An urn contains 12 balls, of which 4 are white. Players A*A*, B*B*, and C*C* successively draw from the urn - A*A* first, then B*B*, then C*C*, then A*A* again, and so on. The winner is the first one to draw a white ball. If the balls that are withdrawn are not replaced, what is the probability, computed up to three places after decimal, that player A*A* wins?

To solve this problem, we need to compute the probability that player A wins the game by drawing the first white ball. The urn contains 12 balls in total, out of which 4 are white and 8 are non-white. Players A, B, and C take turns drawing the balls in sequence, and the first player to draw a white ball wins.

### Total number of outcomes
At each stage, a ball is drawn without replacement. Since no balls are replaced, the total number of possible outcomes (for all balls being drawn) depends on how many turns pass until a white ball is drawn.

### Strategy for Player A to Win
Player A wins if they are the first one to draw a white ball. For this to happen, the white ball must be drawn by player A on their turn. Let's examine each possible scenario:

1. **Player A wins on their first draw (first overall draw):**
   - Probability of player A drawing a white ball on the first draw:
   \[
   P_1 = \frac{4}{12}
   \]

2. **Player A wins on their second draw (fourth overall draw):**
   - For this to happen, the first three draws (by A, B, and C) must all be non-white, and player A must draw a white ball on their second turn.
   - Probability of A, B, and C drawing non-white balls in the first three draws:
   \[
   P(\text{no white balls in first 3 draws}) = \frac{8}{12} \times \frac{7}{11} \times \frac{6}{10}
   \]
   - Then, the probability of A drawing a white ball on their second draw (fourth overall draw):
   \[
   P_4 = \frac{4}{9}
   \]
   - Therefore, the probability that player A wins on their second draw is:
   \[
   P(\text{A wins on fourth draw}) = \frac{8}{12} \times \frac{7}{11} \times \frac{6}{10} \times \frac{4}{9}
   \]

3. **Player A wins on their third draw (seventh overall draw):**
   - For this to happen, the first six draws (by A, B, C, A, B, and C) must all be non-white, and player A must draw a white ball on their third turn.
   - Probability of no white balls in the first six draws:
   \[
   P(\text{no white balls in first 6 draws}) = \frac{8}{12} \times \frac{7}{11} \times \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8} \times \frac{3}{7}
   \]
   - Then, the probability of A drawing a white ball on their third draw (seventh overall draw):
   \[
   P_7 = \frac{4}{6}
   \]
   - Therefore, the probability that player A wins on their third draw is:
   \[
   P(\text{A wins on seventh draw}) = \frac{8}{12} \times \frac{7}{11} \times \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} \times \frac{4}{6}
   \]

4. **Player A wins on their fourth draw (tenth overall draw):**
   - For this to happen, the first nine draws (by A, B, C, A, B, C, A, B, and C) must all be non-white, and player A must draw a white ball on their fourth turn.
   - Probability of no white balls in the first nine draws:
   \[
   P(\text{no white balls in first 9 draws}) = \frac{8}{12} \times \frac{7}{11} \times \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} \times \frac{2}{6} \times \frac{1}{5}
   \]
   - Then, the probability of A drawing a white ball on their fourth draw (tenth overall draw):
   \[
   P_10 = \frac{4}{3}
   \]

#### Final Calculation
We need to add up the probabilities for each scenario where A wins:
\[
P(A \text{ wins}) = P_1 + P_4 + P_7

An urn contains 12 balls, of which 4 are white. Players A, B, and C successively draw from the urn - A first, then B, then C, then A again, and so on. The winner is the first one to draw a white ball. If the balls that are withdrawn are not replaced, what is the probability, computed up to three places after decimal, that player A wins?

Explore the probability that Player A wins a ball-drawing game from an urn containing 12 balls, 4 of which are white. Players A, B, and C draw successively, and the first player to draw a white ball wins. This problem covers key probability concepts, using combinatorics and the multiplication rule of probability.

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