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Introduction game Theory 2008-09. Peter van Emde Boas CHANCE MOVES WHY WOULD YOU EVEN CONSIDER TO PLAY THIS GAME? Peter van Emde Boas ILLC-FNWI-UvA Bronstee.com Software & Services B.V. 2008-09 See: http://staff.science.uva.nl/~peter/teaching/igt09.html

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Introduction game Theory 2008-09. Peter van Emde Boas Sorry; it is in French! Game of Chaos Sorcery Play head or tails against a target opponent. The looser of the game looses one life. The winner of the game gains one life, and may choose to repeat the procedure. For every repetition the ante in life is doubled.

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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

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Introduction game Theory 2008-09. Peter van Emde Boas Flipping a coin HEADSTAILS 1 / -1-1/ 1 1 / -1 hhtt Thorgrim calls head or tails and Urgat flips the coin. Urgat’s move is irrelevant. Nature determines the outcome.

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Introduction game Theory 2008-09. Peter van Emde Boas Probability Theory Set of outcomes: Probability measure : Pow( --> [0,1] Conditions: Non degeneracy: ( ) = 0, ( ) = 1 Additivity: ( A B ) = (A) + (B) when A B = Additivity extends to countable disjoint unions ( A i ) = (A i ), when A i A j = i ≠ j)

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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)

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Introduction game Theory 2008-09. Peter van Emde Boas Probability Theory Example: 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. 1/36 assuming the dice are independent. (A B) = (A) (B) when A (B) involves the first (second) dice only

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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

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Introduction game Theory 2008-09. Peter van Emde Boas Independent Events E F E F (E F) = (E) (F)

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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 the two Lotteries organized in their tribe (Gork’s Whaagh support and Mork’s Wyvern protection campaign). Each ticket costs 1 skull and has a 1 in 2000 chance for a price of exactly the 1000 skulls needed. Failure to pay has the traditional consequences...... Option 1: buy tickets in same lottery Option 2: buy tickets in different lotteries © the Games Workshop

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Introduction game Theory 2008-09. Peter van Emde Boas Conditional Probability E F E F ( E|F ) = (E F) / (F) (E F) = ( E|F ) (F) = = ( F|E ) (E) ( F|E ) = ( E|F ) (F) / (E) BAYES’ RULE

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Introduction game Theory 2008-09. Peter van Emde Boas Multiple Choice Test A Multiple choice test question has k possible answers Students which have prepared themselves always answer right Students 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

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Introduction game Theory 2008-09. Peter van Emde Boas Multiple Choice Test Solving 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=0p=1 k=1 k=3

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Introduction game Theory 2008-09. Peter van Emde Boas Random Variables Random Variable: function X: -> R EExpectation: E X = w X(w) (w) –for continuous W summation becomes integration.... Convergence guaranteed for bounded X ; but who guarantees that X is bounded ???!!

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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....

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Introduction game Theory 2008-09. Peter van Emde Boas Lotteries price prob. $3 1/3 $12 1/6 -$2 1/2 Expectation: 1/2. -2 + 1/6. 12 + 1/3. 3 = 2

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Introduction game Theory 2008-09. Peter van Emde Boas Compound Lottery price prob. $3 1/3 $12 1/6 -$2 1/2 $3 1/2 -$2 1/2 1/54/5 price prob. $3 7/15 $12 1/30 -$2 1/2 In compound lotteries all drawings are assumed to be independent

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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

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Introduction game Theory 2008-09. Peter van Emde Boas Flipping a coin HEADSTAILS 1 / -1-1/ 1 1 / -1 hhtt 1/2 Expectation0 / 0

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Introduction game Theory 2008-09. Peter van Emde Boas Flipping a Biased coin HEADSTAILS 1 / -1-1/ 1 1 / -1 hhtt 2/31/32/31/3 Expectation -1/3 / 1/31/3 / -1/3 Value for Thorgrim -1/31/3

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Introduction game Theory 2008-09. Peter van Emde Boas Lotteries as game Values in chapter 2 Binmore restricts himself to very simple lotteries WL p1 - p p These lotteries are outcomes of games. p Thorgrim prefers p q over q when p > q ; for Urgat the opposite holds

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Introduction game Theory 2008-09. Peter van Emde Boas Comparing Complex Lotteries 010 $0M$1M$5M 0.010.890.10 $0M$1M$5M 0.890.110 $0M$1M$5M 0.900.1 $0M$1M$5M ??

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Introduction game Theory 2008-09. Peter van Emde Boas The St. Petersburg Paradox Flip a fair coin until H appears Price = 2 #T in this sequence so H ---> 1 TH ---> 2 TTH ---> 4 TTTH ---> 8 etc. 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??!

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Introduction game Theory 2008-09. Peter van Emde Boas DUEL The two warriors approach each other. Both are equipped with a javelin which they can throw at their opponent. The probability 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

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Introduction game Theory 2008-09. Peter van Emde Boas Duel p100q99p98q51p50q49q1p0 Distance p100q99p98q51p50q49q1p0 HHHHHHHHMMMMMMMM TTTTTTTT DDD DD D D 100999851504910 U U U U U U U U TT TT T T T T

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Introduction game Theory 2008-09. Peter van Emde Boas Duel p100q99p98q51p50q49q1p0 Distance p100q99p98q51p50q49q1p0 TTTTTTTT DDD DD D D 100999851504910 1-p1001-q991-p981-q511-p501-q491-q11-p0 1 0 (p + q)

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Introduction game Theory 2008-09. Peter van Emde Boas Parcheesi Very 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 = 1 Target may be overthrown waiting allowed

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Introduction game Theory 2008-09. Peter van Emde Boas Parcheesi - Positions Urgat to move Thorgrim to move 12345678 910111213141516 T U

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Introduction game Theory 2008-09. Peter van Emde Boas Parcheesi - Moves and values 12345678 910111213141516 T U 11abcdef 001-a1-b1-c1-d1-e1-f 11abcdef 001-a1-b1-c1-d1-e1-f

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Introduction game Theory 2008-09. Peter van Emde Boas Analyzing a Subgame HT 1/2 MMWW 3 10 U 14 1 a 01-d0 a = 1/2 ( 1 + 1-d ) = 1 - d/2

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Introduction game Theory 2008-09. Peter van Emde Boas Analyzing a Subgame HT 1/2 MMWW 6 10 14 d 01-d0 d = 1/2 (1-d ) ; d = 1/3 a = 5/6 9 0

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Introduction game Theory 2008-09. Peter van Emde Boas Parcheesi - Moves and values 12345678 910111213141516 T U 115/6bc1/3ef 001/61-b1-c2/31-e1-f 115/6bc1/3ef 001/61-b1-c2/31-e1-f 1/2

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Introduction game Theory 2008-09. Peter van Emde Boas Analyzing a Subgame HT 1/2 MMWW 4 12 b 1-b best move on T ? b ≥ 5/6 ==> b = 1/2 (1 + 1/6) ==> b = 7/12 ==> false Hence b ≤ 5/6 b = 1/2 (1 + 1-b) b = 2/3 11 1/6 U

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Introduction game Theory 2008-09. Peter van Emde Boas Analyzing Subgames HT 1/2 MMWW 5 15 c 1-e 14 2/3 U HT 1/2 MMWW 7 13 e 1-c 12 1/3 11 1/6 e ≤ 1/3 ==> c = 1/2 ( 1 + 1-e) c + e/2 = 1 e > 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

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Introduction game Theory 2008-09. Peter van Emde Boas Analyzing Start Position HT 1/2 MMWW 8 16 f 1-f 15 3/4 14 2/3 f ≥ 1/2 ; otherwise nobody moves..... hence f = 1/2 ( 2/3 + 3/4) f = 17/24 which represents the value of the game

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Introduction game Theory 2008-09. Peter van Emde Boas Parcheesi - Moves and values 12345678 910111213141516 T U 115/62/31/3 001/61/31/82/33/47/24 115/62/37/81/31/417/24 001/61/32/3 7/8 1/8 1/4 3/4 17/24 7/24

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