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# 11. Game Theory ( 不是電玩理論 ... ) - PowerPoint PPT Presentation

11. Game Theory ( 不是電玩理論 ... ). 人生充滿著衝突與競爭。 。。 有太多的決策，不是自己說了算 ! 前面所有的分析，決策者都只有一個人，其他因素都是經驗累積而來的數據，或是統計後的或然率。當影響決策不是一個人時。。。.

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11. Game Theory(不是電玩理論 ... )

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11.1 Definition ofPayoff matrix (or table)

• Payoff matrix (or table)

• An m×n matrix is called a payoff matrix of a game if it satisfies,

• There are only two players R and C. (such as 2 persons, 2 companies or 2 nations ...)

• Player R has m choices, and player C has n choices.

• If R chooses alternative Ri, and C chooses alternative Cj then aij denotes the payoff of C to R.

Payoff matrix

-1

-1

-1

2

-1

-1

-1

2

Example 11.1

R與C兩人猜拳，剪刀/石頭/布。勝負之payoffs協議如下: (正值代表R獲利，負值時代表R損失。)

1. 一樣，R賺2元。

2. 不同，R賠1元。

Payoff matrix

11.2 Zero sum game (零和遊戲)

Two persons zero sum game means that the sum of total payoffs of players R and C is zero! On the other words, any one’s gain is the other’s lost.

Column player

y=(y1, y2, ..., yn)是一個probability vector，yj代表C採用Cj策略的機率。所以 ∑j=1 to n(yj) = 1。

Row player

x=(x1, x2, ..., xm)是一個probability vector，xi代表R採用Ri策略的機率。所以 ∑i=1 to m(xi) = 1。

R的立場

III IV

Example 11.1 -續

R與C兩人猜拳，剪刀/石頭/布。勝負之payoffs協議如下: (正值代表R獲利，負值時代表R損失。)

R的probability vector

7-11:

Payoff指全家最少可以吸引到顧客幾千人。。。

- 全家廣告後，其獲利真的就必然是7-11的損失嗎?

Let G=(R, C, A) denote a two persons zero sum game if there is a value VG such that VG = MaxRMinC A(ri, cj) = MinCMaxR A(ri, cj) then VG is said the pure value of G and its location in A is said a saddle point.

3

3

MaxR{MinC A(ri, cj)}

R player的最佳策略: r2

3

VG

The pure value of G

MinC{MaxR A(ri, cj)}

C player的最佳策略: c2

b

c

d

Example 11.3

a12

Theorem 11.1

A 2×2 payoff matrix A= [ ] is a non-strictly determined game if and only if

Max{b, c} < Min {a, d} or

Max{a, d} < Min{b, c}

a12

MinC A(ri, cj)

0

6

1

0

2

3

5

3

5

7

1

-3

6

-4

-4

MaxR A(ri, cj)

6

3

6

7

11.4 Mixed strategy (混合策略)

For a given G=(R, C, A) a two persons zero sum game, any strategy of x or y is said to be a pure strategy if its probability vector exists a component with value 1 (i.e., the others are all zero), otherwise it is a mixed strategy.

MaxR{MinC A(ri, cj)}

R player的最佳策略: r2

[0, 1, 0]

A strictly determined game has pure strategy.

A non-strictly determined game has mixed strategy.

MinC{MaxR A(ri, cj)}

C player的最佳策略: c2

[0, 1, 0, 0]

Example 11.3

R player 混合策略比值:

[6/13, 7/13]

6

7

10

3

C player選擇c2策略後，R player會失去當C player選擇c1策略時，所能帶來的可能payoff之誤差。(這是C player選擇後唯一可以確知的事)

C player 混合策略比值:

[3/13, 10/13]

Example 11.4: 2×3 Payoff matrix

c1: 17/8

c2: -13/8

c3: -13/8

R player

[3/8, 5/8]

C player

[9/10, 1/10]

R player

[9/14, 5/14]

R player

[12/22, 10/22]

Why?

C player選c1時，R payoff的期望值為

-6×(12/22)+7×(10/22) = -2/22

C player

[9/22, 13/22]

C player

[1/14, 13/14]

-6×(9/14)+7×(5/14) = -19/14

Example 11.5: 3×3 Payoff matrix

[r1 odd:r2 odd:r3 odd]，先降行…

= [6:6:48] = [1:1:8]

|10×1－(-2)×(-2)| = 6 ← r1 odd

|6×1－(-2)×(-6)| = 6 ← r2 odd

|6×(-2)－(-6)×(10)| = 48 ← r3 odd

= [38:14:8] = [19:7:4]

• If C takes pure strategy c1 then the expected payoff of R is (6×1+8×1+4×8)/10 = 23/5, and

• the same expected payoff is come out in c2 and c3 cases.

• 2. If R takes pure strategy r1, r2 or r3 then C has the same expected payoff (loss) 23/5 will be figured out.

= |2×(-5)－6×(-8)|

= 38 ← c1 odd

|-2×(-5)－6×4| = 14 ← c2 odd

|-2×(-8)－2×4| = 8 ← c3 odd