P X 50 \u03a6 50 1 2 175 3 p Rounding error is allowed d If Alice wins a

P x 50 φ 50 1 2 175 3 p rounding error is allowed d

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P ( X 50) Φ 50 + 1 / 2 - 175 / 3 p 875 / 36 ! 0 . 0559 . (Rounding error is allowed.) (d) If Alice wins a given round, then what is the expectation of the number on Alice’s die in that round?
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Let X be the number of Alice’s die given that she wins the first round. Then X { 2 , 3 , 4 , 5 , 6 } , and based on Alice’s number X , we know that the number on Bob’s die must be 1 , 2 , ..., X - 1. The distribution of X is P ( X = 2) = 1 4 · 1 6 7 12 = 1 14 , P ( X = 3) = 2 4 · 1 6 7 12 = 1 7 P ( X = 4) = 3 4 · 1 6 7 12 = 3 14 , P ( X = 5) = 4 4 · 1 6 7 12 = 2 7 P ( X = 6) = 4 4 · 1 6 7 12 = 2 7 Therefore the expectation of X (in fact, the conditional expectation) E[ X ] = 6 X k =2 kP ( X = k ) =2 · 1 14 + 3 · 1 7 + 4 · 3 14 + 5 · 2 7 + 6 · 2 7 = 32 7 . 4. (20pts) Suppose Alice and Bob are playing with the same dice as in Problem 3 (Alice with a 6-sided die and Bob with a 4-sided die), but the game rule has changed. In each round they both roll their dice. Whoever gets the first ace is the winner of the game. If they both get aces in the same round, then it is a draw. (a) Suppose they keep on rolling the dice until each of them get their own first ace. Let X and Y be the numbers of rounds required for Alice and Bob to get her and his first ace, respectively. What is the distribution of X and what is the distribution of Y ? X Geo(1 / 6), Y Geo(1 / 4). (b) What are the values of E[ X + Y ] and E[ XY ]? Since E( X ) = 1 / (1 / 6) = 6 and E( Y ) = 1 / (1 / 4) = 4, and X, Y are independent, we have E[ X + Y ] = E( X ) + E( Y ) = 6 + 4 = 10 , E[ XY ] = E( X ) · E( Y ) = 6 · 4 = 24 .
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(c) What is the probability that Alice wins the game? P (Alice wins the game) = P ( X < Y ) = X k =1 P ( X k - 1 | Y = k ) P ( Y = k ) = X k =1 P ( X k - 1) P ( Y = k ) = X k =1 k - 1 X j =1 P ( X = j ) P ( Y = k ) = X k =1 k - 1 X j =1 1 6 5 6 j - 1 1 4 3 4 k - 1 = 1 24 X k =1 3 4 k - 1 1 - ( 5 6 ) k - 1 1 - 5 6 = 1 24 X k =1 6 " 3 4 k - 1 - 3 4 · 5 6 k - 1 # = 1 4 " X k =1 3 4 k - 1 - X k =1 5 8 k - 1 # = 1 4 1 1 - 3 4 - 1 1 - 5 8 = 1 3 .
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5. (20pts) Two points each. No explanations needed.
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  • Spring '13
  • Mattingly
  • Probability, Probability theory, Alice, Dice, white ball

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