Singapore Mathematics Project Festival
2003
Project title
Team Members
Quah Wan Fen Nicole
Quah Wan Jen Cheryl
Sim Jingwei
Tan Wei Lin
Project Supervisors
Mrs. Sia Wai Leng
Mr. Kenneth Lui
School
Raffles Girls’ School (Secondary)
Project Category
Junior Section
Contents
Summary………………………………………………………………………..
3
Introduction……………………………………………………………………...
5
Aims & Objectives..……………………………………………………………..
6
Theoretical Background…………………………………………………………
7
Procedure of Investigations…………………………………..………………….
15
Investigation I...…………………………………………………………………
16
Investigation II...………………………………………………………………...
21
Investigation III…..……………………………………………………………...
24
Conclusion…………….………………………………………………………...
28
References & Acknowledgements………………………………………………
29
Matrices in Game Strategy
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Summary
This project is an investigation on Bridge, the common card game for four players
who team up to play, pair against pair (the teams are called “North-South” or “EastWest”, according to their seating positions at the table). It is played using the usual
52-card deck. At each round, each player is to play (or put down in full view) one of
his cards. The player who plays the highest card amongst the four wins the round or
“trick” for his team.
The introduction defines briefly game theory and matrices, as well as gives the
connection between Bridge and these two main aspects of our investigation.
Our original aim for this project was to propose an optimal strategy for Bridge,
through investigating various situations and models, where strategies and payoffs
were approached differently.
We adopted the following methodology in our investigation. Various methods
involving matrices were used to solve the examples we came up with for each game
model. The optimal strategies for Investigations I and III yielded patterns or
generalizations. However, there was insufficient time to complete Investigation II.
We realized that Bridge was too complex to investigate within the time frame that we
had. Instead, we came up with several simplified number games that retained a few of
the methods of play from Bridge, namely that the four players played two-against-two
and that each player had to put down one card for each round, with the player playing
the highest-valued card out of all four cards winning the round for his team.
Our first investigation involved dealing three cards to each player. We assumed
‘telepathy’ between the partners where each player would have a definite knowledge
of his partner’s hand. This allowed for the strategy, which we investigated and found
an optimal solution to, to be the cards that the partners should play as a round
together. It was found that the largest payoff would result from East-West team
playing Min [max(East), max(West)] and Min of player holding {max [max(East),
max(West)]}, while the North-South team used the same strategy, playing Min
Matrices in Game Strategy
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[max(North), max(South)] and Min of player holding {max [max(North),
max(South)]}.
After Investigation I was completed, we decided to investigate other game models,
which had randomly chosen structured payoffs, resulting in more variation in the
payoff matrix. In Investigations II and III, each player had 4 cards, and as we found it
difficult to retain the 13-cards-within-a-suit property of card games, we modified
them into number games (i.e. assigning an increasing numerical value to all cards
required).
We came up with a varied payment that would allow for more variation in the payoff
matrices; the winning pair gains the basic $0.30 from the losing pair for every play.
Extra money was gained from the losing pair based on the summed value of their
cards (see value-and-payment table in Investigation II report). We then mapped out
every way of playing a single distribution of hand and the average was taken for all
the ways of playing for each set. Optimal strategies were developed for our two
examples, but we could not develop a full generalization.
As for Investigation III, it was created as a furthering of Investigation I by including
such a payment system like Investigation II. It was approached using “telepathy” and
the pairing of cards, and it was found that Investigation I’s generalization could be
applied in solving the first step of optimal strategies required. The following payoff
matrices were constructed such that the strategy was to find which pair of cards to
start with, and though no generalization was reached, we had a general assumption
that when a team had matching rows/columns, their strategy would be ¼ for all the
rest and the matching ones would be the same such that it needed only one of them to
make the strategy equal 1. We found that there were also easier ways of finding the
value of the game by looking at the payoff matrix alone.
The various investigations carried out in this project have allowed us to better
understand how to develop our own investigative models from existing games. It is
hoped that with our first steps of investigation, a generalization of the actual game of
Bridge can be proposed in the future, taking into account the more minute (but
complex, in the real gambling world) details such as bidding and trumps.
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Introduction
The project
This project involves an investigation of several game models that
have been simplified and adapted from the card-game bridge. We chose to work on
this because we took an interest in a different form of mathematics – matrices. Using
matrices to solve games of strategy was easier to relate to than other diverse fields
where matrix mechanics are involved, e.g. forest management, chaos etc.
The matrix
When a set of regular numbers is written as an oblong arrangement of
terms consisting of m rows and n columns and thus treated independently, this forms
the basis of a different expression of mathematics – matrices. A matrix is a concise
way of uniquely representing and working with linear transformations.
The theory
Game theory is the formal analysis of strategic behaviour, or the
relation between interdependent agents. It takes into consideration the consequences
of actions, and the interplay of competition and cooperation. Any situation with two
or more people requiring decision-making can be classified as a game. For many
games, continual playing trains the player to think strategically and determine the best
moves to maximize his chances of winning through basic logic. Mathematics can also
be used to map out a more structured format of any game. When it is possible to
represent the outcomes of different strategies in the form of a payoff matrix, the
optimal strategies for each player and the expected value of the game can be solved.
The investigation
We aimed to find the optimal strategies of various number
games that were simplified and modified from the game of Bridge. We applied
different techniques of matrix solving to find the optimal strategy for each player, to
either minimize his loss or maximize his gain.
The Bridge basis
We originally intended to base our investigation on the 4-player
card game of bridge. The game of bridge requires comparison of players’ cards with
originally assigned values to them, and weighing numbers against one another is also
a part of calculating optimal strategies of competitors. However, due to the complex
nature of the actual card game Bridge, we later came up with several simplified game
models, which we based our investigation on instead.
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Aims
 Research segment
To find out how matrices are applied in games of
strategy to the advantage of the player and to solve for optimal strategies and
values of games.
To find out how matrix mechanics applied in solving
various game models and approaches differ.
 Investigative segment
Our original aim was to find an optimal strategy or
generalization for the card game Bridge. We later modified this to finding optimal
strategies or generalizations to various simplified game models, using
experimentation through matrices, tabulations and formulae.
Objectives
 To come up with several simplified game models from Bridge that require various
approaches to solve. These models still retain a few Bridge rules:
a) There are four players who team up to play, pair against pair;
b) The player who plays the highest card amongst the four played wins the round
or “trick” for his team.
 To use various methods and forms of matrices to solve and propose the optimal
strategies of these models.
 To come up with generalizations for each of these models based on the results of
our investigation.
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Theoretical Background
(I) Bridge
In Bridge, the standard deck of 52 cards is used. 4 players compete in pairs – players
East and West (aptly named for their seating position at the play-table) play against
North and South. In every round, the pair that plays the card of the highest value of
those from the suit established beforehand wins the round, or “trick”, as it is called in
Bridge.
(II) Optimal Strategies & Saddle Points
To demonstrate the use of a payoff matrix, we use the example of the spinning wheel,
a carnival game:
Row wheel
Player R
Figure 1.1
Column wheel
Player C
The payoff matrix is then fashioned, showing how much row-wheel player has to pay
column wheel player when there are different combinations. Note: The payoffs in the
matrix are arbitrary. An example of such a matrix would look like this:
Region Nos.
C falls on 1
C falls on 2
C falls on 3
R falls on 1
-$3
$2
-$6
R falls on 2
-$5
-$4
$5
R falls on 3
$2
$3
$0
R falls on 4
$1
$4
-$3
Figure 1.2
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This table shows how much Player R gains for any move. Should R’s move fall on
region 4, and Player C’s in region 2, R will receive $4 from C. Positive numbers
shown are thus R’s gains and C’s losses, and vice-versa for the negative numbers. The
following table shows the division of regions in each player’s game wheel, and
consequently, the probabilities of landing in each section.
Region
Fraction for R’s
Fraction for C’s
No.
wheel
wheel
1
1
/4
1
/2
2
1
/4
1
/3
3
1
/6
1
/6
4
1
/3
N.A.
The expected payoff to Player R is the weighted average of the payoffs to him, where
each payoff is weighted according to the probability of its occurrence. This is
expressed by the equation
a11p1q1 + a12p1q2 + … + a1np1qn + a21p2q1 + … + amnpmqn
(1)
Figure 1.3
which is obtained through multiplying Player R’s strategy, the row vector:
p = [p1 p2 … pm]
by Player C’s strategy, the column vector:
q=
q1
q2
q3
q4
Here, pi = probability that Player R makes move i (i = 1, 2, …, m)
qj = probability that Player C makes move j (j = 1, 2, …, n)
and by the payoff matrix (refer to example of Figure 1.2), which can then be
represented similarly this way
A=
a11
a12 … a1n
a21 a22 … a2n
.
.
.
.
.
.
.
.
.
am1 am2 … amn
Considering that piqj is the probability that for any one play of the game, Player R
makes move i and Player C makes move j. For such a pair of moves, the payoff to
Player R would then be aij. Multiplying each possible payoff (A) by its corresponding
Matrices in Game Strategy
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probability (pq) and summing over all possible payoffs, equation (1) is then obtained.
Alternatively, it can be written in matrix form as:
a11
E (p, q) = [p1 p2 … pm]
a12 … a1n
q1
q2
q3
q4
a21 a22 … a2n
.
.
.
.
.
.
.
.
.
am1 am2 … amn
Where E (p, q) is the expected payoff to Player R, and – E (p, q) is thus the expected
payoff to Player C. Thus, for the game described above, the expected payoff to Player
R is denoted by.
1
1
1
1
/4 /4 /6 /3
-3 2 -6
-5 -4 5
2 3 0
1 4 -3
1
/2
1
/3
1
/6
= -0.4305
Therefore, it can be expected that Player C will gain an average of 43 cents (this
number is tended towards over many times of playing only) from the row wheel
player every play.
In this spinning wheel game, the outcome is based on chance and the players have no
control over their moves, thus the above is just the calculation of predetermined
strategy. However, based on the payoff matrix, the players could find optimal
strategies for themselves by manipulating the areas assigned to each number on their
wheel.
Optimal strategy is defined as Player R’s strategy to maximize his gain, denoted as
p*, and Player C’s strategy to minimize his loss, denoted as q*. The fundamental
theorem of zero-sum games states that there exist strategies p* and q* such that E(p*,
q) ≥ E(p*, q*) ≥ E(p, q*) for all strategies p and q. Also, v = E(p*, q*), where v is the
value of the game, and where v = 0, the game is termed fair.
One way of finding an optimal strategy is by using the saddle point of a payoff
matrix. A saddle point is an entry ars in a payoff matrix A where ars is (i) the smallest
entry in its row, and (ii) the largest entry in its column. 60 is the saddle point in the
following matrix:
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-50
-5
30
90
75
60
60 -30 -10
0
Therefore, p* = [0 1(rth entry) 0] and q* =
0
1
sth entry
Player R’s optimal strategy is thus to always make the rth move, and Player C’s the
sth move. This is known as a pure strategy, contrary to what is known as a mixed
strategy – more than one move to be made for the optimal strategy. A game with a
payoff matrix that has a saddle point is known as strictly determined, and its value is
the payoff in the saddle point entry. Also, a payoff matrix may have several saddle
points, but the uniqueness of the value of a game guarantees that the numerical values
of all saddle points are the same.
For non-strictly determined, 2 X 2 matrix games, the optimal strategies of Players R
and C can be found using the theorem:
a22 – a12
p* =
a22 – a21
a11 – a12
a11 + a22 – a21 – a21
a11 + a22 – a21 – a21
q*=
a11 + a22 – a21 – a21
a11 – a21
a11 + a22 – a21 – a21
a11a22 – a12a21
and v =
a11 + a22 – a21 – a21
(III) Optimal Strategies & Assignment Matrices
Operations research requires assigning tasks to facilities on a one-to-one basis, e.g.
distributing machines to building sites. The assignment problem, requires the
number of facilities and tasks (taken to be n of each) to be equal. Therefore, there are
(n)(n-1)(n-2)…(3)(2)(1) = n!
possible ways to assign all n tasks. Among the n! assignments, a next task is given to
find one that is optimal in some sense. To define an optimal assignment, we
introduce the following term which takes on the appropriate unit of the problem:
cij = cost of assigning the ith facility the jth task
where
i, j = 1,2,…,n
The cost matrix of the problem refers to the n X n matrix:
Matrices in Game Strategy
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C=
c11
c12 … c1n
c21
.
.
.
cn1
c22 … c2n
.
.
.
.
.
.
cn2 … cnn
As each task can be assigned one unique facility, and each facility can be assigned to
one unique task, no two corresponding cij’s can come from the same row or column.
Therefore, given any n X n cost matrix C, an assignment is a set of n entry positions,
no two of which lie in the same row nor column (i.e. no facility can be assigned
twice). The sum of the n entries of the assignment is called the problem’s cost (there
will be n! costs in this case). The assignment with the smallest possible cost is the
optimal assignment. The assignment problem seeks to find an optimal assignment in
a given cost matrix. In the following example of a cost matrix, the various possible
assignments are highlighted in yellow. The final optimal assignment is blue:
50 35 40
5 65 30
2 3 25
50 35 40
5 65 30
2 3 25
50 35 40
5 65 30
2 3 25
50 + 65 + 25 = 140
50 + 30 + 3 = 83
35 + 5 + 25 = 65
50 35 40
5 65 30
2 3 25
50 35 40
5 65 30
2 3 25
50 35 40
5 65 30
2 3 25
35 +30 + 2 = 67
40 + 5 + 3 = 48
40 + 65 + 2 = 107
However, as the value of n in C increases, it becomes impractical to solve any
assignment problem. There is a theorem that if a number is added to or subtracted
from all entries of any one row or column of a cost matrix, then an optimal
assignment for the resulting matrix is also the optimal assignment for the original cost
matrix. The Hungarian Method applies this such that the resulting matrix contains
an assignment entirely of zero entries (called optimal assignment of zeros; obviously
optimal because it is impossible to get a cost below zero from a matrix of nonnegative
entries). The original numbers at the position of the zero entries form the optimal
assignment for the original problem. The last three steps are applied iteratively as
many times as necessary to generate an optimal assignment of zeros:
1. Subtract the smallest entry in each row from all row entries. With this, each
row will have at least one zero entry, and all other entries will be nonnegative.
2. Where a column still has only non-zero entries, subtract the smallest entry in
each column from all column entries. With this, each row and column will
Matrices in Game Strategy
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have at least one zero entry, and all entries of the resulting matrix will be
nonnegative.
3. Draw lines through appropriate rows and columns such that all zero entries are
now covered and with the minimum number of lines. This may be done
through trial and error, but for larger numbers, algorithms suitable for
computer coding are available.
4. To define the optimality in this case,
(a) If the minimum number of covering lines is n (i.e. the number of rows
or columns in this n X n cost matrix), an optimal assignment of zeros is
possible and the Hungarian Method is complete. Search for set of n
zero entries, no two of which lie on the same row or column. If more
than one set exists, validate optimal set.
(b) If the minimum number of covering lines is less than n, proceed to
Step 5 as an optimal assignment of zeros is not present.
5. Determine the smallest entry not covered by any line. Subtract this entry from
all uncovered entries and then add it to all entries covered by both a horizontal
and a vertical line (i.e. twice). Then, return to Step 3 and reiterate until 4(a) is
reached.
The Hungarian Method can only be used for minimization. The problem of
maximizing the sum of entries of a cost matrix is easily converted to one of
minimizing the sum by multiplying each entry of the cost matrix by –1.
(IV) Optimal Strategies & Linear Programming
For larger, non-strictly determined matrices, their optimal strategies are found using
linear programming. The Simplex method, used to solve linear programming
problems, can be applied to manually calculate the optimal strategies and the value of
the game, but the more practical way is to key a list of constraints into one of the
many
Simplex
applets
available
online,
for
example
at
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/simplex.html.
Matrices in Game Strategy
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To make all the elements in a payoff matrix positive, a constant r is added, which can
later be subtracted from the value of the game. Player R seeks to find v, such that it is
the minimum of the expected payoffs no matter what Player C plays.
If C plays column 1
a11p1 + a21p2 + … + am1pm ≥ v
If C plays column 2
a12p1 + a22p2 + … + am2pm ≥ v
If C plays column n
a1np1 + a2np2 + … + amnpm ≥ v
.
.
.
.
.
.
Thus, Player R seeks to find p1, p2, . . . , pm such that v is a maximum
subject to
a11p1 + a21p2 + … + am1pm – v ≥ 0
a12p1 + a22p2 + … + am2pm – v ≥ 0
.
.
.
.
.
.
a1np1 + a2np2 + … + amnpm – v ≥ 0
p1 + p2 + … + pm = 1
p1 ≥ 0, p2 ≥ 0, …, pm ≥ 0,
v ≥0
pi
Assuming that v > 0, since all entries of A are positive, and letting yi = v ,
we
1
get
y1 + y2 +…+ ym =
Thus v is a maximum only if y1 + y2 +…+ ym is a
v
minimum.
The problem is restated as
Minimize y1 + y2 + … + ym
subject to
a11y1 + a21y2 + … + am1ym ≥ 1
a12y1 + a22y2 + … + am2ym ≥ 1
.
.
.
.
.
.
.
.
.
a1ny1 + a2ny2 + … + amnym
.
.
.
≥1
y1 ≥ 0, y2 ≥ 0, …, ym ≥ 0,
These constraints are used to find the Player R’s optimal strategy and the value of the
game. Alternatively, Player C’s strategy is found by doing the reverse, which is as
follows
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Maximize x1 + x2 + … + xn
subject to
a11x1 + a12x2 + … + a1nxn ≥ 1
a21x1 + a22x2 + … + a2nxn ≥ 1
.
.
.
.
.
.
.
.
.
.
.
.
am1x1 + am2x2 + … + amnxn ≥ 1
x1 ≥ 0, x2 ≥ 0, …, xn ≥ 0
Large m X n matrices can also be simplified and then solved by eliminating
dominant/recessive rows and columns. If each element of the rth row of A is ≤ the
corresponding element of the sth row of A, then the rth row is recessive to the sth row,
and the sth row is said to dominate the rth row. And if each element of the rth column
of A ≥ the corresponding element in the sth column of A, then the rth column is called
recessive and the sth column is said to dominate the rth column. The recessive
columns and rows can be removed to reduce the size of the matrix.
A note on the types of games: Games where a payoff matrix is involved that is based
on the moves of two players R and C are known as two-person games or matrix
games. A second matrix representing payoffs of Player C can be constructed, but in
the class of games called constant-sum games, the sum of the payoff to R and the
payoff to C is constant for all mn pairs of RC moves. Thus a special kind of constantsum game is the zero-sum game, where the amount won by one player is exactly the
amount lost by the other player, and it is sufficient to study only the payoff matrix for
R. The fundamental theorem of matrix games is that every matrix game has a solution
(i.e. there are optimal strategies for R and C, and moreover, v = v').
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Procedure of Investigations
♦ We realized that Bridge was too complex to investigate within the time frame that
we had. Instead, we came up with several simplified number games that retained a
few of the rules of bridge, as stated in the objectives.
♦ Our first investigation involved starting out with a small sample of 3 cards per
player. To simplify the game, we assumed a so-called ‘telepathy’ between the
partners, where they would be able to discuss which cards they were going to
play. This allowed for the strategy, which we planned to investigate and find an
optimal solution to, to be the cards that the partners should pair up and play.
♦ After Investigation I was completed, we decided to investigate another game
model, which had randomly chosen payoffs, resulting in more variation in the
payoff matrix. In Investigations II and III, each player has 4 cards, and as we
found it difficult to retain the 13-cards-within-a-suit property of card games, we
modified them into number games.
♦ In Investigation II, the ‘telepathy’ has been removed and thus the game is quite
different from that in Investigation I, as the objective is changed to finding the
optimal strategy for the starting 2 cards.
♦ Investigation III was carried out at the same time as Investigation II. It was
approached the same way as Investigation I, but it involved the kind of varied
payoff matrix of Investigation II.
♦ Various forms of matrices were used to solve the examples we came up with for
each game model. The optimal strategies for investigations I and III yielded
patterns or generalizations. However, there was insufficient time to come up with
a generalization for Investigation II.
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Findings of Investigations
Investigation (I)
In the first of our investigations, we dealt three cards (within a deck of 3 to Ace) to
each of the four players, as this number was easier to work with within the grounds of
one suit. East and West pair up to play against North and South. Each player was to
play one card for each round; should the value of his card be higher than all other
cards played, he would win that trick for his team. The following example was
extracted from our portfolio, trial-played with the following cards:
North – Q, 8, 4
South – K, J, 6
East – A, 10, 3
West – 9, 7, 5
To simplify the game, we assumed that partners could discuss their strategy, thus
incorporating a type of ‘telepathy’. This allowed us to look into the pairing of each
team’s cards. We mapped out in a table how the cards could be distributed and the
chances of winning of each possible pair against every possible move of the opponent.
This was with the hypothesis that the number of tricks each team can win is related to
the way the cards are distributed and paired up. The first trick possibilities are as
follows; scoring was 1 point for each trick won and -1 point if the team loses a trick
(the table below shows scoring from East-West’s point of view. This zero-sum
scoring is equivalent to scoring {win +1, lose -1}, then netting off both teams’ scores
East-West Card Pairs
at the end of each game:
A9
A7
A5
10 9
10 7
10 5
39
37
35

QK QJ
1
1
1
1
1
1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
3
3
North-South Card Pairs
Q6 8K 8J 86 4K 4J 46
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
-1
-1 -1
1 -1 -1
1
-1
-1 -1
1 -1 -1
1
-1
-1 -1
1 -1 -1
1
-1
-1 -1
1 -1 -1
1
-1
-1 -1 -1 -1 -1
1
-1
-1 -1 -1 -1 -1 -1
3
3
3
-5
3
3

9
9
9
-5
-5
-5
-5
-7
-9
-7
Figure 2.1
The expected value for payoffs for all given openings are tabulated in a Microsoft
Excel spreadsheet below.
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E-W
N-S
1 -1
9
-1 -2 0 -10
-3 -4 -2 -12
-5 -6 -4 -14
Figure 2.2
At first it was felt that the
sum of net wins would
indicate a pattern, so we
formed the small table
above. We deducted net
wins of the North-South
team from those of EastWest, as the first table
reflecting
first
trick
possibilities presents the
game from East-West’s
point of view. It appears
that the saddle point of
this (in blue), when the
reference headings (in
green) are compared with
the document, it can be
seen that the cards with
the
most
frequent
occurrence of these values
actually form the optimal
first-trick strategy.
In this Microsoft Excel
spreadsheet, we divided
each net win by nine. This
was to weight the net wins
according
to
their
probability of occurrence
(only one card pair should
be played out of the
opponent’s
9
card
choices). The expected
values (EV) of the games
(each beginning with
different card pairs) were
then
compared.
The
largest EVs (indicated
with a tick) were then
recorded
for
further
investigation.
Matrices in Game Strategy
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As it turned out, the optimal first-trick strategy for each team would be to pair and
play the combination:
E-W Team to play:
N-S Team to play:
Min [max(East), max(West)]
Min [max(North), max(South)]
Min of player holding
Min of player holding
{max [max(East), max(West)]}
{max [max(North), max(South)]}
In our example, E-W Team would open with Min[A,9]=9 (the latter card being
West’s highest card). East obviously holds the highest of all 6 original cards and is
thus obliged to play Min{E hand=A,10,3}=3. In other words, this card trick would be
made up of the lower of the two highest cards from the team’s two hands and the
lowest from the second hand. Intuitively, this high-low combination is to be expected
as it implies less wastage. The next best trick is achieved by playing the highest card
of all eight and the lowest card from the second hand (E-W: A and 5). This is
followed by playing the middle-valued card of each hand (E-W: 10 and 7).
1. The table below shows the original “cost matrix” for the possible plays,
reflecting the number of tricks (total, not netted off) that are possible for EastWest:
W plays 5
W plays 7
W plays 9
E plays 3
0
-1
-2
E plays 10
-2
-2
-2
E plays A
-9
-9
-9
Figure 2.3
2. Subtract minimum of each row from respective row – subtract -9 from Row 1,
subtract -9 from Row 2, subtract -0 from Row 3:
W plays 5
W plays 7
W plays 9
E plays 3
9
8
7
E plays 10
7
7
7
E plays A
0
0
0
Figure 2.4
3. Since Column 3 fully contains zeros, it is settled for the time being. We
subtract the minimum of each column from respective column – subtract 7
from Column 1, subtract 7 from Column 2:
W plays 5
W plays 7
W plays 9
E plays 3
2
1
0
E plays 10
0
0
0
E plays A
0
0
0
Figure 2.5
Matrices in Game Strategy
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4. Minimum number of lines to cover all null entries is 3; optimal assignment of
zeros is possible, and reflected as follows:
E plays 3
2
1
0 
W plays 5
W plays 7
W plays 9
E plays 10
0
0 
0
E plays A
0 
0
0
Figure 2.6
5. Therefore, when we refer to the original table, it follows that this:
E plays 3
0
-1
-2
W plays 5
W plays 7
W plays 9
E plays 10
-2
-2
-2
E plays A
-9
-9
-9
Figure 2.7
is the optimal strategy for the original cost matrix. Note that our hypothesized
strategy for the first card (in this case East’s 3 and West’s 9) is also sustained
here.
To put this information in a diagram (where separate rows are of separate players’
cards arranged from the lowest-valued to the largest):
3
10
A
4
8
Q
5
7
9
6
J
K
3-card Hands
East-West Card Pairs
North-South Card Pairs
Figure 2.8
We then did the same for 4 cards and 5 cards per player, attempting to find a
generalization. It was successful for both cases (provided the deck is labeled 1-16 and
1-20 respectively, in accordance to card value). Here are 2 examples:
1
5
12
13
2
7
11
1
4
8
9
16
3
6
10
15
4-card Hands
East-West Card Pairs
North-South Card Pairs
Figure 2.9
Matrices in Game Strategy
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2
6
7
14
17
1
3
13
10
18
5
9
11
15
20
4
8
12
16
19
5-card Hands
East-West Card Pairs
North-South Card Pairs
Figure 2.10
So far, we have assumed a simple game in which each team member knows whether
his partner is playing his high, medium or low card, and the opposing team exhibits
random behaviour. There is a rational strategy to be derived. However, over a large
enough number of games, either team is as likely to draw a given hand. The EV of the
game is therefore 0 (“fair” game). [We could go one step further to look at the 2ndand 3rd-trick plays, which involve 4 possible combinations from each team.]
In the next section, we will modify the rules to explore more complex game scenarios.
Matrices in Game Strategy
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Findings of Investigation
Investigation (II)
In this second game model, we decided to come up with a varied payment that would
allow for more variation in the payoff matrices. We also adopted a different approach
from Investigation I by doing away with assumed collaboration between partners.
Value
North
East
South
West
1
1
2
3
4
2
8
7
6
5
3
9
10
11
12
4
16
15
14
13
Figure 3.1 One random example of a distribution
The game is created such that the winning pair gains a basic $0.30 from the losing
pair for every play. They then gain extra money from the losing pair based on the
summed value of the numbers they have been dealt:
Value
Payment
2
$1.80
3
$1.50
4
$1.20
5
$0.90
6
$0.60
7
$0.30
8
$0
Figure 3.2
Thus, the players strive to win each ‘trick’ with as small a summed up value as
possible. North always plays his card first, followed by East, South and then West.
We decided to define the strategy for this game as the cards which North and East
decided to play at the very beginning of each game, as the strategy in Investigation I
(how the partners should pair up their cards and play) would not work here without
‘telepathy’. We then mapped out every single way of playing a single distribution of
hand, e.g. Figure 3.4, and the average was taken for all the ways of playing for each
set, i.e. each way that North and East could start.
Matrices in Game Strategy
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Hand val
Payoff
Basic 03
2
1.8
2.1
3
1.5
1.8
4
1.2
1.5
5
0.9
1.2
6
0.6
0.9
7
0.3
0.6
8
0
0.3
Figure 3.3 Payoff system based on hand value




E
2
7
10
15
S
3
6
11
14




W
4
5
12
13




NS
Largest Won
4
0
8
1
12
0
16
1




EW
Won NS Val
1
0
0
4
1
0
0
8
Net
EW
Val NS Pts
2
0
0
1.5
6
0
0
0.3
Total
1.8
-1.2
Figure 3.4
The average of each set was written in a payoff matrix, figure 3.5, and we checked for
a saddle point. No saddle point was found so there was no pure optimal strategy and
we used the Simplex Method as described in the theoretical background to solve this
4x4 payoff matrix.
Starting cards of East
Starting cards
of North
N
1
8
9
16
2
7
10
15
1
0.198
-0.64
-0.002
0.110
8
0.331
0.499
-0.022 -0.189
9
0.255
0.408
-0.007 -0.020
16 -0.163 -0.151
0.206
0.713
Figure 3.5 Payoff matrix
Maximize p = x1 + x2 + x3 + x4
subject to
1.198x1 + 0.936x2 + 0.998x3 + 1.110x4 + s <= 1
1.331x1 + 1.499x2 + 0.978x3 + 0.811x4 + t <= 1
1.255x1 + 1.408x2 + 0.993x3 + 0.980x4
+ u <= 1
0.837x1 + 0.849x2 + 1.206x3 + 1.713x4
+ v <= 1
Which yields
p = 190/207; x1 = 20/69, x2 = 0, x3 = 130/207, x4 = 0
1
p = x1 + x2 + x3 + x4 = v
v = 1 / (190 / 207) = 207 / 190
q1 = x1v = (207 / 190)(20 / 69) = 6 / 19
q2 = x2v = (207 / 190)(0) = 0
q3 = x3v = (207 / 190)(130 / 207) = 13 / 19
q4 = x4v = (207 / 190)(0) = 0
Matrices in Game Strategy
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EW
Pts
-2.1
0
-0.9
0
-0.3
Therefore EW’s optimal strategy is
1 ≈ 0.089 (3 sig fig).
6/19
0
13/19
0
and the value of the game is 17/190 –
For Player R –
Minimize p = y1 + y2 + y3 + y4
subject to
1.198y1 + 1.331y2 + 1.255y3 + 0.837y4 + s >= 1
0.936y1 + 1.499y2 + 1.408y3 + 0.849y4 + t >= 1
0.998y1 + 0.978y2 + 0.993y3 + 1.206y4
+ u >= 1
1.110y1 + 0.811y2 + 0.980y3 + 1.713y4
+ v >= 1
Which yields
p = 190/207, y1 = 0, y2 = 205/437, y3 = 0, y4 = 346/771
p = y1 + y2 + y + y4 = 1
v = 1 / (190 / 207) = 207 / 190 v
p1 = y1v = (207 / 190)(0) = 0
p2 = y2v = (207 / 190)(205 / 437) = 369/722
p3 = y3v = (207 / 190)(0) = 0
p4 = y4v = (207 / 190)(346 / 771) = 353/722
Therefore NS’s optimal strategy is
0 369/722 0 353/722
We applied the same method to another distribution of cards, figure 3.6. The optimal
strategy was solved using the Simplex Method as well, and the answer is shown
below.
North
East
South
West
1
2
5
1
4
2
7
6
3
10
3
9
8
12
11
4
16
15
13
14
Figure 3.6
Starting cards of East
5
Starting cards
of North
Value
6
8
15
-0.212
2
-0.304 -0.224
0.001
7
-0.011
-0.334 -0.398
9
-0.195 -0.158
0.022
0.011
-0.376
16 -0.206 -0.364 -0.400
0.263
Figure 3.7 Payoff matrix
v ≈ -0.180 (3 sig fig)
NS [109/399 511/2000 1417/6251 4687/19152]
EW
654/3629
787/2500
2437/10000
4687/17396
We intended to investigate similar examples further, in order to propose a
generalization or pattern in the optimal strategies based on our findings. However,
due to time constraints, we only managed to solve two such examples.
Matrices in Game Strategy
Page 23
Findings of Investigation
Investigation (III)
As a continuation with our investigations, we furthered Investigation I and tested our
generalization (reflected in the diagram in Investigation I), by combining it with the
payoffs involved in Investigation II. The payoffs follow the same system as in
Investigation II but are different exponents of 10. However, upon application of the
same ranking style used in Investigation II, we found that it yielded tables like this
N,S / E,W
3, 15
9, 10
11, 5
16, 4
2, 14
-120
120
120
-120
7, 8
-120
-120
-120
-120
12, 6
-120
120
120
-120
13, 1
-120
120
120
-120
Figure 4.1
which were equivalent to what we already had and brought us nowhere. So we
applied a different ranking system, with the numbers 1-4 as Rank 1, 5-8 as Rank 2 etc.
N
E
S
W
2
3
1
4
7
9
6
5
12
11
8
10
13
16
14
15
Figure 4.2
∑
1620
1650
90
-1920
90
270
450
-960
90
270
450
-960
-1920
-1440
-960
-480
∑
NS/EW
3,4
3,5
3,10
3,15
9,4
9,5
9,10
9,15
11,4
11,5
11,10
11,15
16,4
16,5
16,10
16,15
-1800
2, 1
-210
-180
-150
-120
-150
-120
-90
-60
-150
-120
-90
-60
-120
-90
-60
-30
-1050
2, 6
180
180
-150
-120
-150
-120
-90
-60
-150
-120
-90
-60
-120
-90
-60
-30
-1050
2, 8
180
180
-150
-120
-150
-120
-90
-60
-150
-120
-90
-60
-120
-90
-60
-30
540
2, 14
120
120
120
-120
120
120
120
-60
120
120
120
-60
-120
-90
-60
-30
-1050
7, 1
180
180
-150
-120
-150
-120
-90
-60
-150
-120
-90
-60
-120
-90
-60
-30
-1110
7, 6
150
150
-150
-120
-150
-120
-90
-60
-150
-120
-90
-60
-120
-90
-60
-30
-1110
7, 8
150
150
-150
-120
-150
-120
-90
-60
-150
-120
-90
-60
-120
-90
-60
-30
270
7, 14
90
90
90
-120
90
90
90
-60
90
90
90
-60
-120
-90
-60
-30
810
12, 1
150
150
150
-120
150
150
150
-60
150
150
150
-60
-120
-90
-60
-30
540
12, 6
120
120
120
-120
120
120
120
-60
120
120
120
-60
-120
-90
-60
-30
540
12, 8
120
120
120
-120
120
120
120
-60
120
120
120
-60
-120
-90
-60
-30
0
12, 14
60
60
60
-120
60
60
60
-60
60
60
60
-60
-120
-90
-60
-30
540
13, 1
120
120
120
-120
120
120
120
-60
120
120
120
-60
-120
-90
-60
-30
270
13, 6
90
90
90
-120
90
90
90
-60
90
90
90
-60
-120
-90
-60
-30
270
13, 8
90
90
90
-120
90
90
90
-60
90
90
90
-60
-120
-90
-60
-30
-270
13, 14
30
30
30
-120
30
30
30
-60
30
30
30
-60
-120
-90
-60
-30
Figure 4.3 Table constructed from given hand in figure 4.1
Matrices in Game Strategy
Page 24
The optimal strategy in this case would be to find the four pairs of cards each team
should play, such that they gain the maximum payoff. By observing the total payoff
for all ways of using a pair, and taking into account that each card could only be
played once, we found that by applying the generalization for Investigation I, we were
still able to deduce the optimal strategy for both players. Thus, the pairs of cards N&S
would play are 2,14; 7,8; 12,6; 13,1 and for E&W they would play 3,15; 9,10; 11,5;
16,4. These strategies intersected to give the table below.
NS/EW
3,15
9,10
11,5
16,4
2, 14
-120
120
120
-120
7, 8
-120
-90
-120
-120
12, 6
-120
120
120
-120
13, 1
-120
120
120
-120
Figure 4.4
The eight pairs of cards had 24 different ways of being played in all (order not being
taken into account), which was found by taking 4! = 4 X 3 X 2 X 1 = 24. We then
created a payoff matrix, which showed the average outcome when starting with a
certain pair (once again, the order was not taken into account). It looked as such –
NS/EW
3,15
9,10
11,5
16,4
2, 14
-150
-80
-70
-150
7, 8
0
-210
-240
0
12, 6
-150
-80
-70
-150
13, 1
-150
-80
-70
-150
Figure 4.5
Using recessive and dominant rows and columns, the matrix can be reduced to
NS/EW
3, 15
9, 10
11, 5
2, 14
-150
-80
-70
7, 8
0
-210
-240
Figure 4.6
The pairs 2, 14 and 3, 15 are free to be changed with the eliminated rows and column
respectively, as they are actually equal. Applying the Simplex Method, we arrived at
NS’s strategy of [275/518 0 15/32 275/518] and EW’s strategy of
where the identical rows and columns had equal proportions and it
sufficed to look at one.
¾ ,
¼
¾
¾
The value of this game is –112.5. Looking at NS’s strategy more closely, we
discovered that 275/518 and 15/32 did not add up to one, but instead 0.99963803. By
Matrices in Game Strategy
Page 25
going back to the Simplex applet and solving the matrix again, but using decimals
instead of fractions for the tableaus, we found that the slight inaccuracy was attributed
to the rounding done during the conversion from decimal to fraction. More accurately,
NS’s strategy would be [17/32 0 15/32 17/32].
We managed to tabulate the strategies of six other randomly distributed hands, which
are as follow.
N
E
S
W
N
E
S
W
N
E
S
W
3
5
2
1
5
1
4
2
1
4
3
8
2
6
5
11
6
10
4
7
8
7
6
3
11
9
8
9
10
11
12
13
14
15
15
16
14
13
10
9
7
12
12
16
13
14
15
16
v = -30
v = 97.5
NS [¼ ½ ½ ¼ ]
EW
NS [335/1192 335/714 ¼ 0 ]
EW
¼
¼
½
½
N
E
S
W
8
1
12
5
11
4
9
7
13
6
3
14
15
10
2
16
v = 45
¼
¼
½
½
NS [0 130/513 5/18 15/32]
EW
¼
½
½
¼
½
½
¼
¼
N
E
S
W
N
E
S
W
3
5
1
2
6
2
5
1
11
10
4
6
10
4
8
3
12
14
7
9
12
7
9
11
15
16
8
13
13
14
16
15
v = -120
NS [½ ¼ ½ ¼]
EW
v = -90
NS [70/249 0 15/32 ¼]
EW
½
¼
½
¼
v=0
NS [160/569 ¼ 160/341 0]
EW
¼
¼
½
0
Figure 4.7
Note: The same problem as stated above occurred in some of our hands, but we were
unable to correct them by looking at the decimals. However, they are all within an
accuracy of 0.0005.
From the examples we created, we saw that whenever a team had matching
rows/columns, the strategy would be to the effect of the non-matching rows/columns
being ¼ each, while the matching ones would all get the same value such that only
adding one of them would give the correct value of 1. However, our first hand
contradicted this, so a general assumption would be that for most cases where a team
Matrices in Game Strategy
Page 26
sees matching rows/columns, a strategy can be applied as above, though it is not
always true.
Another interesting pattern observed was that the total payoff of each row and column
was always the same. Furthermore, Value of Game = Average of payoffs in matrix =
Average of 24 ways of play = 4 X Average of payoffs in 2nd table. We found that this
was both an easy and a quick way of enabling the players to know the eventual
outcome of their game and whose side it leaned towards.
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Conclusion
Generalizations could be found for Investigation I and III but not completely for
Investigation II due to time constraints.
In Investigation I, common number sense (backed slightly by saddle points and the
Hungarian Method) showed the general optimal strategy for that game model: the EW team played the cards Min [max(East), max(West)] and Min of player holding
{max [max(East), max(West)]}, while the N-S team used the same strategy, playing
Min [max(North), max(South)] and Min of player holding {max [max(North),
max(South)]}. This high-low combination can be reasoned out logically as it implies
less wastage.
In Investigation II, the simplex method was applied to two hands that we mapped out
in order to obtain the payoff matrices. Due to a lack of time, we did not manage to
map out more distributions of hands to come up with more examples of optimal
strategies for this particular game model. However, given more time, we would have
tried out more examples to come up with a generalization or pattern for the optimal
strategy as in Investigation I and Investigation III.
In Investigation III, we came up with several examples and distributions, and using
the Simplex Method to solve them, we saw that for 6 out of the 7 cases, whenever a
team had matching rows/columns, the strategy would be to the effect of the nonmatching rows/columns being ¼ each, while the matching ones would all get the
same value such that only adding one of them would give the correct value of 1. Thus
a general assumption would be that for most cases where a team sees matching
rows/columns, a strategy can be applied as above, though it is not always true.
From our project, it is possible to see that the optimal strategies to most game models
can be solved using matrices. We can also conclude that, once enough examples have
been worked out, one can usually find a pattern or general assumption about the
optimal strategy of that game model.
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Page 28
References
 Anton, Howard & Chris Rorres. (2000). Elementary Linear Algebra Applications
Version. Canada: Von Hoffman Press, Inc.
 Eric Weissten’s World of Mathematics. (1999-2002). Matrix - from MathWorld.
 http://mathworld.wolfram.com/Matrix.html
 http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/tutorialsf4/frames4_
4.html
 http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/tutorialsf4/frames4_3.html
Acknowledgements
♥ Our teacher-in-charge, Mrs. Sia Wai Leng, for her guidance and support.
♥ Our other teacher advisor, Mr. Kenneth Lui, for his suggestions of related
literature.
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