Balancing the Color Proportions: An Analysis of Three Urns

Consider three urns, one colored red, one white, and one blue. The red urn contains 1 red and 4 blue balls; the white urn contains 3 white balls, 2 red balls, and 2 blue balls; the blue urn contains 4 white balls, 3 red balls, and 2 blue balls. At the initial stage, a ball is randomly selected from the red urn and then returned to that urn.

At every subsequent stage, a ball is randomly selected from the urn whose color is the same as that of the ball previously selected and is then returned to that urn. In the long run, what proportion of the selected balls are red? What proportion are white? What proportion are blue? 

In this blog post, we will analyze a problem involving three urns, each containing different combinations of red, white, and blue balls. The goal is to find the long-run proportions of red, white, and blue balls that will be selected in the process.

To start, we are given the contents of each urn:

The red urn contains 1 red ball and 4 blue balls

The white urn contains 3 white balls, 2 red balls, and 2 blue balls

The blue urn contains 4 white balls, 3 red balls, and 2 blue balls

At the initial stage, a ball is randomly selected from the red urn and then returned to that urn. At every subsequent stage, a ball is randomly selected from the urn whose color is the same as that of the ball previously selected and is then returned to that urn.

To find the long-run proportions of each color, we can use the steady-state probabilities. The steady-state probabilities of being in each urn are proportional to the number of balls of the selected color in each urn. The proportion of red balls selected is proportional to the number of red balls in each urn. Hence, the proportion of red balls selected is (1 + 2 + 3)/(1 + 3 + 4) = 6/8 = 3/4. The proportion of white balls selected is proportional to the number of white balls in each urn. Hence, the proportion of white balls selected is (3 + 2 + 4)/(1 + 3 + 4) = 9/8 = 3/4. The proportion of blue balls selected is proportional to the number of blue balls in each urn. Hence, the proportion of blue balls selected is (4 + 2 + 2)/(1 + 3 + 4) = 8/8 = 1.

Thus, we have found that in the long run, the proportions of red and white balls selected are equal and both equal to 3/4, while the proportion of blue balls selected is equal to 1.

In conclusion, we have used the steady-state probabilities to find the long-run proportions of red, white, and blue balls selected in the process involving three urns. This problem can be a good illustration of the principles of Markov chains and steady-state probabilities.