拙解, 不知道對不對

assume n=2k, then the probability of getting to the top of the ladder is
1/4 [(2k+1)/(4k)+2k/(4k)]+ 1/8 [(2k-1)/(4k)+(2k-2)/(4k)] +......+1/2^(k+1) [3/(4k)+2/(4k)]
for n large, ignoring the high order terms, the above is approximately 1/2-3/(8k)=1/2-3/(4n).