1
Mathematical Modelling
Worksheet 12
Gambling
1.
In a coin matching game two players, A and B simultaneously select heads or
tails and compare their choices.
If both choose heads A will pay B 5p, if both choose tails A will pay B 1p, but
if there is not a match B will pay A 3p.
We can try to determine if this game is fair by looking at the expected outcomes of
the two players.
Let
p = number of times A selects heads.
q = number of times B selects heads.
Player A
Player B
Heads
Heads
pq
Tails 1  p q
Tails
p1  q 
1  p 1  q 
The expected outcome of the two players A and B can be written as follows:
 A  5 pq  3 p1  q   3q1  p   1  p 1  q 
 12 pq  4 p  4q  1
 B  5 pq  3 p1  q   3q1  p   1  p 1  q 
 12 pq  4 p  4q  1
The maximum expected outcome of  A and  B can be determined by considering
the partial derivative with respect to p and q.
2
 A
1
 12q  4  0  q 
p
3
 A
1
 12 p  4  0  p 
q
3
Also
 B
 12q  4  0  q 
p
 B
 12 p  4  0  p 
q
1
3
1
3
Having found the values of p and q, which maximise the expected outcome for
players A and B, we can now put these values into  A and  B to determine which
player gains the most.
1
 1  1   1   1 
 A  12    4   4   1 
3
 3  3   3   3 
1
 1  1   1   1 
 B  12    4   4   1  
3
 3  3   3   3 
Therefore it appears that the game is unfair because player A is expected to win more
than player B does.
2.
Two players, A and B each choose one of the three dice coloured red, green
and blue. In the ensuing game, each player rolls the chosen die, the one
showing the higher number being the winner.
Red die:
Green die:
Blue die:
5, 7, 8, 9, 10, 18
2, 3, 4, 15, 16, 17
1, 6, 11, 12, 13, 14
Since the total on each die is 57 the game would appear fair, however is it?
In order to determine if this game is fair, the coloured dice can be played against each
other and the winning die can be recorded in a matrix, this will enable us to see if any
of the die have an advantage against the others.
3
First Match: Red vs. Green
Red die
Green die
2
5
R
7
R
8
R
9 10 18
R R R
3 R R R R
4 R R R R
15 G G G G
16 G G G G
17 G G G G
R
R
G
G
G
R
R
R
R
R
7
12
5
Probability of Green die winning 
12
Probability of Red die winning 
The results of the first match indicate that the Red die is more favourable and has a
higher probability of winning against the Green die.
Second Match: Red vs. Blue
Blue die
Red die
1
5
R
7
R
8
R
9 10 18
R R R
6
11
12
13
14
B
B
B
B
B
R
B
B
B
B
R
B
B
B
B
R
B
B
B
B
R
B
B
B
B
R
R
R
R
R
5
12
7
Probability of Blue die winning 
12
Probability of Red die winning 
The results of the second match indicate the Blue die is more favourable having the
greater probability of winning against the Red die.
4
Third Match: Blue vs. Green
Blue die
Green die
2
3
4
15
16
17
1
G
6 11 12 13 14
B B B B B
G B B
G B B
G G G
G G G
G G G
B
B
G
G
G
B
B
G
G
G
B
B
G
G
G
5
12
7
Probability of Green die winning 
12
Probability of Blue die winning 
The results of the third match indicate that the Green die is more favourable and has
the greater probability of winning against the Blue die.
Therefore from each of the three matches it has been seen that the die are not evenly
matched producing a biased game, because one of the dice is more favourable.
In order to make the game fair the players could use random selection of the dice, so
that they don’t know which die they are choosing and hence which one is most
favourable. Alternatively if the players are allowed to view the dice, then the players
opponent should be asked to choose first, so that the player can gain the opportunity
of choosing a more favourable die than the opponents.
3.
Horses compete in a single race in which there are n horses. If the probability
n
that horse i wins is Pi where 1 Pi  1 , then the odds against horse i winning
are
1  Pi 
Pi
to 1 .
If the total stake money on horse i reflects the probability of horse i winning
and the bookmaker takes 20% for profit and tax, we can find the bookmakers
odds.
Let X denote amount of money paid out on horse i before bookmakers 20%.
And Yi denote the total money staked on horse i.
Yi  Pi
 Yi  kPi
(1)
5
Bookmakers odds 
Total win for punters
to 1
Total stake on i
Total win for punter = Money paid out – stake
1
= X  X  Yi
5
4
= X Y i
5
4 n
= 1 Yi  Yi
5
4
n
= k 1 Pi  kPi
5
4

= k   Pi 
(2)
5

4

k   Pi 
5
 to 1
Bookmakers odds = 
kPi
4
 Pi
to 1
= 5
Pi
4.
In a two horse race, the amount of money bet on the tote (on course) is
currently £600 on horse A (the favourite) and £200 on horse B. These amounts
may be taken to reflect the relative chances of the horses winning.
Just before the race starts, conspirator X bets £400 on horse B and the tote
works out the odds paid for winning bets on the basis of the total amount £600 for each horse – after allowing for a 20% cut for tax and profits.
Off course bets do not affect the odds paid by the tote, but off course
bookmakers pay the odds calculated by the tote (starting price). Just before the
race starts, conspirator Y places an off course bet of £10,000 on horse A.
We can workout how much we would expect the conspirators to gain form this
illegal action.
In the beginning the probabilities that the horses win are:
Pi  A 
Stake on A
600 3


Total staked on race 800 4
Pi B   1  Pi  A 
1
4
6
4
 Pi
5
to 1
Bookmakers odds =
Pi
4 3

5
4 1 ,
Bookmakers odds on A =
3
15
4
4 1

5
4  11
Bookmakers odds on B =
1
5
4
Then conspirator X places £400 on horse B. The amounts placed on the horses reflects
the relative chances of them winning.
Therefore the new probabilities are:
Pi  A 
600 1
  Pi B 
1200 2
Conspirator X having placed £400 on horse B has resulted in evening the relative
chances of the horses winning.
This will have resulted in the Bookmakers odds changing;
4 1

5
2  3 = Bookmakers odds on B
Bookmakers odds on A =
1
5
2
Now conspirator Y places an off course bet of £10,000 on horse A, which is taken at
the new Bookmakers odds on A.
In working out how much the conspirators stand to gain from this action there are two
possibilities to consider:
(1) Expected gain if horse A wins.
(2) Expected gain if horse B wins.
3 3 
3

Expected gain if horse A wins = 10000      400    £4200
5 4 
4

1 3 
1

Expected gain if horse B wins =  400     10000    £2440
4 5 
4

 Overall expected gain = £4200  £2440
= £1760
7
This system would not work in practice because usually there are more than 2 horses
in a race. Therefore it would be more difficult to accurately predict the winner and
affect its starting price by betting on less favourable opponents. This would create a
higher risk when placing very large bets off course because there would be no
guarantee that the horse will win, also with several horses in a race their prices will be
fluctuating constantly making it difficult to keep track of the new prices. Most likely
the authorities will no doubt keep track of bets and would be alerted by the placement
of unusually large bets.
5.
If N denotes the number of premium bonds eligible for prizes in a particular
month and s denotes the number of prizes awarded, we can obtain an
expression for the probability that a holder of r bonds (r<<s) will not receive
a prize. Note that a bond cannot win more than 1 prize in any month. Hence
we can find the expected number of consecutive draws in which the holder
will not win a prize.
N = Total number of bonds.
r = Number of bonds holder has.
s = Number of prizes awarded.
N – r = Number of bonds which the holder doesn’t own.
N  r

 is the number of ways of picking s bonds from the total the holder doesn’t
 s 
own.
N  r


s 

 Probability that a holder of r bonds will not receive a prize =
 N 


 s 
Hence the number of consecutive draws in which the holder will not receive a prize
N  r


s 

=
= 0.5
 N 


s


n
If one prize is drawn for every 10800 eligible bonds:
8
(ii)
We can find the probability that a holder of 100 bonds will win no prizes in 5
consecutive years.
N = 10800
s =1
r = 100
n = 60 months i.e. 5 years.
Inserting this data into the equation for the number of consecutive draws in which the
holder will not receive a prize, we get:
10700 


 1 
10800 


 1 
(iii)
60
 10700 


 10800 
60
 0.57
We can find the expected number of draws before the holder wins a prize.
N  r


 s  = 0.5 is the expected number of consecutive draws in which the holder will
 N 


 s 
win a prize.
n
N = 10800
s =1
r = 100
n
 10700 

  0.5
 10800 
 10700 
n ln 
  ln 0.5
 10800 
n
ln 0.5
 10700 
ln 

 10800 
n  75 months