Expected Balls

Solution

Let the initial contents of the two bags be \[ (B_1,W_1) \quad \text{and} \quad (B_2,W_2), \] where \(B_i\) is the number of black balls in bag \(i\), and \(W_i\) is the number of white balls in bag \(i\). The rule says that at each step, you remove a white ball from one of the bags, whenever one is available. No black ball is ever removed. So throughout the entire process, \[ \text{number of black balls in bag 1} = B_1, \qquad \text{number of black balls in bag 2} = B_2. \] When the process ends, all white balls are gone, but every black ball is still there. Therefore the number of black balls remaining is not random at all: \[ B_1+B_2. \] So the expected number of black balls remaining is simply \[ \mathbb{E}[\text{black balls remaining}] = B_1+B_2. \] The answer depends on the initial composition only through the initial number of black balls. The white-ball counts affect how long the process lasts, but they do not affect the final number of black balls. So if initially the bags contain \[ (B_1,W_1),\;(B_2,W_2), \] then the expected number of black balls remaining at the end is always \[ B_1+B_2. \] In other words, changing the initial white counts changes nothing about the final black total, while changing the initial black counts changes the answer exactly by the same amount.