Gambling  What are the Odds?
By the end of this Unit you will be able to:
 Understand the odds involved in throwing a die
 Calculate what is involved in setting up a Dice game to raise money at a school fete
 Understand more about other aspects of gambling
 Odds involved in throwing a die
 Setting up a dice game
 Evolution to roulette
Topic 1: The odds involved in throwing a die
Gambling with dice (plural of die) is, as we have seen, a very ancient pastime that was considered an essential part of young boys’ education in classical Roman times. Let us now look at the mathematics involved in this activity.
Jargon: language that is specific to a certain topic or subject
Statistics: analysis of empirical or factual data and the appropriate interpretation of such analysis
Exclusive: restricted to one thing or event
Unbiased: not favouring any particular side
Conundrum: difficult question to which there is no obvious solution
Consider a normal six sided die. When we throw the die a number of different things or events can happen. In statistical jargon, the number of possible, mutually exclusive events (no chance of getting two of them together; they are quite separate) that may occur, when throwing the die once are:
{get a 1}; {get a 2}; {get a 3}; {get a 4}; {get a 5}; {get a 6}
Such occurrences are often referred to as simple events. Moreover, one can pretty much assume, that if the die is fairly well balanced the chance (or probability denoted by P) of any of these simple events occurring and hence the chance or probability of getting any particular simple event is the same and equal to 1/6 – we say you have a one in six (1/6) chance or probability (P) of getting a 1,2, 3, 4, 5, or 6.
A compound event is one like {get less than 3} and would comprise the two simple events {{get a 1} or {get a 2}}
To obtain the probability of any event (including a simple event) one adds up the number of simple events and divides by the total number of possible simple events.
Thus, for example,
P ({get a 2}) = {number of simple events favourable to getting a two} / {total number of simple events }
= 1/6 (since there are 6 (equally probable) outcomes)
or, P{get a number less than 3} = {number of simple events favourable to getting less than 3 } / {total number of simple events }
= 2/6 = 1/3
Expressed in terms of odds, the odds of getting any number when throwing a die is 1:5 (1 to 5), that is one chance of getting any prespecified number and 5 chances of not getting it. By similar reasoning, the odds of getting a number less than 3 is 2:4 which is the same as 1:2.
Pairs
In your pairs, read through the following statemenst and then discuss the questions that follow.
Let’s think about shapes that could serve the function of a die and why we’ve settled on a die.
Well, if we’re only interested in two options the coin is hard to beat and pretty unbiased.
Three options would require a threesided (surfaced) shape which is pretty much impossible if the surfaces are flat. How about 4 sides?
Questions:
Can we construct any other regularsided 3d shape to act as a die? How many sides do these shapes have? The pictures below will help you.
When you have answered this question, read through the text that follows and see how close you were to being on the right track.
For a foursided shape we’re in business, with the pyramid. Then things start to get tricky. For a fivesided shape, we need a square based pyramid but if we have this, then four of the surfaces are triangles and one is a square. For a six sided shape, things get easier, in that we can use a normal die or we can use two pyramids stuck back to back to make six regular triangular surfaces. Of course, it will tend to land in an acceptable way after being thrown, but a big problem that would arise is that we would have a problem seeing the number! For a normal die we take the number on top, now we would have to probably take a numbered surface.
We then get to the issue of seven. Well a pentagon on top and bottom and five squares around the side might be fine apart from when we try to find out which side we’ve landed on. For eight sides we’ve got our squarebased pyramid backtoback
Things only get really promising when we get to twelve sides. We can use a pentagon on top and bottom and ten pentagons around the side – a regular twelve sided figure which would act well as a die and where the number would be on top. Similarly we can have a 20sided shape made up of triangles which is symmetrical and regular and can act as a pretty good 20sided die should we require such a thing!
Fact File: Technical Terms
Sides  Shape  Notes  
4  tetrahedron  Each face has three numbers: they are arranged such that the upright number (which counts) is the same on all three visible faces. Alternatively, all of the sides have the same number in the lowest edge and no number on the top. This die does not roll well and thus it is usually thrown into the air instead.  
6  cube  A common die. The sum of the numbers on opposite faces is seven.  
8  octahedron  Each face is triangular; looks something like two Egyptian pyramids attached at the base. Usually, the sum of the opposite faces is 9.  
10  pentagonal trapezohedron  Each face is kiteshaped; five of them meet at the same sharp corner (as at the top of the diagram in this row), and five at another equally sharp one; about halfway between them, a different group of three faces converges at each of ten blunter corners. The ten faces usually bear numbers from zero to nine, rather than one to ten (zero being read as "ten" in many applications), and often all odd numbered faces converge at one sharp corner, and the even ones at the other.  
12  dodecahedron  Each face is a regular pentagon.  
20  icosahedron  Faces are equilateral triangles. Typically, opposite faces add to twentyone. A 2nd century CE Roman icosahedron die is in the collection of the British Museum, though the game it was used for is not known.[7] 
Adapted from Wikipedia online encyclopaedia
(Small groups of 5)
In your small groups attempt to construct shapes with 4, 6, 8, 10 and 12 regular sides. Number each side according to the number of sides and see if the shapes will roll like a die. (You will need paper, scissors and stickytape)
Topic 2: Setting up a Dice Game
What would you say to a group of parents who felt that it was “immoral” to have gambling at a School function? Do you think you would have a better chance of convincing them that it was morally acceptable if the money was going to a good cause, like a bursary fund for needy learners? Do you personally think it makes a difference morally, for what purpose the money is being used? Why or why not?
Let us imagine that we want to set up a dice game to raise money at a School fete. We are going to set up a game, based on one throw of the die, where people may bet on the outcome.
Assuming for the moment that we are only considering betting on a particular number, consider a game where people may bet on which number will come up in a particular throw of the die.
Let’s consider the following scenario first. Let us imagine that a teacher sets up a game in his or her class where learners bet on the outcome of the throw of a die. For the time being the teacher, let us call him Enoch, will act as the “house”; that is he will take learners’ bets and pay out according to certain odds. What are these odds? Well, if the game is played in such a way that neither any player of the game nor the organiser of the game who takes the bets (Enoch in this case) has any advantage we call the game a balanced game. In the case where the odds are set so that the game is balanced, then neither the player nor the house would be expected to win in the long run.
We need to determine the odds for a bet. That is, if one bets on a one with R1 and the one comes up what should the payout be? We will first set the odds so that the game is balanced i.e. over the longterm, noone who makes any bet on the game should win or lose. The best way to test this is to run the “balanced game test”. This test says that if you bet on every possible outcome, then for a balanced game you should neither win nor lose.
So let’s consider betting R1 on all the numbers. That is, making six bets of R1; in other words betting R1 on each number. Well our outlay is R6 each time; for the game to be balanced. I must thus get paid R6 from the game each time. That is, since only one number can come up and I have bet on each number, the winning number must get paid out R6 (including the original bet) for my outlay to equal my return. Thus giving payoutodds of 5:1 on each number (paying out five times the amount that is bet plus returning the original bet) will ensure that the game can be defined as balanced.
We mentioned before that this is all very well for demonstrating the game to his learners, but if Enoch and his class are to run a stall at the annual school fete, it may not be a good idea for him to pay people out in a way that makes the game balanced. Can you think of why this is so?
Well he might get unlucky, someone could bet R1000 on the number five and five might then just come up. If this were to happen he would have to find R5000 – this could break the bank; well the school’s bank at any rate, and more than likely lose Enoch his job.
So what should Enoch do to run the game at the fete and guarantee the game makes some money? A good idea would be to make the payouts (the amount a person wins) an amount which gives the stall a “house advantage”; for example, pay out R4.50 and not R5 for a winning number. Let us examine what happens now if we apply the balanced game test. Say R1 is bet on each outcome (the six numbers) – that is R6 is bet. How much is paid out? Well if we set the winning payout on a number to be R4.50 then R4.50 plus the R1 bet on the winning number will be paid out; in other words, doing the necessary calculations, we can see that each throw yields a profit to the house of R0.50 cents. In gaming or gambling jargon, the handle or total amount staked (played or bet) for each play is R6 (and remember here we are assuming that R1 is bet on each possible outcome) but the win (the amount paid out) is only R5.50.
It will be useful at this stage to introduce some gaming terms: Win percentage and House Advantage. The Win % is the percentage of a stake a player can expect to win back. If the game is balanced the Win% is 100%; that is the player (and the house) do not operate at any advantage. The house advantage is the difference between 100% and the Win%. Hence if the Win% is 97% (as is approximately the case for South African Roulette games in Casinos), the house advantage is 3%. As we discussed above, we can calculate the Win% by working out the return from an array of bets which cover all possible outcomes.
Mathematically we can express this as follows:
Win% = 100 * Win/Handle (all outcomes covered) = 100 * (5.5/6) = 91.67% (to 2 decimal places) or 11/12 expressed as a fraction.
Therefore:
The House advantage = 100 – Win% = 8.33% (to 2 decimal places)
Note that, expressed as a fraction, the house advantage = 0.5/6 = 1/12
We summarise as follows:
Does this mean Enoch’s stall will make lots of money on the day of the fete? Well the odds have been tipped in Enoch’s favour but it depends on the throw of the die!!!” (This is where the true sense of “gambling” comes in – remember in the first Unit we discussed that gambling always involves an outcome that we cannot predict.)
Although over a long, long period Enoch would become increasingly sure that he would secure a profit, on school fete day he might just be unlucky.
That is, as we mentioned above, if a couple of people placed some big bets and he lost them he could still get financially wiped out. We would say therefore that the risk of loss is quite high unless the game is played a very large number of times.
Can you think of any solutions to this perplexing question or conundrum? Well we could “limit the risks” – let us look at what this means:
If we limit the stake (the amount someone can bet on a particular play) then because we know that the longer the game is played the more likely we are to turn a profit, it will reduce the risk. If we limit the bet to say R5, then our maximum payout (excluding the original bet) for any particular person and their particular bet is R5 multiplied by 4.5 which equals R22.50. If we include the original bet the payout is R27.50.
Activity 2: What are the odds?
Individual
Work on your own to figure out this scenario.
We could try and assess what chance Enoch has of losing money under really unlucky circumstances, assuming he sets up the game to have the house advantage of discussed above.
Imagine a case where only one person is playing at the stall and bets the maximum bet of R5 on the number {4} each time. Say she gets lucky five times in a row and wins each time.
 What is the chance of this happening? Express your answer as a percentage. (Hint : What is the probability that she obtains a four once (1/6); what is the probability that she obtains a four twice – can you see this would be (1/6)*(1/6) = 1/36. Now continue the train of logic ... )
 How many Rands will Enoch lose?
If you have worked this out correctly you will see that Enoch has a very low chance of losing, and will any event not lose too much money.
[You should have calculated that Enoch has about a 1 in 10 thousand (0.0129%) chance of losing R112.50 (R22.50 x 5 = R112.50) so maybe he could relax.]
We can see from this Activity above that having a house advantage will help Enoch make some money at the fair and the losses look manageable.
However, he could limit his losses further. What else could we try and do to reduce the risk of a big loss? Well we could try and encourage people to make “offsetting bets”. Let us explain what is meant by this term. If we had a lot of people betting R5 on the number three and we could encourage people to bet an equally large amount on some other number we would cancel or nullify our potential loss on three since only one number can win. In other words the potential large payout will have been “offset” against the other bets. This will tend to happen naturally when large numbers of people play, as they’re unlikely to all bet on the same number! This means that the larger the number of people playing, the more chance of benefiting from the house advantage. Enoch and his class must pack the people into his stall! if they want to make money!
Based on what we have examined in this Topic, do you think we can now see what the three things that professional gambling organisations (like casinos) try to do?
1. They set the payout odds so that they give themselves a house advantage.
2. They limit the size of any particular bet
3. They encourage, where possible, offsetting bets. Remember that generally simply having as large a number of people as possible playing a particular game will result in offsetting effects.
So these are the things that Enoch must consider when he sets his game up. He can only get a large number of people to play the game if it is indeed attractive for players to play. We will now proceed to consider the game from the player’s perspective or point of view to find out what makes it attractive and how a player can sensibly manage his bets and hence his risk.
Averse: against
Conglomerate: grouped together
Consistent: remaining the same or constant
The perspective of the player
We will look at the psychology of gambling in detail in the next Unit. For now we just need to know that people play games like the die game above for the excitement or thrill of having a chance of winning something. They are prepared to accept their expected loss (the house’s advantage) because the act of playing and the associated excitement is entertaining. As we saw above the school stall at the fair (or the gambling house or the Casino) would always prefer an even spread of bets because they are then least at risk of having a large loss. The player him/her self, being the person who does the betting, can also manage the process of what is generally referred to as player risk. They can, for example, put the maximum bet on a single number or they might want to spread their risks and put a small amount of money on say three numbers – this gives them a larger chance of winning something (what casinos call the “Hit percentage”) but less chance of winning a large amount.
Enoch starts thinking about these things and considers that at the fair he might be dealing with both riskaverse (against risk) players, who are players not keen to risk too much and who would rather win something small fairly frequently and risktaking players who put “all their eggs in one basket” and just bet on a single number. He decides that what he will do is have four extra categories of bets at his stall to accommodate riskaverse players. He will set it up so that in addition to betting on just one single number, players can also bet on the following four categories:
 The numbers {1, 2, 3} – these he calls [numbers less than or equal to three]
 The numbers {4, 5, 6} – these he calls [numbers greater than or equal to four]
 Odd numbers {1, 3, and 5}
 Even numbers {2, 4, and 6}
If the game were balanced Enoch knows that for each of the bets above he would pay out the amount staked (plus the original bet) but how does he calculate what he will have to pay out if he wants the same house advantage that he had with the single number bets?
The way to work it out is as follows: Let’s consider the odd and even bets. Say someone bets on both (R1 each) so that the amount staked on all possible outcomes is R2. If the game were balanced they would simply always get their money back (R2) Remember that paying out R4.50 on a bet of R5 on single number gave a house advantage of 1/12 (8.33%) or equivalently a win fraction of 11/12.
Can we use the same house advantage for the conglomerate bets?
Keeping the win fraction as before at 11/12, we must have
We think you can work out or calculate that for the even/odd bets and on the {1, 2, 3} and {4, 5, 6} bets, the payout must be R 2 x 11/12 = R1.83 (including the original bet). Remember, you will now only need R2 to cover all possibilities.
So Enoch must pay out 83c on the winning even/odd bets and similarly on the {1, 2, 3} and {4, 5, 6} bets to maintain a consistent house advantage (Win%) of 8.33%.
Enoch starts to worry about all the change he has to keep. Dealing with 83c all the time is a real pain. He wonders whether they should simply make it 80c – you can probably see that this gives a house advantage of exactly 10% (payout of R1.80 on a bet of R2). This makes a lot of sense to Enoch as these numbers are nice, rounded numbers. Then he thinks – how will this affect the other single number bets? Will these still be rounded off?
Well we work it out in the same way:
The stake for all possible outcomes is six (each of the 6 numbers). So the payout must be R5.40 (R5.40/R6 must equal 9/10) –well, thinks Enoch, this is not too bad and means there are no strange decimals. So for each winning number Enoch pays out R5.40 (including the original bet) – or R4.40 excluding the original bet.
We are now going to move on to looking at some more complex issues about dice and the growth of other forms of gambling like Roulette.The Evolution to Roulette and dead numbers
Enoch is pondering again. He tests out the game on his friend Ayanda, but Ayanda thinks the payoffs are wrong. If I put R1 in evens and I get an even then I must win a R1 says Ayanda, not 80c. Enoch tries to explain about house advantage but Ayanda finds this confusing.
So Enoch has another idea. He carves a sevensided die from a piece of wood; it has a top and a bottom in the shape of a pentagon and five square sides around its middle. It has the numbers 1 through to 6 on it as well as 0. Then he says he will use his sevensided die at the fair but will call 0 a “dead number”. That is, it won’t be included in the odd or even numbers or {numbers less than or equal to three} or . {numbers greater than or equal to four}. You can bet on it and it will just pay what the other numbers pay but it won’t be included in the odds or evens or other grouped bets.
Enoch decides to pay R5 on the numbers (including the dead {0}) for a R1 bet (R6 including the bet) and R1 on the odds and even, just as Ayanda wanted. If {0} comes up then the bets on {0} get paid out but bets on the odds and evens and the {1, 2, 3} and {4, 5, 6} don’t get paid out.
This seems much simpler in terms of handling the money and Ayanda thinks this is a much better arrangement.
What is the house advantage on this new game? Well the Win percentage is:
This means that the House advantage is 14.29%, (100% minus 85.71%) which is quite a lot more than before!
Can we check that paying out R1 on the odds and evens also gives the same house advantage?
This is a bit trickier. Let’s consider the scenario where the game is played seven times. We would expect to get one of each of the seven numbers. Moreover, say each time we bet R1 on (each of) odds, evens, {1, 2, 3}, {4, 5, 6} and R1 on {0}, so that each time we would have to win something. (If you are betting on every possible outcome then one of your bets has to be correct and you must win.)
The seven plays would cost us R35; our winnings (including the stake) are R2 x 3 times on the {1, 2, 3}, R2 x 3 times on the {4, 5, 6} and R6 once on the {0}. This would total R30 (R6 multiplied by 5) , so the Win fraction is 30/35 = 6/7. This is exactly the same fraction as that which we would get with the single number payoffs. Hence the house advantage remains 14.29% for all the potential bets.
Enoch is really happy – a simple arrangement with no fractions and with payoffs that everyone seems to find more acceptable and logical. So we see that including a dead number has fixed his problems – he’s ready for launch. The only remaining problem is that 14.29% seems a pretty steep house advantage – Enoch could expect in the longer term to be getting a return of 14.29% on the money bet each time the game is played. Financial investors, who put their money in interest bearing ventures, would think that 14.29% is a very good return for a whole year! Maybe noone will play the game after all after they’ve worked out Enoch’s house advantage?
The history of roulette is in itself vague and uncertain. With so many myths and legends intertwined with fantasylike stories, there is no real way to find out how exactly the game was created. Keeping this in mind, even though the invention of the game is uncertain, it is normally attributed to the French mathematician and philosopher, Blaise Pascal. In the 17th century, Pascal was known for his fascination with perpetual motion and some say this fascination of his led to the invention of the first real Roulette table.
In French, the term ‘roulette’ means ‘small wheels’. In all probability, Roulette was invented by Pascal in the 17th century and by the early 18th century, there were already a few more versions of the game that were being played in England. However, it wasn’t until the mid 19th century that Roulettes’ French connection was created.
In the year 1842, two French gentlemen Louis and François Blanc invented the first real modern Roulette table. This invention consisted of single zero tables that are still used today. However, at this time gambling was considered to be an illegal activity in France. So, the Blanc brothers introduced this game to all the gambling halls all over Europe and the United States. Hereafter, Roulette rapidly gained devoted followers and has ever since, become one of the most popular table gambling games. In those beginning days, the game was played a lot by royalty and the game was soon nicknamed the ‘King” (or by some, the “Queen”) of all casino games!
Adapted from: www.toproulettesystems.com/short_history_of_roulette.html
Words:
Myths: not strictly true stories
Intertwined: mixed together
Attributed: seen as being responsible for
Perpetual motion: constant movement
Enoch starts thinking … yet again!
He remembers a game called Roulette and that he has an old Roulette wheel. He digs it out and sees it has thirty six (36) red and black numbers (1 through to 36) as well as a green {0}. It works on the same ideas he had developed for the die game: A dead 0 !! However because there are a lot more numbers (36 rather than 6 on a die) the house advantage is much more reasonable.
The parallels between Roulette and Enoch’s sevensided die with the dead number are extraordinary. Roulette also has only one dead number (green {0}) and pays out in a way that makes sense to everyone, – you receive R2 (per R1 bet) when you win on the odds and evens (as well as the equally split reds and blacks), you receive R36 (per R1 bet) when you win on an individual number but because there are a lot more numbers, the win percentage is much less. Moreover, there are a number of other betting options – you can bet on {1 – 12}; {13 – 24} or {25 – 36}; this pays out R2 (excluding the original bet) on a R1 bet.
Enoch starts reading about Roulette; he discovers the version he has is the original European Roulette that they use at the famous Monte Carlo Casino. In the USA they use a different Roulette wheel with a single 0 AND a double 0. This has a much larger house advantage – very close to double the European version.
Anyway, Enoch has to prepare for the fete. His sevensided die with the dead number looks decidedly strange so he ponders how to fix that problem. How about a spinning disc marked into seven equal portions – that’s much simpler than a peculiar looking die and works just like a Roulette wheel. So he takes his disc Roulette wheel and carefully pastes a cardboard marker alongside the disc. When the disc is spun the arrow will end up pointing to one of seven numbers.
Enoch’s Roulette Disc
Pairs
In your pairs do the following calculations based on Enoch’s Roulette Disc above. .
The first 5 plays of the wheel yield the following outcomes
{2}, {4}, {3}, {0}, {3}
Outcomes
Play 
Odds 
Evens 
{1,2,3} 
{4,5,6} 
1 
2 
3 
4 
5 
6 
0 
1 

X 
X 


X 





2 

X 

X 



X 



3 
X 

X 



X 




4 










X 
5 
X 

X 



X 




If you bet R1 on the number {4} for each of the 5 plays what would your payoffs be for each of the 5 plays?
From the above you should be able to see the main differences and similarities between dice and Roulette odds and how these are calculated
Small Groups
Working in your small groups, write short definitions for the following terms. Avoid simply repeating what is written in the text. Try to use your own words as far as possible. When you have completed your “Dictionary” transfer it to a large sheet of paper. Choose from all group’s efforts, a few posters to display that you think best explain this particular set of gambling jargon.
Terms to define:
Gambling
Odds (as in “what are the odds of this happening?”)
House advantage
Balanced Game
Unbiased die
Win fraction
Self Assessment
I can now understand the odds involved in throwing a die
I can now calculate what is involved in setting up a Dice game
I can now understand more about other aspects of gambling
What I enjoyed most about this Unit was............
What I enjoyed least about this Unit was...............
What I would like to know more about is...................