回複：i am just trying to help

let Ld=Li dided.Hd= Huang dided.
P(L^Hd)=.3
P(Ld^H)=(1-.3)*.5=.35
P(L^H)=(1-.3)*(1-.5)=.35---P(L^Hd)=.3,P(Ld^H)=(1-.3)*.5=.35,P(L^H)=(1-.3)*(1-.5)=.35-----.........
So,P(L^Hd)=.3+.3*.35+.3*.35*.35+.3*.35*.35*.35+...=.3/(1-.35)=.4615
P(Ld^H)=1-.4615=.5385.Huang is better.

所有跟帖： 

• Told you Huang is better. Infinite series are involved. -北鷹- ♂ (0 bytes) () 01/29/2010 postreply 20:10:12

• 回複：Told you Huang is better. Infinite series are involved. -jinjing- ♀ (45 bytes) () 01/29/2010 postreply 20:21:19