GEK 1544
Mathematics of Games
Your lecturer: Assoc. Prof. Leung Man Chun
Dept. of Mathematics
S- 17- 06 - 01
Tel. No. 6516-2758
e-mail: [email protected]
IVLE 
Expect to find lecture PowerPoint files
(simply scroll down the front page) and other
information on the course.
Lectures will be webcasted. Not under my
control. Please be mindful of poor quality and
delays on the webcast.
Lectures:
Mon & Thur 6 – 8 PM in
U-Town Auditorim 3 (Alice & Peter Tan Colleage)
Tutorials: Start on the third week.
Assessment:
•One Open Book Test without Internet or
Mobile Communication (about 1 hour; 20%).
Date of the Test: 16 March 2015 (Monday;
6:15pm – 7:15pm in U-Town Auditorim 3 ).
•Final Exam (80%): can bring in one (1) handwritten (on both sides) A-4 size help sheet.
Date of the Exam: 27-April-2014
(Mon evening).
Please double check the
IVLE timetable.
Refer to the Library for
past
exam. Papers.
Stata Centre in MIT
(funded partially by Bill Gate)
designed by the famed architect Frank Gehry,
an example of “deconstructionist” architecture.
In the movie "21", when Ben is celebrating
his birthday, the cake `says’
1, 1, 2, 3, 5, 8, 13, 21 ...
These are known as Fibonacci numbers
which follow the relation
cn
,
cn 1  cn  cn 1
( e. g.
21  13  8).
http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibnat.html
Classic TV game show Let's Make A Deal:
1
The better strategy is to switch doors.
How do we calculate probability?
Suppose there are countable (finite) number of
events, separated from one another, and each one
has equal likelihood, then (roughly speaking)
# of events in your favour
P
.
total # of events
Dealer
Low cards,
2, 3, 4, 5 & 6.
Player
High cards,
A, K, Q, J & 10,
A high concentration of Aces and Tens give a higher
chance of Blackjacks (which pay at 3:2 unless the
dealer also has Blackjack).
Low cards are good for the dealer, because the dealer
must hit stiff hands (that is, 12 – 16 points), unlike the
player who can hit or stand according to the strategy.
Tens bust all stiff hands and will increase the chance
the dealer will lose.
Basic Hi - Lo system:
-1 for each dealt of Ten and Ace,
+1 one for any value 2 - 6.
Values like 7, 8 & 9 do not affect the count.
Lord of the Dies.
Chevalier de Mere (1607-1684)
To get at least one `six’ in 4 rolls.
In keeping with his Salon methods, Gombaud enlisted
two famous `mathematicians’, Pascal and Fermat to
solve the problems. In a series of letters they laid the
foundation for the modern theory of probability
Why at this time and
why at this place.
Let’s start with the renowned mathematician
Descartes. We’ve heard
of his famous motto:
"Cogito;
ergo sum"
“I think,
therefore I am.”
René Descartes
(1596-1650)
Legend tells us that Descartes often woke
up late and spent the cold mornings in a
stove-heated room, contemplating
mathematics and philosophy. One day he
came out and got the idea of
doubt
According to Descartes, we should apply
our mind to question everything that is
not firmly rooted in “reason”.
Thus, to doubt is to think:
“I doubt, therefore I am.”
This is referred as Descartes’
doubt.
Believe
Idea
First action
Applying reason to
study chances.
Pascal (1623 – 1662)
Fermat (1601-1665)
A `six’ in four rolls:direct method.
D1
D2
D3
D4
1
P
D

(
)
.
1
 A six in the first roll.
6
 A six in the second roll. P( D2 )  16 .
 A six in the third roll. P( D3 )  1 .
6
 A six in the fourth roll. P( D4 )  1 .
6
1 1 1 1 4
At least one `six’ in four rolls      ?
6 6 6 6 6
What’s wrong? (Profit is based on 1 : 1.)
A six in four rolls:indirect method.
C1 
C2 
C3 
C4 
5
P
C

(
)
.
Not a six in the first roll.
1
6
5
P
C

(
)
.
Not a six in the second roll.
2
6
5
Not a six in the third roll. P(C3 )  .
6
Not a six in the fourth roll. P(C4 )  5 .
6
Not a `six’ in four rolls  C1  C2  C3  C4 .
P(C1  C2  C3  C4 )  ?
Together.
P ( Not [C1  C2  C3  C4 ] )  ?
A together B  A then B .
P( A  B )  P( A) * P( B ).
if and only if events A and B are independent.
That is, the results of the first experiment A have no
effect on the results of the second experiment B, and
vis versa.
Pascal became interested in problem that dates to medieval times,
if not earlier, the problem of the points. Suppose two players
agree to pay a certain number of games, and whoever be the first
to score five wins is the one who wins.
Unfortunately, they are interrupted before they can finish. How
should the stake be divided among them if, say, one has won four
games and the other has won three? Given that in each game the
two players have equal chance of winning.
Player A
Won 4 games.
Player B
Won 3 games.
Game 8
Game 9
Overall
A wins
A wins
B wins
B wins
A wins
B wins
A wins
B wins
A wins
A wins
A wins
B wins
1
Each case has equal probability 
2
for A & B .
Generalizations, binomial expansion and
expectation.
Blank page for you to write notes.
Flipping two fair coins
simultaneously.
Results:
H
H
T
Cases.
H
T
T
Only one case as
both coins have to be
head.
Two cases:
Ha
Tb
Hb
Ta
Only one case as
both coins have to be
tail.
When you toss two pennies there are three
possibilities,
Two Heads
Two Tails
One of each
=
So does that mean there is a 1/3 chance of getting
a double head?
Everyone learning statistics knows that is wrong.
In fact, if we label the coins A and B, there are
four possibilities,
Two Heads
Two Tails
A=head, B=tail
A=tail, B=head
So the chance of getting a double head is 1/4 not
1/3.
One day Bose was demonstrating the 'ultra-violet
catastrophe' to his students, that is, he was showing
them that theory predicts a curve far different from
experiments.
He made the equivalent error on predicting that two
pennies come down double heads one time in three –
Surprisingly, the calculation came out in accordance
with experiment.
An embarrassing mistake if you set out to prove that it
didn't. An even more embarrassing error when he
realized he made a “kindergarten” mistake.
In desperation, he wrote to Einstein, who saw at once
what it meant:
Two photons (light particles) are fundamentally
indistinguishable.
Therefore the mathematics should make no
distinction between photon A and photon B,
just treat them as 'two photons'.
Dice.
Two dice totals
Die 2
Die 1
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
11
12
10
Total
Number of
combinations
Probability
2
1
2.78%
3
2
5.56%
4
3
8.33%
5
4
11.11%
6
5
13.89%
7
6
16.67%
8
5
13.89%
9
4
11.11%
10
3
8.33%
11
2
5.56%
12
1
2.78%
Total
36
100%
Cards and probabilities.
Four French suits: diamonds (♦), spades (♠),
hearts (♥) and clubs (♣),
* Probability of being dealt a pair in two cards is 3/51.
* The probability of being dealt 5 Spades (called a Spade flush) in 5
cards is
13 12 11 10 9
1
. . . . 
.
52 51 50 49 48 2000
Later in the course, we will see that:
~0.0000154
~0.00024
~0.0014
~0.00197
~0.004
Excluding
Straight flush