Best Answer m00li, 06 May 2014 - 01:38 AM

Spoiler for no aha here(I) Lets look at a scenario where all blacks are finished first and only k whites are finished. No. of balls remaining are (w-k). The probability of this happening is

^{b-1+k}C_{k}(b*(b-1)*(b-2)..*1*w*(w-1)*(w-2)..(w-(k-1)) / (w+b)*(w+b-1)*...(w-(b+k-1))=

^{b-1+k}C_{k}(b!*k!^{w}C_{k}) / (b+k)!^{w+b}C_{b+k}= (b/(b+k))*(

^{w}C_{k}/^{w+b}C_{b+k})=(1/

^{w+b}C_{b})*(b/(b+k))*(^{b+k}C_{b})=(1/

^{w+b}C_{b})*^{b+k-1}C_{b-1 }therefore, expected number of balls remaining = (1/

^{w+b}C_{b})*(S_{0->w-1}[^{b+k-1}C_{b-1}(w-k)] + S_{0->b-1}[^{w+k-1}C_{w-1}(b-k)]) .... where S_{x->y }denotes sum of series from x to y

(II) Alternative(I) could have also be interpreted as:probability that w-k whites remain = (arranging b blacks and k whites such that last ball picked is a black)/(arranging b black and w white balls)

= (arranging b-1 blacks and k whites such that last ball picked is a black)/(arranging b black and w white balls)

= (

^{b+k-1}C_{b-1}/^{w+b}C_{b)}therefore, expected number of balls remaining = (1/

^{w+b}C_{b})*(S_{0->w-1}[^{b+k-1}C_{b-1}(w-k)] + S_{0->b-1}[^{w+k-1}C_{w-1}(b-k)]) .... where S_{x->y }denotes sum of series from x to y

... gotta rush.. will continue later

continuing...

S_{0->w-1}[^{b+k-1}C_{b-1}(w-k)] = w*S_{0->w-1}(^{b+k-1}C_{k}) - S_{0->w-1}(^{b+k-1}C_{k*}k) ...(A)

Using ^{n}C_{r + }^{n-1}C_{r-1} = ^{n+1}C_{r }: w*S_{0->w-1}(^{b+k-1}C_{b-1}) = w*^{b+w-1}C_{w-1}

_{Using }S_{0->n}^{s+k}C_{k}k = (s+1)^{s+n+1}C_{n-1: }S_{0->w-1}(^{b+k-1}C_{k*}k) = b*^{b+w-1}C_{w-2}

Hence, (A) = w*^{b+w-1}C_{w-1 - }b*^{b+w-1}C_{w-2 }= ^{b+w}C_{w-1}

Similarly S_{0->b-1}[^{w+k-1}C_{w-1}(b-k)] = b*^{w+b-1}C_{b-1 - w}*^{w+b-1}C_{b-2 }= ^{b+w}C_{b-1}

_{Therefore, expected number of balls remaining = }(^{b+w}C_{w-1} + ^{b+w}C_{b-1})/^{b+w}C_{b} = **w/(b+1) + b/(w+1)**