思路跟sum of dice throwing类似

Label the squares using their horizontal and vertical indices. Let P(i,j) be the probability of reaching square (i,j) before returning to the center (3, 3). However, as far as probability is concerned, there are only 6 symmetrically different positions. For example,
P(1,1) = P(1,5) = P(5,1) = P(5,5)
P(1,2) = P(1,4) = P(2,5) = P(4,5) = P(5,4) = P(5,2) = P(4,1) = P(2,1)
P(1,3) = ...
P(2,2) = ...
P(2,3) = ...
P(3,3)

You can write several equations like
P(1,3) = 2P(2,3)/6 + 2P(1,2)/3
because there are 2 squares equivalent to (2,3) each having 1/6 of chance of reaching (1,3) in one move, and there are 2 squares equivalent to (1,2) each having 1/3 of chance of reaching (1,3) in one move.

Solve the equations and you get P(1,1) = 1/12.

所有跟帖： 

• If p(1,1)=1/12,p(1,1)=1/6p(2,3)+1/6p(3,2),p(2,3)=1??? -jinjing- ♀ (184 bytes) () 03/31/2010 postreply 17:18:21

• 检查一下你的推导：p(1,1)=1/6p(2,3)+1/6p(3,2), p(2,3)=1/4 -空指针异常- ♂ (66 bytes) () 04/01/2010 postreply 12:25:40

• you should be know:4p(23)=1,as p(23)=p(32)=p(24)=p(43) -jinjing- ♀ (120 bytes) () 04/01/2010 postreply 14:53:52

• 回复：you should know:4p(23)=1,as p(23)=p(32)=p(24)=p(43) -jinjing- ♀ (0 bytes) () 04/01/2010 postreply 14:55:22