Problem No. 11. Urns and balls

There are 7 urns with 30 balls each. In each urn there is equal number of white and black balls. Randomly, we take out one ball from the first urn and put it into the second urn. Then, we take one ball from the second urn and put it into the third urn. We continue this process, until we take a ball from the seventh urn and put it into the first urn. Then, we take a ball from the first urn. What are the chances that the ball is black?

Student tip

First, find the probabilities for taking a black ball from each urn in first 7 steps.

Solution

Each urn consists of 15 white and 15 black balls, so the probability of taking out a black ball from the first urn is 1/2. After putting a ball into the second urn, the probability of taking out a black ball from the second urn is conditioned by the color of the ball that was put into the second from the first urn. If the ball was white (which can happen with probability of 1/2), then the probability of taking out a black ball from the second urn is 15/31 (since the urn is consisted of 31 balls, among which 15 are black). If the ball was black (which can happen with probability of 1/2), then the probability of taking out a black ball from the second urn is 16/31 (since the urn is consisted of 31 balls, among which 16 are black). So, in total, the probability of taking out a black ball from the second urn is $$\frac{1}{2} \cdot \frac{15}{31} + \frac{1}{2} \cdot \frac{16}{31} = \frac{1}{2} \cdot (\frac{15}{31} + \frac{16}{31}) = \frac{1}{2}.$$

If we continue the reasoning in the similar manner, we will find that the probability of taking out a black ball from each of the seven urns is 1/2.

Finally, we should find the probability of taking out a black ball from the first urn (after putting a ball from the 7th to the 1st urn, which according to the previous results can be white or black with the same probability of 1/2). But now, before putting a ball from the 7th to the 1st urn, we deal with an uncertain content of the 1st urn, namely the 1st urn contains 15 white and 14 black balls with probability of 1/2, and 14 white and 15 black balls with probability of 1/2. To this uncertain content we can add white or black with the same probability of 1/2. So, the content of the 1st urn before taking out the ball for the second time is

16 white and 14 black balls, with the probability of 1/4,

15 white and 15 black balls, with the probability of 1/2,

14 white and 16 black balls, with the probability of 1/4.

So, in total, the probability of taking out a black ball from the first urn (the second time) is $$\frac{1}{4} \cdot \frac{14}{30} + \frac{1}{2} \cdot \frac{15}{30} + \frac{1}{4} \cdot \frac{16}{30} = \frac{1}{2} \cdot (\frac{7}{30} + \frac{15}{30} + \frac{8}{30}) = \frac{1}{2}.$$