Universal probability for k number of containers

Question

We have k containers, in each of them a number of white and b number of black balls. We select one random ball from the first, throw it into the second. Then we pick a random ball from it (i.e. from the second container), throw it into the third, etc. What is the probability that we will draw a white ball from the last container?

I tried first for only 2 containers: We have four possible outcomes: $(A_1A_2),(A_1B_2)(B_1A_2),(B_1B_2)$ Where $A_1A_2$ means we have drawn a white ball to the second basket and we have pulled a white ball from the second basket.

So for two baskets the probability would be $\frac b {a+b}$

But I don't know how to extend it to k containers. Any help?

Answer 1

Let there be $w_i$ white and $t_i$ total balls in container $i$ – initially. Label only the white balls by their initial container. Now ask "what is the probability of pulling a white ball from the last container that originated from the first?" Multiply the following factors:

• That ball must have been picked from the first container and thrown into the second: $\frac{w_1}{t_1}$
• It is the only ball from the first container in the second now, and yet it was picked and thrown into the third: $\frac1{t_2+1}$
• It is the only ball from the first container in the third now, and yet it was picked and thrown into the fourth: $\frac1{t_3+1}$
• etc. etc. until it is picked from the final container with probability $\frac1{t_k+1}$.

We can repeat this reasoning for balls starting in any container (but note that the first fraction for a ball starting in container $i>1$ has denominator $t_i+1$, since some ball has already been thrown into it). Since the events of picking up white balls from the last container that started from different containers are mutually exclusive and exhaustive, we add the probabilities to get the final result: $$\sum_{i=1}^k\frac{w_i}{\prod_{j=i}^k(t_j+[j>1])}$$ Here $[\cdot]$ is the Iverson bracket, which is $1$ if whatever inside is true and $0$ otherwise.