Introduction game Theory 2007-08. Peter van Emde Boas

CHANCE MOVESWHY WOULD YOUEVEN CONSIDERTO PLAY THIS GAME?

Peter van Emde Boas

ILLC-FNWI-UvABronstee.com Software & Services B.V.

2006-07

See: http://staff.science.uva.nl/~peter/teaching/igt08.html

Introduction game Theory 2007-08. Peter van Emde Boas

Sorry; it is in French!Game of Chaos

Sorcery

Play head or tails against atarget opponent. The looserof the game looses one life.The winner of the game gainsone life, and may choose torepeat the procedure. For everyrepetition the ante in life isdoubled.

Introduction game Theory 2007-08. Peter van Emde Boas

CHANCE MOVES

• Chance moves controled by another player (Nature) who is not interested in the result

• Nature is bound to choose her moves fairly with respect to commonly known probabilities

• Resulting outcomes for active players become lotteries

Introduction game Theory 2007-08. Peter van Emde Boas

Flipping a coin

HEADS TAILS

1 / -1 -1/ 1 -1 / 1 1 / -1

h ht t

Thorgrim calls head or tails and Urgat flips the coin. Urgat’s move is irrelevant. Nature determines the outcome.

Introduction game Theory 2007-08. Peter van Emde Boas

Probability Theory

Set of outcomes: Probability measure : Pow( --> [0,1]

Conditions:Non degeneracy: ( ) = 0 , ( ) = 1Additivity: ( A B ) = (A) + (B) when A B = Additivity extends to countable disjoint unions ( Ai) = (Ai) , when Ai Aj = i ≠ j)

Introduction game Theory 2007-08. Peter van Emde Boas

Interpretation of probability• Frequentionalist

– If the experiment is repeated many many times, the proportion of instances where the event occurs converges to its probability

• Subjective– Related to the odds against which someone is

willing to place a fair bet on the outcome of the experiment:odds a against b ( a:b ) denotes probability

p = b/(a+b)

Introduction game Theory 2007-08. Peter van Emde Boas

Probability TheoryExample: Fair Dice

Set of outcomes: { 1, 2, 3, 4, 5, 6 }Probability measure (A) = #A / 6

with 2 dices the set of outcomes is. Each outcome has prob. 1/36assuming the dice are independent.(A B) = (A) (B) when A (B)involves the first (second) dice only

Introduction game Theory 2007-08. Peter van Emde Boas

Possible sets • is finite : {1,2,3,4,5,6}

• all computations finite (but tedious)• no problem with existence

• is countably infinite: {0,1,2,3,....}• few problems with existence (convergence)• few computations possible (analytic)

• is continuous: [0, 1]• problems with existence (not all events

measurable)• hard to compute

Introduction game Theory 2007-08. Peter van Emde Boas

Independent Events

E

F

E F

(E F) = (E) (F)

Introduction game Theory 2007-08. Peter van Emde Boas

Paying off an old debt

Azagh and Bolgh owe Urgat the amount of 1000 skulls.They only have 2 skulls. They decide to gamble in thetwo Lotteries organized in their tribe (Gork’s Whaagh support and Mork’s Wyvern protection campaign). Eachticket costs 1 skull and has a 1 in 2000 chance for a priceof exactly the 1000 skulls needed. Failure to pay has thetraditional consequences......Option 1: buy tickets in same lotteryOption 2: buy tickets in different lotteries

© the Games Workshop © the Games Workshop

Introduction game Theory 2007-08. Peter van Emde Boas

Conditional Probability

E FE

F

( E|F ) = (E F) /(F)

(E F) = ( E|F ) (F) == ( F|E ) (E)

( F|E ) = ( E|F ) (F) / (E)BAYES’ RULE

Introduction game Theory 2007-08. Peter van Emde Boas

Multiple Choice TestA Multiple choice test question has k possible answersStudents which have prepared themselves always answer rightStudents which went to the House Party always guess the answer

Only a fraction p of the students have prepared themselves

What is the probability that the right answer was obtained by guessing?

( unprepared | right ) = (right | unprepared) (unprepared) / (right)

(prepared | right ) = (right | prepared) (prepared) / (right)

q := 1 / (right) ( unprepared | right ) + (prepared | right ) = 1

Introduction game Theory 2007-08. Peter van Emde Boas

Multiple Choice TestSolving the Equations for X = ( unprepared | right ) :

X = (1-p).q /k ; 1-X = p.q : so 1 = ( (1-p)/k + p).q . Therefore

q = k/(1-p+pk) and X = (1-p)/(1-p+pk) = (k/k-1).1/(1+(k-1)p) - 1/(k-1)

X=1

X=0

p=0 p=1

k=1

k=3

Introduction game Theory 2007-08. Peter van Emde Boas

Random Variables

• Random Variable: function X: -> R

• Expectation: EE X = w X(w) (w)– for continuous W summation becomes

integration....

• Convergence guaranteed for bounded X ; but who guarantees that X is bounded ???!!

Introduction game Theory 2007-08. Peter van Emde Boas

Lotteries• A Lottery is just a stochastic variable

– outcomes < --- > draws– function value < --- > price collected

• Expectation of lottery < --- > expected price < --- > expected cost for participating

• Compound lottery : Price is free participation in another lottery....

Introduction game Theory 2007-08. Peter van Emde Boas

Lotteries

priceprob.

$31/3

$121/6

-$21/2

Expectation:1/2 . -2 + 1/6 . 12 + 1/3 . 3 = 2

Introduction game Theory 2007-08. Peter van Emde Boas

Compound Lottery

priceprob.

$31/3

$121/6

-$21/2

$31/2

-$21/2

1/5 4/5

priceprob.

$37/15

$121/30

-$21/2

In compound lotteries all drawings are assumed to be independent

Introduction game Theory 2007-08. Peter van Emde Boas

Game Values• Without chance moves

– Game value <---> Best result player can obtain from game

– Computed by Zermelo’s algorithm (Backward induction)

• With chance moves– Game value becomes a lottery– if final pay-off is numeric expectation can

be computed

Introduction game Theory 2007-08. Peter van Emde Boas

Flipping a coin

HEADS TAILS

1 / -1 -1/ 1 -1 / 1 1 / -1

h ht t1/21/2 1/21/2

Expectation 0 / 0 0 / 0

Introduction game Theory 2007-08. Peter van Emde Boas

Flipping a Biased coin

HEADS TAILS

1 / -1 -1/ 1 -1 / 1 1 / -1

h ht t2/31/3 2/31/3

Expectation -1/3 / 1/3 1/3 / -1/3

Value forThorgrim -1/3 1/3

Introduction game Theory 2007-08. Peter van Emde Boas

Lotteries as game Values

• in chapter 2 Binmore restricts himself to very simple lotteries

W L

p 1 - ppp

These lotteries are outcomes of games.Thorgrim prefers pp

over q q when p > q ;for Urgat the oppositeholds

Introduction game Theory 2007-08. Peter van Emde Boas

Comparing Complex Lotteries

0 1 0

$0M $1M $5M

0.01 0.89 0.10

$0M $1M $5M

0.89 0.11 0

$0M $1M $5M

0.9 0 0.1

$0M $1M $5M

??

??

Introduction game Theory 2007-08. Peter van Emde Boas

The St. Petersburg ParadoxFlip a fair coin until H appears

Price = 2 #T in this sequence

so H ---> 1 TH ---> 2 TTH ---> 4 TTTH ---> 8etc.

Expectation: 1/2 * 1 + 1/4 * 2 + 1/8 * 4 + 1/16 * 8 + ..... = 1/2 + 1/2 + 1/2 + 1/2 + ..... = But how much are you willing to pay for participation??!

Introduction game Theory 2007-08. Peter van Emde Boas

DUEL

The two warriors approach each other. Both are equippedwith a javelin which they can throw at their opponent. Theprobability of hitting the opponent is a decreasing function of the distance. The warrior who throws first and has missed his throw remains unarmed and is therefore doomed to die.

© the Games Workshop

Lord Tyrion

© the Games Workshop

Sir Urquard

Introduction game Theory 2007-08. Peter van Emde Boas

Duel

p100 q99 p98 q51 p50 q49 q1 p0

Distance

p100 q99 p98 q51 p50 q49 q1 p0

H HHHHHHHM M M M M M M M

T TTT TTT TDDDDD DD

100 99 98 51 50 49 1 0

U

UU

U

UU U

UT T

TT

T

T

TT

Introduction game Theory 2007-08. Peter van Emde Boas

Duel

p100 q99 p98 q51 p50 q49 q1 p0

Distance

p100 q99 p98 q51 p50 q49 q1 p0

T TTT TTT TDDDDD DD

100 99 98 51 50 49 1 0

1-p100 1-q99 1-p98 1-q51 1-p50 1-q49 1-q1 1-p0

1

0

(p + q)

Introduction game Theory 2007-08. Peter van Emde Boas

ParcheesiVery simplified version of traditional board game, like Mens erger je Niet,.....Players move tokens from start to target over trajectory.Distance moved determined by chance.Opponent’s tokens are reset to start when hit by your’s.

Urgat start

Thorgrim start

Target

Chance = Coin Flip:Heads = 2 Tails = 1Target may be overthrownwaiting allowed

Introduction game Theory 2007-08. Peter van Emde Boas

Parcheesi - PositionsUrgat to move

Thorgrim to move

1 2 3 4 5 6 7 8

9 10 11 12 13 14 15 16

TT

UU

Introduction game Theory 2007-08. Peter van Emde Boas

Parcheesi - Moves and values

1 2 3 4 5 6 7 8

9 10 11 12 13 14 15 16

TT

UU

1 1 a b c d e f

0 0 1-a 1-b 1-c 1-d 1-e 1-f1 1 a b c d e f

0 0 1-a 1-b 1-c 1-d 1-e 1-f

Introduction game Theory 2007-08. Peter van Emde Boas

Analyzing a Subgame

H T1/2 1/2

M MW W

3

10 10

UU

141

a

0 1-d 0

a = 1/2 ( 1 + 1-d ) = 1 - d/2

Introduction game Theory 2007-08. Peter van Emde Boas

Analyzing a Subgame

H T1/2 1/2

M MW W

6

10 1014

d

01-d 0

d = 1/2 (1-d ) ; d = 1/3a = 5/6

9 0

Introduction game Theory 2007-08. Peter van Emde Boas

Parcheesi - Moves and values

1 2 3 4 5 6 7 8

9 10 11 12 13 14 15 16

TT

UU

1 1 5/6 b c 1/3 e f

0 0 1/6 1-b 1-c 2/3 1-e 1-f1 1 5/6 b c 1/3 e f

0 0 1/6 1-b 1-c 2/3 1-e 1-f

1/2

1/2

1/2

1/2

1/21/2

1/21/2

Introduction game Theory 2007-08. Peter van Emde Boas

Analyzing a Subgame

H T1/2 1/2

M MW W

4

12 12

b

1-b 1-b

best move on T ?

b ≥ 5/6 ==>b = 1/2 (1 + 1/6) ==>b = 7/12 ==> false

Hence b ≤ 5/6b = 1/2 (1 + 1-b) b = 2/3

11 1/6

UU

Introduction game Theory 2007-08. Peter van Emde Boas

Analyzing Subgames

H T1/2 1/2

M MW W

5

15 15

c

1-e 1-e14 2/3

UU

H T1/2 1/2

M MW W

7

13 13

e

1-c 1-c12 1/311 1/6

e ≤ 1/3 ==> c = 1/2 ( 1 + 1-e) c + e/2 = 1e > 1/3 ==> c = 1/2 ( 1 + 2/3 ) = 5/6 ==> e = 1/2 ( 1/3 + 1/6) = 1/4 ==> false

1-c = e/2 ≤ 1/6 ==> e = 1/2 ( 1/6 + 1/3) = 1/4 & c = 7/8

Introduction game Theory 2007-08. Peter van Emde Boas

Analyzing Start Position

H T1/2 1/2

M MW W

8

16 16

f

1-f 1-f15 3/414 2/3

f ≥ 1/2 ; otherwise nobodymoves.....

hencef = 1/2 ( 2/3 + 3/4) f = 17/24which representsthe value ofthe game

Introduction game Theory 2007-08. Peter van Emde Boas

Parcheesi - Moves and values

1 2 3 4 5 6 7 8

9 10 11 12 13 14 15 16

TT

UU

1 1 5/6 2/3 1/3

0 0 1/6 1/3 1/8 2/3 3/4 7/241 1 5/6 2/3 7/8 1/3 1/4 17/24

0 0 1/6 1/3 2/37/8

1/81/4

3/417/24

7/24