HW4Solutions (dragged) 2

# HW4Solutions (dragged) 2 - P n,m = n-m n m,n ≥ m...

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We will assume all sequences are equally likely. To calculate P 2 , 1 , the only sequence where A leads B for all counts are { AAB } . The total number of sequences are ( n + m )! n ! m ! = (2+1)! 2!1! = 3, thus, P 2 , 1 = 1 3 . To calculate P 3 , 1 , the only sequences where A leads B for all counts are { AAAB, AABA } . The total number of sequences are ( n + m )! n ! m ! = (3+1)! 3!1! = 4, thus, P 3 , 1 = 2 4 = 1 2 . To calculate P 4 , 1 , the only sequences where A leads B for all counts are { AAAAB, AAABA, AABAA } . The total number of sequences are ( n + m )! n ! m ! = (4+1)! 4!1! = 5, thus, P 3 , 1 = 3 5 . To calculate P 3 , 2 , the only sequences where A leads B for all counts are { AAABB, AABAB } . The total number of sequences are ( n + m )! n ! m ! = (3+2)! 3!2! = 10, thus, P 3 , 1 = 1 5 . Similarly, P 4 , 2 = 1 3 and P 4 , 3 = 1 7 . Theoretical Exercise 3.21 b. P n, 1 = P (A receives first 2 votes) = n n + 1 n - 1 n = n - 1 n + 1 P n, 2 = P (A receives first 2 votes and at least 1 of the next 2 = n n + 2 n - 1 n + 1 1 - 2 n 1 n - 1 = n - 2 n + 2 Theoretical Exercise 3.21 c.
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Unformatted text preview: P n,m = n-m n + m ,n ≥ m Theoretical Exercise 3.21 d. P n,m = P (A always ahead) = P (A always ahead | A receives last vote) n n + m + P (A always ahead | B receives last vote) m n + m = n n + m P n-1 ,m + m n + m P n,m-1 Theoretical Exercise 3.21 e. The conjecture of (c) is true when n + m = 1 ( n = 1 ,m = 0). Assume by induction that P n,m = n-m n + m when n + m = k . Now suppose that n + m = k + 1. By (d) and the induction hypothesis, we have that P n,m = n n + m n-1-m n-1 + m + m n + m n-m + 1 n + m-1 = n-m n + m (1) which completes the proof. Theoretical Exercise 3.22 If the weather is dry on January 1st, then P = P (dry on Jan 1st) = 1 and, conditioning 3...
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