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The Jelly Beans problem has inspired me to write the following problem:There are two pockets, a left pocket and a right pocket. Let (L(0),R(0)) denote the initial numbers of beans in each pocket respectively. Let the sequence (L(i),R(i)), where i=1,2,3,..., denote the number of beans in left and right pockets respectively after a single bean is removed from ONLY one of the two pockets on the i-th removal. For the i-th removal of a bean, either the left pocket is selected with probability p(i)={L(i-1)+1}/{L(i-1)+R(i-1)+2} or the right pocket is selected with probability q(i)={R(i-1)+1}/{L(i-1)+R(i-1)+2}. This process ends when an empty pocket is selected for the removal of a bean. When this happens, let X denote the number of beans in the other pocket. Find E[X].You may want to do this problem for L(0)=R(0)=4.

Let L(0)=m, R(0)=n, i.e. we start at (m,n).E(X)=Sum_i {i*P(we reach (i,0))*P(we reach out into right pocket)}+Sum_ j {j*P(we reach (0,j))*P(we reach out into left pocket)}Note that the probability of each path from (m,n) to (i,0)*P(we reach out into right pocket) is the same: (the last fraction being the probability that we reach out into the empty right pocket, with i beans in left pocket)And, the number of paths from (m,n) to (i,0) is So