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Unformatted text preview: STAT 455/855 – Midterm Solutions Fall, 2007 1. (15 marks) (a) (6 marks) Let p 1 be the probability that player 1 wins when he rolls first and let p 1 be the probability that player 1 wins if he doesn’t roll first. By symmetry, p 1 doesn’t depend on which of players 2 , . . . , 6 rolls first. Conditioning on the outcome of the first roll, we obtain p 1 = (1) 1 6 + p 1 5 6 and p 1 = p 1 1 6 + p 1 4 6 + (0) 1 6 . From the second equation we obtain p 1 = 1 2 p 1 . Plugging this into the first equa tion, we obtain p 1 = 1 6 + 5 12 p 1 , or p 1 = 2 / 7. (b) (3 marks) The number of times the die is rolled until a winner is declared is a Geometric random variable with parameter p = 1 / 6. Therefore, the expected number of times the die is rolled is 6. (c) (6 marks) Now we condition on 3 possible outcomes for the first roll: a 1 is rolled, a 2 is rolled, or neither a 1 or 2 is rolled. By symmetry, given that neither a 1 or 2 was rolled first, the conditional probability that player 1 wins is the same whether...
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This note was uploaded on 09/15/2010 for the course STAT 455/855 taught by Professor Glentakahara during the Fall '09 term at Queens University.
 Fall '09
 GLENTAKAHARA

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