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Introduction game Theory 2008-09. Peter van Emde Boas
CHANCE MOVESWHY WOULD YOUEVEN CONSIDERTO PLAY THIS GAME?
Peter van Emde Boas
ILLC-FNWI-UvABronstee.com Software & Services B.V.
2008-09
See: http://staff.science.uva.nl/~peter/teaching/igt09.html
Introduction game Theory 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. Peter van Emde Boas
Independent Events
E
F
E F
(E F) = (E) (F)
Introduction game Theory 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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 2008-09. 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