Thats a random walk on a bridge right? e.g. a drunk man stands on position 7 of a bridge that has length 20, and takes one step left or right every turn. If he falls of the closer side A wins.I've seen this problem but I don't know an easy solution technique...maybe someone could point one out?The only solution technique I know is this: considering the gambling problem:Let P(1) be the chance for A to win when B has 1 dollarLet P(2) be the chance for A to win when B has 2 dollars...Let P(N) be the chance for A to win when B has N dollars, which is the total,so P(N) = 0Let P(0) is the chance for A to win when B has 0 dollars, which means A won, so P(0) = 1Then we can get N equations:P(0) = 1P(1) = 0.5*1 + 0.5*P(2)P(2) = 0.5( P(1) + P(3) )P(3) = 0.5( P(2) + P(4) )...P(N) = 0We could invert this matrix, but these equations always specify P(i) as the midpoint of P(i+1) and P(i-1), so solution is a line, with P(N)=0, P(0)=1, giving P(i) = (N-i)/(N)(OK, notation should have been better).That is, the chance for A to win when B has i dollars and there's a total of N dollar on the table is (N-i)/(N) = (20-13)/20 = 7/20Is there an easier way????(Sorry before I had 8/21 on the last line...that was from a bad calculation)
Last edited by cdmurray80
on October 30th, 2006, 11:00 pm, edited 1 time in total.