maths questions. The best way to use these videos is to pause the video at the start of the question, attempt it, then watch the video to check the answer and make notes on the method. Click on the top left hand corner to change the video being shown in the playlist.
The playlists below combine these worked solutions with the syllabus content videos, all grouped into the relevant syllabus strands:
Playlist 1, Algebra 1:
Sequences, Binomial, Logs, Induction, Permutations, Gaussian elimination:
Playlist 2, Complex numbers:
Converting from Cartesian to Polar, De Moivre’s Theorem, Roots of Unity:
Playlist 3: Functions:
Sketching graphs, Finding Inverses, Factor and Remainder Theorem, Sketching 1/f(x), sketching absolute f(x), translating f(x):
The Telephone Numbers – Graph Theory
The telephone numbers are the following sequence:
1, 1, 2, 4, 10, 26, 76, 232, 764, 2620, 9496…
(where we start from n=0).
This pattern describes the total number of ways which a telephone exchange with n telephones can place a connection between pairs of people.
To illustrate this idea, the graph below is for n=4. This is when we have 10 telephones:
Each red line represents a connection. So the first diagram is for when we have no connections (this is counted in our sequence). The next five diagrams all show a single connection between a pair of phones. The last three diagrams show how we could have 2 pairs of telephones connected at the same time. Therefore the 4th telephone number is 10. These numbers get very large, very quickly.
Finding a recursive formula
The formula is given by the recursive relationship:
T(n) = T(n-1) + (n-1)T(n-2)
This means that to find (say) the 5th telephone number we do the following:
T(5) = T(5-1) + (5-1)T(5-2)
T(5) = T(4) + (4)T(3)
T(5) = 10 + (4)4
T(5) = 26
This is a quick way to work out the next term, as long as we have already calculated the previous terms.
Finding an nth term formula
The telephone numbers can be calculated using the nth term formula:
This is going to be pretty hard to derive! I suppose the first step would start by working out the total number of connections possible between n phones – and this will be the the same as the graphs below:
These clearly follow the same pattern as the triangular numbers which is 0.5(n² +n) when we start with n = 1. We can also think of this as n choose 2 – because this gives us all the ways of linking 2 telephones from n possibilities. Therefore n choose 2 also generates the triangular numbers.
But then you would have to work out all the permutations which were allowed – not easy!
Anyway, as an example of how to use the formula to calculate the telephone numbers, say we wanted to find the 5th number:
We have n = 5. The summation will be from k = 0 and k = 2 (as 5/2 is not an integer).
Therefore T(5) = 5!/(20(5-0)!0!) + 5!/(21(5-2)!1!) + 5!/(22(5-4)!2!)
T(5) = 1 + 10 + 15 = 26.
Finding telephone numbers through calculus
Interestingly we can also find the telephone numbers by using the function:
y = e0.5x2+x
and the nth telephone number (starting from n = 1) is given by the nth derivative when x = 0.
So when x = 0, the third derivative is 4. Therefore the 3rd telephone number is 4.
The fifth derivative of the function is:
So, when x =0 the fifth derivative is 26. Therefore the 5th telephone number is 26.
If you liked this post you might also like:
Fermat’s Theorem on the Sum of two Squares – A lesser known theorem from Fermat – but an excellent introduction to the idea of proof.
Unbelievable: 1+2+3+4…. = -1/12 ? A result that at first glance looks ridiculous – and yet can be shown to be correct. How?
Circular inversions II
There are some other interesting properties of circular inversions. One of which is that they preserve the “angle” between intersecting circles. Firstly, how can circles have an angle between them? Well, we draw 2 tangents to both the circles at the point of intersection, and then measure the angle between the 2 tangents:
Therefore we can see that the “angle” between these 2 circles is 59.85 degrees. If we then carry out a circular inversion we see the following:
The inversion has been done with regards to the black circle centred around the origin. The red and blue circles are mapped from outside the the black circle onto circles inside the black circle. Now if we do the same as before – by finding the 2 tangents at the point of intersection, we find that the angle has remained the same – it is still 59.85 degrees.
It is also possible to find circles which remain unchanged under the inversion. This happens when a circle is orthogonal (at a 90 degree angle) to the circle with which the inversion is being carried out.
The small circle has an angle of 90 degrees with the large circle, and therefore when we invert with respect to the large circle, we map the small circle onto itself.
The question is, why is all this useful? Well, an entire branch of mathematics (non-Euclidean geometry) is concerned with being able to map points in our traditional Euclidean worldview (the geometry of high school triangles, parallel lines and circle theorems) to different geometrical systems entirely. Circular inversion is a good introduction to this concept.
Also, circular inversion can sometimes make studying mathematical shapes easier to understand and explain. For example, (from Wolfram):
It would be very difficult to explain mathematically how the shape above is generated – whilst there are patterns, it is not obvious how to explain them. However, if we invert this shape through a circular inversion (with the circle at centre of the image) then we get the following:
This is the image inside the circle – and now we can clearly see the pattern behind the generated image. So, inversion has a lot of potential for simplifying geometrical problems.
Modelling Infectious Diseases
Using mathematics to model the spread of diseases is an incredibly important part of preparing for potential new outbreaks. As well as providing information to health workers about the levels of vaccination needed to protect a population, it also helps govern first response actions when new diseases potentially emerge on a large scale (for example, Bird flu, SARS and Ebola have all merited much study over the past few years).
The basic model is based on the SIR model – this is represented by the picture above (from Plus Maths which has an excellent and more detailed introduction to this topic). The SIR model looks at how much of the population is susceptible to infection, how many of these go on to become infectious, and how many of these go on to recover (and in what timeframe).
Another important parameter is R0, this is defined as how many people an infectious person will pass on their infection to in a totally susceptible population. Some of the R0values for different diseases are shown above. This shows how an airbourne infection like measles is very infectious – and how malaria is exceptionally hard to eradicate because infected people act almost like a viral storage bank for mosquitoes.
One simple bit of maths can predict what proportion of the population needs to be vaccinated to prevent the spread of viruses. The formula is:
VT = 1 – 1/R0
Where VT is the proportion of the population who require vaccinations. In the case of something like the HIV virus (with an R0 value of between 2 and 5), you would only need to vaccinate a maximum of 80% of the population. Measles however requires around 95% vaccinations. This method of protecting the population is called herd immunity
This graphic above shows how herd immunity works. In the first scenario no members of the population are immunised, and that leads to nearly all the population becoming ill – but in the third scenario, enough members of the population are immunised to act as buffers against the spread of the infection to non-immunised people.
The equations above represent the simplest SIR (susceptible, infectious, recovered) model – though it is still somewhat complicated!
dS/dt represents the rate of change of those who are susceptible to the illness with respect to time. dI/dt represents the rate of change of those who are infected with respect to time. dR/dt represents the rate of change of those who have recovered with respect to time.
For example, if dI/dt is high then the number of people becoming infected is rapidly increasing. When dI/dt is zero then there is no change in the numbers of people becoming infected (number of infections remain steady). When dI/dt is negative then the numbers of people becoming infected is decreasing.
The constants β and ν are chosen depending on the type of disease being modelled. β represents the contact rate – which is how likely someone will get the disease when in contact with someone who is ill. ν is the recovery rate which is how quickly people recover (and become immune.
ν can be calculated by the formula:
D = 1/ν
where D is the duration of infection.
β can then be calculated if we know R0 by the formula:
R0 = β/ν
So, for example, with measles we have an average infection of about a week, (so if we want to work in days, 7 = 1/ν and so ν = 1/7). If we then take R0 = 15 then:
R0 = β/ν
15 = β/0.14
β = 2.14
Therefore our 3 equations for rates of change become:
dS/dt = -2.14 I S
dI/dt = 2.14 I S – 0.14 I
dR/dt = 0.14 I
Unfortunately these equations are very difficult to solve – but luckily we can use a computer program to plot what happens. We need to assign starting values for S, I and R – the numbers of people susceptible, infectious, recovered (immune) from measles. Let’s say we have a total population of 11 people – 10 who are susceptible, 1 who is infected and 0 who are immune. This gives the following outcome:
This shows that the infection spreads incredibly rapidly – by day 2, 8 people are infected. By day 10 most people are immune but the illness is still in the population, and by day 30 the entire population is immune and the infection has died out.
An illustration of just how rapidly measles can spread is provided by the graphic above. This time we start with a population of 1000 people and only 1 infected individual – but even now, within 5 days over 75% of the population are infected.
This last graph shows the power of herd immunity. This time there are 100 susceptible people, but 900 people are recovered (immune), and there is again one infectious person. This time the infection never takes off in the community – those who are already immune act as a buffer against infection.
If you enjoyed this post you might also like:
Differential Equations in Real Life – some other uses of differential equations in modelling predator-prey relationships between animal populations.
Circular Inversion – Reflecting in a Circle
This topic is a great introduction to the idea of mapping – where one point is mapped to another. This is a really useful geometrical tool as it allows complex shapes to be transformed into isomorphic (equivalent) shapes which can sometimes be easier to understand and work with mathematically.
One example of a mapping is a circular inversion. The inversion rule maps a point P onto a point P’ according to the rule:
OP x OP’ = r2
To understand this, we start with a circle radius r centred on O. The inversion therefore means that the distance from O to P multiplied by the distance from O to P’ will always give the constant value r2
This is an example of the circular inversion of the point A to the point A’.
We have the distance of OA as √2 and the radius of the circle as 2. Therefore using the formula we can find OA’ by:
OA’ = r2 / OA = 4/√2
OA’ = 21.5. This means that the point A’ is a distance of 21.5 away from O on the same line as OA.
We can check that the Geogebra plot is correct – because this point A’ is plotted at (2,2) – which is indeed (using Pythagoras) a distance of 21.5 from O.
A point near to the edge of the circle will have an inversion also close to the circle
A point near to the centre will have an inversion a long way from the circle. The point (0,0) will be undefined as no point outside the circle will satisfy the inversion equation.
So, that is the basic idea behind circular inversion – though it gets a lot more interesting when we start inverting shapes rather than just points.
Circles through the origin map onto straight lines to infinity (see above).
Circles centred on the origin map to other circles centred on the origin (above).
Ellipses create these shapes (above).
The straight line through A B maps to a circle through the origin (above).
The solid triangle ABC maps to the pink region (above).
The solid square ABCD maps to the pink region (above).
These shapes can all be explored using the reflect object in circle button on Geogebra.
It is possible to extend the formula to 3 dimensions to give spherical inversion:
The above image is a 3D human head inverted in a sphere (from the Space Symmetry Structures website. There’s lots to explore on this topic – it’s a good example of how art can be mathematically generated, as well as introducing isomorphic structures.
If you enjoyed this post you might also like:
The Riemann Sphere : Another form of mapping using spheres is the Riemann Sphere – which is a way of mapping the entire complex plane to the surface of a sphere.
Fractals, Mandelbrot and the Koch Snowflake. Using maths to model infinite patterns.
Graphically Understanding Complex Roots
If you have studied complex numbers then you’ll be familiar with the idea that many polynomials have complex roots. For example x2 + 1 = 0 has the solution x = i and -i. We know that the solution to x2 – 1 = 0 ( x = 1 and -1) gives the two x values at which the graph crosses the x axis, but what does a solution of x = i or -i represent graphically?
There’s a great post on Maths Fun Facts which looks at basic idea behind this and I’ll expand on this in a little more detail. This particular graphical method only works with quadratics:
You have a quadratic graph with complex roots, say y = (x – 1)2 + 4. Written in this form we can see the minimum point of the graph is at (1,4) so it doesn’t cross the x axis.
Reflect this graph downwards at the point of its vertex. We do this by transforming y = (x – 1)2 + 4 into y = -(x – 1)2 + 4
We find the roots of this new equation using the quadratic formula or by rearranging – leaving the plus or minus sign in.
-(x – 1)2 + 4 = 0
(x-1) = ± 2
x = 1 ± 2
Plot a circle with centre (1,0) and radius of 2. This will touch both roots.
We can now represent the complex roots of the initial equation by rotating the 2 real roots we’ve just found 90 degrees anti-clockwise, with the centre of rotation the centre of the circle.
The points B and C on the diagram are a representation of the complex roots (if we view the graph as representing the complex plane). The complex roots of the initial equation are therefore given by x = 1 ± 2i.
It’s relatively straightforward to show algebraically what is happening:
If we take the 2 general equations:
1) y = (x-a)2 + b
2) y = -(x-a)2 + b (this is the reflection at the vertex of equation 1 )
(b > 0 ). Then the first equation will always have complex roots. The roots of both equations will be given by:
1) a ±i√b
2) a ±√b
So we can think of (2) as representing a circle of radius √b, centred at (a,0). Therefore multiplying √b by i has the effect of rotating the point (√b, 0 ) 90 degrees anti-clockwise around the point (a,0). Therefore the complex roots will be graphically represented by those points at the top and bottom of this circle. (a, √b) and (a, -√b)
Graphically finding complex roots of a cubic
There is also a way of graphically calculating the complex roots of a cubic with 1 real and 2 complex roots. This method is outlined with an algebraic explanation here
We plot a cubic with 1 real and 2 complex roots, in this case y = x3 – 9x2 + 25x – 17.
We find the line which goes through the real root (1,0) and which is also a tangent to the function.
If the x co-ordinate of the tangent intersection with the cubic is a and the gradient of the tangent is m, then the complex roots are a ± mi. In this case the tangent x intersection is at 4 and the gradient of the tangent is 1, therefore the complex roots are 4 ± 1i.
Maths Studies IA Exploration Topics:
This is the British International School Phuket’s IB maths exploration page. This list is primarily for Maths Studies students – though may also be of use to SL and HL students interested in statistics and probability. If you are doing a Maths SL, HL exploration then go to this page instead.
Make sure you read the Maths Studies guidance from the IB prior to starting your IA maths exploration – this linked site gives the full list of assessment criteria you will be judged against and also gives 9 full examples of investigations students have done.
Given the assessment criteria it’s probably easiest to conduct a data analysis investigation, though you can choose to explore other parts of the syllabus instead. To get good marks make sure you carefully follow the marking criteria points given by the IB and try and personalise your investigation as much as possible. Be innovative, choose something you are interested in and enjoy it!
Primary or Secondary data?
The main benefit of primary data is that you can really personalise your investigation. It allows you scope to investigate something that perhaps no-one else has ever done. It also allows you the ability to generate data that you might not be able to find online. The main drawback is that collecting good quality data in sufficient quantity to analyze can be time consuming. You should aim for an absolute minimum of 50 pieces of data – and ideally 60-100 to give yourself a good amount of data to look at.
The benefits of secondary data are that you can gain access to good quality raw data on topics that you wouldn’t be able to collect data on personally – and it’s also much quicker to get the data. Potential drawbacks are not being able to find the raw data that fits what you want to investigate – or sometimes having too much data to wade through.
Secondary data sources:
1) The Census at School website is a fantastic source of secondary data to use. If you go to the random data generator you can download up to 200 questionnaire results from school children around the world on a number of topics (each year’s questionnaire has up to 20 different questions). Simply fill in your email address and the name of your school and then follow the instructions.
2) If you’re interested in sports statistics then the Olympic Database is a great resource. It contains an enormous amount of data on winning times and distances in all events in all Olympics. Follow links at the top of the page to similar databases on basketball, golf, baseball and American football.
3) If you prefer football, the the Guardian stats centre has information on all European leagues – you can see when a particular team scores most of their goals, how many goals they score a game, how many red cards they average etc. You can also find a lot of football stats on the Who Scored website. This gives you data on things like individual players’ shots per game, pass completion rate etc.
4) The Guardian Datablog has over 800 data files to view or download – everything from the Premier League football accounts of clubs to a list of every Dr Who villain, US gun crime, UK unemployment figures, UK GCSE results by gender, average pocket money and most popular baby names. You will need to sign into Google to download the files.
5) The World Bank has a huge data bank – which you can search by country or by specific topic. You can compare life-expectancy rates, GDP, access to secondary education, spending on military, social inequality, how many cars per 1000 people and much much more.
6) Gapminder is another great resource for comparing development indicators – you can plot 2 variables on a graph (for example urbanisation against unemployment, or murder rates against urbanisation) and then run them over a number of years. You can also download Excel speadsheets of the associated data.
7) Wolfram Alpha is one of the most powerful maths and statistics tools available – it has a staggering amount of information that you can use. If you go to the examples link above, then you can choose from data on everything from astronomy, the human body, geography, food nutrition, sports, socioeconomics, education and shopping.
Example Maths Studies IA Investigations:
Some of these ideas taken from the excellent Oxford IB Maths Studies textbook.
1) Is there a correlation between hours of sleep and exam grades?
Studies have shown that a good night’s sleep raises academic attainment.
2) Is there a correlation between height and weight?
The NHS use a chart to decide what someone should weigh depending on their height. Does this mean that height is a good indicator of weight?
3) Is there a correlation between arm span and foot height?
This is also a potential opportunity to discuss the Golden Ratio in nature.
4) Is there a correlation between the digit ratio and maths ability?
Studies show there is a correlation between digit ratio and everything from academic ability, aggression and even sexuality.
5) Is there a correlation between smoking and lung capacity?
6) Is there a correlation between GDP and life expectancy?
Run the Gapminder graph to show the changing relationship between GDP and life expectancy over the past few decades.
7) Is there a correlation between numbers of yellow cards a game and league position?
Use the Guardian Stats data to find out if teams which commit the most fouls also do the best in the league.
8) Is there a correlation between Olympic 100m sprint times and Olympic 15000m times?
Use the Olympic database to find out if the 1500m times have go faster in the same way the 100m times have got quicker over the past few decades.
9) Is there a correlation between sacking a football manager and improved results?
A recent study suggests that sacking a manager has no benefit and the perceived improvement in results is just regression to the mean.
10) Is there a correlation between time taken getting to school and the distance a student lives from school?
11) Does eating breakfast affect your grades?
12) Is there a correlation between stock prices of different companies?
Use Google Finance to collect data on company share prices.
13) Does teenage drinking affect grades?
A recent study suggests that higher alcohol consumption amongst teenagers leads to greater social stress and poorer grades.
14) Is there a correlation between unemployment rates and crime?
If there are less work opportunities, do more people turn to crime?
15) Is there a correlation between female participation in politics and wider access to further education?
16) Is there a correlation between blood alcohol laws and traffic accidents?
17) Is there a correlation between height and basketball ability?
18) Is there a correlation between stress and blood pressure?
19) Is there a correlation between Premier League wages and league positions?
1) Are a sample of student heights normally distributed?
We know that adult population heights are normally distributed – what about student heights?
2) Are a sample of flower heights normally distributed?
3) Are a sample of student weights normally distributed?
4) Are a sample of student reaction times normally distributed?
Conduct this BBC reaction time test to find out.
5) Are a sample of student digit ratios normally distributed?
6) Are the IB maths test scores normally distributed?
IB test scores are designed to fit a bell curve. Investigate how the scores from different IB subjects compare.
7) Are the weights of “1kg” bags of sugar normally distributed?
Other statistical investigations
1) Does gender affect hours playing sport?
A UK study showed that primary school girls play much less sport than boys.
2) Investigation into the distribution of word lengths in different languages.
The English language has an average word length of 5.1 words. How does that compare with other languages?
3) Do bilingual students have a greater memory recall than non-bilingual students?
Studies have shown that bilingual students have better “working memory” – does this include memory recall?
4) Investigation about the distribution of sweets in packets of Smarties. A chance to buy lots of sweets! Also you could link this with some optimisation investigation.
5) 22) Using Chi Squared to crack codes – Chi squared can be used to crack Vigenere codes which for hundreds of years were thought to be unbreakable. Unleash your inner spy!
6) Which times tables do students find most difficult to learn? – Are some calculations like 7×8 harder than others? Why?
Modelling using calculus
1) How can you optimise the area of a farmer’s field for a given length of fence?
A chance to use some real life maths to find out the fence sides that maximise area.
2) Optimisation in product packaging.
Product design needs optimisation techniques to find out the best packaging dimensions.
Probability and statistics
1) The probability behind poker games
2) Finding expected values for games of chance in a casino.
3) Birthday paradox:
The birthday paradox shows how intuitive ideas on probability can often be wrong. How many people need to be in a room for it to be at least 50% likely that two people will share the same birthday? Find out!
4) Which times tables do students find most difficult?
A good example of how to conduct a statistical investigation in mathematics.
5) Handshake problem
With n people in a room, how many handshakes are required so that everyone shakes hands with everyone else?
If you want to do an investigation with a bit more mathematical content then have a look at this page for over 100 ideas for Maths SL and HL students.
This post is based on the fantastic PlusMaths article on bluffing– which is a great introduction to this topic. If you’re interested then it’s well worth a read. This topic shows the power of mathematics in solving real world problems – and combines a wide variety of ideas and methods – probability, Game Theory, calculus, psychology and graphical analysis.
You would probably expect that there is no underlying mathematical strategy for good bluffing in poker – indeed that a good bluffing strategy would be completely random so that other players are unable to spot when a bluff occurs. However it turns out that this is not the case.
As explained by John Billingham in the PlusMaths article, when considering this topic it helps to really simplify things first. So rather than a full poker game we instead consider a game with only 2 players and only 3 cards in the deck (1 Ace, 1 King, 1 Queen).
The game then plays as follows:
1) Both players pay an initial £1 into the pot.
2) The cards are dealt – with each player receiving 1 card.
3) Player 1 looks at his card and can:
(b) bet an additional £1
4) Player 2 then can respond:
a) If Player 1 has checked, Player 2 must also check. This means both cards are turned over and the highest card wins.
b) If Player 1 has bet £1 then Player 2 can either match (call) that £1 bet or fold. If the bets are matched then the cards are turned over and the highest card wins.
So, given this game what should the optimal strategy be for Player 1? An Ace will always win a showdown, and a Queen always lose – but if you have a Queen and bet, then your opponent who may only have a King might decide to fold thinking you actually have an Ace.
In fact the optimal strategy makes use of Game Theory – which can mathematically work out exactly how often you should bluff:
This tree diagram represents all the possible outcomes of the game. The first branch at the top represents the 3 possible cards that Player 2 can be dealt (A,K,Q) each of which have a probability of 1/3. The second branch represents the remaining 2 possible cards that Player 1 has – each with probability 1/2. The numbers at the bottom of the branches represent the potential gain or loss from betting strategies for Player 2 – this is calculated by comparing the profit/loss relative to if both players had simply shown their cards at the beginning of the game.
For example, Player 2 has no way of winning any money with a Queen – and this is represented by the left branch £0, £0. Player 2 will always win with an Ace. If Player 1 has a Queen and bluffs then Player 2 will call the bet and so will have gained an additional £1 of his opponents money relative to a an initial game showdown (represented by the red branch). Player 1 will always check with a King (as were he to bet then Player 2 would always call with an Ace and fold with a Queen) and so the AK branch also has a £0 outcome relative to an initial showdown.
So, the only decisions the game boils down to are:
1) Should Player 1 bluff with a Queen? (Represented with a probability of b on the tree diagram )
2) Should Player 2 call with a King? (Represented with a probability of c on the tree diagram ).
Now it’s simply a case of adding the separate branches of the tree diagram to find the expected value for Player 2.
The right hand branch (for AQ and AK) for example gives:
1/3 . 1/2 . b . 1
1/3 . 1/2 . (1-b) . 0
1/3 . 1/2 . 0
So, working out all branches gives:
Expected Value for Player 2 = 0.5b(c-1/3) – c/6
Expected Value for Player 1 = -0.5b(c-1/3) + c/6
(Player 1’s Expected Value is simply the negative of Player 2’s. This is because if Player 2 wins £1 then Player 1 must have lost £1). The question is what value of b (Player 1 bluff) should be chosen by Player 1 to maximise his earnings? Equally, what is the value of c (Player 2 call) that maximises Player 2’s earnings?
It is possible to analyse these equations numerically to find the optimal values (this method is explained in the article), but it’s more mathematically interesting to investigate both the graphical and calculus methods.
Graphically we can solve this problem by creating 2 equations in 3D:
z = 0.5xy-x/6 – y/6
z = -0.5xy+x/6 + y/6
In both graphs we have a “saddle” shape – with the saddle point at x = 1/3 and y = 1/3. This can be calculated using Wolfram Alpha. At the saddle point we have what is known in Game Theory as a Nash equilibrium – it represents the best possible strategy for both players. Deviation away from this stationary point by one player allows the other player to increase their Expected Value.
Therefore the optimal strategy for Player 2 is calling with precisely c = 1/3 as this minimises his loses to -c/6 = -£1/18 per hand. The same logic looking at the Expected Value for Player 1 also gives b = 1/3 as an optimal strategy. Player 1 therefore has an expected value of +£1/18 per hand.
We can arrive at the same conclusion using calculus – and partial derivatives.
z = 0.5xy-x/6 – y/6
For this equation we find the partial derivative with respect to x (which simply means differentiating with respect to x and treating y as a constant):
zx = 0.5x – 1/6
and also the partial derivative with respect to y (differentiate with respect to y and treat x as a constant):
zy = 0.5y -1/6
We then set both of these equations to 0 and solve to find any stationary points.
0 = 0.5x – 1/6
0 = = 0.5y -1/6
x = 1/3 y = 1/3
We can then see that this is a saddle point by using the formula:
D = zxx . zyy – (zxy)2
(where zxx means the partial 2nd derivative with respect to x and zxy means the partial derivative with respect to x followed by the partial derivative with respect to y. When D < 0 then we have a saddle point).
This gives us:
D = 0.0 – (0.5)2 = -0.25
As D < 0 then we have a saddle point – and the optimal strategy for both players is c= 1/3 and b = 1/3.
We can change the rules of the game to see how this affects the strategy. For example, if the rules remain the same except that players now must place a £1.50 bet (with the initial £1 entry still intact) then we get the following equation:
Player 2 Expected Value = b/12(-1+7c) – 3c/12
This has a saddle point at b = 3/7, c = 1/7. So the optimal strategy is 3/7 bluffing and 1/7 calling. If Player 2 calls more than 3/7 then Player 1 can never bluff (b = 0), leaving Player 2 with a negative Expected Value. If Player 2 calls less than 3/7 then Player 1 can always bluff (b = 1).
If you enjoyed this you might also like:
The Gambler’s Fallacy and Casino Maths – using maths to better understand casino games
Game Theory and Tic Tac Toe – using game theory to understand games such as noughts and crosses
The Riemann Sphere
The Riemann Sphere is a fantastic glimpse of where geometry can take you when you escape from the constraints of Euclidean Geometry – the geometry of circles and lines taught at school. Riemann, the German 19th Century mathematician, devised a way of representing every point on a plane as a point on a sphere. He did this by first centering a sphere on the origin – as shown in the diagram above. Next he took a point on the complex plane (z = x + iy ) and joined up this point to the North pole of the sphere (marked W). This created a straight line which intersected the sphere at a single point at the surface of the sphere (say at z’). Therefore every point on the complex plane (z) can be represented as a unique point on the sphere (z’) – in mathematical language, there is a one-to-one mapping between the two. The only point on the sphere which does not equate to a point on the complex plane is that of the North pole itself (W). This is because no line touching W and another point on the sphere surface can ever reach the complex plane. Therefore Riemann assigned the value of infinity to the North pole, and therefore the the sphere is a 1-1 mapping of all the points in the complex plane (and infinity).
So what does this new way of representing the two dimensional (complex) plane actually allow us to see? Well, it turns on its head our conventional notions about “straight” lines. A straight line on the complex plane is projected to a circle going through North on the Riemann sphere (as illustrated above). Because North itself represents the point at infinity, this allows a line of infinite length to be represented on the sphere.
Equally, a circle drawn on the Riemann sphere not passing through North will project to a circle in the complex plane (as shown in the diagram above). So, on the Riemann sphere – which remember is isomorphic (mathematically identical) to the extended complex plane, straight lines and circles differ only in their position on the sphere surface. And this is where it starts to get really interesting – when we have two isometric spaces there is no way an inhabitant could actually know which one is his own reality. For a two dimensional being living on a Riemann sphere, travel in what he regarded as straight lines would in fact be geodesic (a curved line joining up A and B on the sphere with minimum distance).
By the same logic, our own 3 dimensional reality is isomorphic to the projection onto a 4 dimensional sphere (hypersphere) – and so our 3 dimensional universe is indistinguishable from a a curved 3D space which is the surface of a hypersphere. This is not just science fiction – indeed Albert Einstein was one to suggest this as a possible explanation for the structure of the universe. Indeed, such a scenario would allow there to be an infinite number of 3D universes floating in the 4th dimension – each bounded by the surface of their own personal hypersphere. Now that’s a bit more interesting than the Euclidean world of straight lines and circle theorems.
If you liked this you might also like:
Imagining the 4th Dimension. How mathematics can help us explore the notion that there may be more than 3 spatial dimensions.
The Riemann Hypothesis Explained. What is the Riemann Hypothesis – and how solving it can win you $1 million
Are You Living in a Computer Simulation? Nick Bostrom uses logic and probability to make a case about our experience of reality.