Understanding the Urn Problem

The problem involves tracking the number of red (R) and blue (B) balls in an urn over two steps, starting with 1R and 1B.

The process rule: Pick a ball, return it along with another ball of the same color. The total number of balls increases by 1 at each step.

Step 1 Analysis

• Initial State: 1 Red, 1 Blue (Total = 2 balls).
• Probability of picking Red: $P(R_1) = 1/2$.
• If Red is picked, 1 Red is added. The urn becomes: 2 Red, 1 Blue (Total = 3 balls).
• Probability of picking Blue: $P(B_1) = 1/2$.
• If Blue is picked, 1 Blue is added. The urn becomes: 1 Red, 2 Blue (Total = 3 balls).

Step 2 Analysis

We need the state after Step 2 to have an equal number of red and blue balls (R = B).

• Case 1: State after Step 1 is (2R, 1B) (This happened with probability $P(R_1) = 1/2$)

  • Probability of picking Red now: $P(R_2 | 2R, 1B) = 2/3$. Urn becomes (3R, 1B). (R ≠ B)
  • Probability of picking Blue now: $P(B_2 | 2R, 1B) = 1/3$. Urn becomes (2R, 2B). (R = B)
  • The probability of reaching the equal state (2R, 2B) via this path is $P(R_1) \times P(B_2 | 2R, 1B) = (1/2) \times (1/3) = 1/6$.

• Case 2: State after Step 1 is (1R, 2B) (This happened with probability $P(B_1) = 1/2$)

  • Probability of picking Red now: $P(R_2 | 1R, 2B) = 1/3$. Urn becomes (2R, 2B). (R = B)
  • Probability of picking Blue now: $P(B_2 | 1R, 2B) = 2/3$. Urn becomes (1R, 3B). (R ≠ B)
  • The probability of reaching the equal state (2R, 2B) via this path is $P(B_1) \times P(R_2 | 1R, 2B) = (1/2) \times (1/3) = 1/6$.

Calculating Total Probability

The total probability of having an equal number of red and blue balls after two steps is the sum of probabilities from the successful paths:

Total Probability = Probability (Path 1) + Probability (Path 2)

Total Probability = $1/6 + 1/6 = 2/6 = 1/3$.