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).

所有跟帖： 

• 飛魚兄很厲害呀，在哪裏做quant? -康MM- ♀ (0 bytes) () 07/20/2009 postreply 16:10:44

• 回複：飛魚兄很厲害呀，在哪裏做quant? -雪山飛魚- ♂ (279 bytes) () 07/20/2009 postreply 19:17:00

• Can 康MM explain... -Commentate- ♂ (50 bytes) () 07/21/2009 postreply 09:53:18

• 回複：Can 康MM explain... -康MM- ♀ (123 bytes) () 07/22/2009 postreply 09:24:57

• 哇，這麽費解，還麵試題呢。 -Commentate- ♂ (0 bytes) () 07/23/2009 postreply 16:03:19