| [ QuizWit ] in KIDS 글 쓴 이(By): parsec ( 먼 소 류 ) 날 짜 (Date): 2003년 3월 22일 토요일 오전 10시 14분 28초 제 목(Title): Re: random walk. it's too hard a math for me. counting 0,1,2,3, .. is all i can do, with pains. but let me try it with this tedious nogada. the chance to be at position y after n steps will be the sum of all chances that steps to right(say j) minus steps to left(say k) equals y. j+k=n, j-k=y. 2j = n+y, 2k = n-y. thus the answer will be C(n,(n+y)/2) * p^((n+y)/2) * (1-p)^((n-y)/2) for n-y ∈ 2N, and 0 for n-y ∈ 2N+1. my little gray stuff smokes... /* - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ... May the source be with you! - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - */ |