HW4Solutions (dragged)

# HW4Solutions (dragged) - p A The probability A rolls the...

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Home Work 4 Solutions IEOR 172, Fall 2015 Problem 3.74 For player A to roll a 9, they must roll { 6 , 3 } , { 3 , 6 } , { 5 , 4 } , or { 4 , 5 } , thus, the probability A will roll a 9 is p A = 4 36 = 1 9 . For player B to roll a 6, they must roll { 1 , 5 } , { 5 , 1 } , { 2 , 4 } , { 4 , 2 } , or { 3 , 3 } , thus, the probability B will roll a 6 is p B = 5 36 . Let E be the event that A wins. Let us condition on what happens on the first roll. P ( E ) = P ( E | A 9) P (A 9) + P ( E | A not 9) P (A not 9) = 1 · 1 9 + h P ( E | A not 9 , B 6) P (B 6 | A not 9) + P ( E | A not 9 , B not 6) P (B not 6 | A not 9) i (1 - 1 9 ) = 1 9 + 0 · 5 36 + P ( E ) · (1 - 5 36 ) (1 - 1 9 ) = 1 9 + P ( E ) · 31 36 · 8 9 Solving for P ( E ) we find that P ( E ) = 1 9 1 - 31 36 · 8 9 = 9 19 For another way, we consider the probability the game stops with A as the final roller with n rolls. The probability A rolls the final roll on the first roll is
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Unformatted text preview: p A The probability A rolls the fnal roll on an even roll is zero The probability A rolls the fnal roll on the third roll is (1-p A )(1-p B ) p A The probability A rolls the fnal roll on the f±th roll is (1-p A ) 2 (1-p B ) 2 p A The probability A rolls the fnal roll on the (2 n-1) st roll is (1-p A ) n-1 (1-p B ) n-1 p A Then, P ( E ) = 1 X n =1 (1-p A ) n-1 (1-p B ) n-1 p A = p A 1 X n =0 (1-p A ) n (1-p B ) n = p A ✓ 1 1-(1-p A )(1-p B ) ◆ = p A ✓ 1 1-(1-1 9 )(1-5 36 ) ◆ = 9 19 1...
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