GEK1544 The Mathematics of Games
                                  Tutorial 2 – Suggested Solutions




1.    Using the pr...
Then
                                    1       1       1    1
                       P (A ∪ B ∪ C ∪ D) = 4 ·
           ...
Hence
                           24                  24
                    35                    35
          R 1−       ...
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S 2

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S 2

  1. 1. GEK1544 The Mathematics of Games Tutorial 2 – Suggested Solutions 1. Using the property on probability : P (A ∪ B) = P (A) + P (B) − P (A ∩ B) . (1) show that P (A∪B∪C) = P (A)+P (B)+P (C)−P (A∩B)−P (A∩C)−P (B∩C)+P (A∩B∩C) . (2) Guess the formula for (no need to prove it) P (A ∪ B ∪ C ∪ D) . (3) Using the formula for (3), argue in the positive direction that the probability of ob- 671 taining at least one six in four rolls of a single dice is 1296 , exactly the same result as Pascal’s argument in the negative direction. Suggested solution. Let Z = A ∪ B . We infer that P (A ∪ B ∪ C) = P (Z ∪ C) = P (Z) + P (C) − P (Z ∩ C) = P (A ∪ B) + P (C) − P (C ∩ Z) = P (A) + P (B) − P (A ∩ B) + P (C) − P (C ∩ (A ∪ B)) = P (A) + P (B) + P (C) − P (A ∩ B) − P ([C ∩ A] ∪ [C ∩ B]) = P (A) + P (B) + P (C) − P (A ∩ B) − {P (C ∩ A) + P (C ∩ B) − P ([C ∩ A] ∩ [C ∩ B])} = P (A) + P (B) + P (C) − P (A ∩ B) − {P (A ∩ C) + P (B ∩ C) − P (A ∩ B ∩ C)} = P (A) + P (B) + P (C) − P (A ∩ B) − P (A ∩ C) − P (B ∩ C) + P (A ∩ B ∩ C) . Based on the “ - + -” order, we conjecture that P (A ∪ B ∪ C ∪ D) = P (A) + P (B) + P (C) + P (D) − P (A ∩ B) − P (A ∩ C) − P (A ∩ D) − P (B ∩ C) − P (B ∩ D) − P (C ∩ D) + P (A ∩ B ∩ C) + P (A ∩ B ∩ D) + P (B ∩ C ∩ D) + P (C ∩ D ∩ A) − P (A ∩ B ∩ C ∩ D) . Set A = {obtain a 6 in 1st roll ; doesn’t matter what numbers in the other three rolls } , B = {obtain a 6 in 2nd roll ; doesn’t matter what numbers in the other three rolls } , C = {obtain a 6 in 3rd roll ; doesn’t matter what numbers in the other three rolls } , D = {obtain a 6 in 4th roll ; doesn’t matter what numbers in the other three rolls } .
  2. 2. Then 1 1 1 1 P (A ∪ B ∪ C ∪ D) = 4 · −6· 2 +4· 3 − 4 6 6 6 6 3 3 4·6 −6 +4·6−1 864 − 216 + 24 − 1 671 = 4 = 4 = . 6 6 1296 Observe also that 4 671 5 = 1− 1296 6 4 1 1 1 1 1 = 1− 1− =1− 1−4· +6· 2 −4· 3 + 4 6 6 6 6 6 4! 2 binomial expansion : (1 − x)4 = 1 − 4x + x − 4x3 + x4 2!2! 1 1 1 1 = 4· −6· 2 +4· 3 − 4 . 6 6 6 6 2. Suppose you are dealt 3, 4, 5, as the first three cards of a poker hand, and there are two subsequent cards to be served to you. What is the probability of getting a straight (i.e., getting “A, 2, 3, 4, 5,” ; “2, 3, 4, 5, 6” or “3, 4, 5, 6, 7”) ? Suggested solution. Let P ({A} then {2}) denotes the probability of first getting an A, then 2 . Likewise for other notations. We have 4 4 P ({A} then {2}) = ∗ . 49 48 It follows that P (getting a straight) = P ({A} then {2}) + P ({2} then {A}) + P ({2} then {6}) + P ({6} then {2}) + P ({6} then {7}) + P ({7} then {6}) 4 4 = 6∗ ∗ . 49 48 Note that each case in step two above have empty intersection with any other cases. 3. After Chevalier de Mere’s found out the probability of obtaining one or more double sixes in 24 rolls of a pair of dice, and saw that it is less than half, he stopped playing the game. Define a new house odd R : 1 so that de Mere’s would think it is a fair game and consider playing it again. Suggested solution. To be a fair game, we need to find R so that R ∗ P (win) + [ −1] ∗ P (not win) = 0 . Here 24 35 P (not win) = (not a pair of ‘six’ in anyone of the 24 rolls). 36
  3. 3. Hence 24 24 35 35 R 1− + (−1) =0 36 36 That is 24 35 36 0.5086 R= 24 ≈ ≈ 1.035 . 1− 35 0.4914 36

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