Let's solve the problem step by step using the given hint about geometric sums. The key idea is to calculate the probability that each player is the first to roll a six.

Step 1: Understanding the Problem

The players roll the die in sequence: A, B, C, and repeat. The game continues until someone rolls a six. When a player rolls a six, they are eliminated, and the remaining two players continue.

We are asked to find the probability that each player (A, B, C) is the first to exit the game by rolling a six.

Step 2: Calculating the Probabilities

Let's denote the probability that a player rolls a six as $p = \frac{1}{6}$. The probability that a player does not roll a six is $q = \frac{5}{6}$.

1. Probability that A exits first:

   • A rolls a six on the first try: $p = \frac{1}{6}$.
   • A rolls a number other than six and B also doesn't roll a six, and C doesn't roll a six, then it comes back to A, and the game repeats. The probability of this sequence (where no one rolls a six in one complete cycle) is $q^3 = \left(\frac{5}{6}\right)^3$.

   The probability that A exits first can be expressed as a sum of an infinite geometric series: $P_A = \frac{1}{6} + \left(\frac{5}{6}\right)^3 \cdot \frac{1}{6} + \left(\frac{5}{6}\right)^6 \cdot \frac{1}{6} + \dots$ This is a geometric series with the first term $a = \frac{1}{6}$ and common ratio $r = \left(\frac{5}{6}\right)^3$.

   The sum of this geometric series is: $P_A = \frac{a}{1 - r} = \frac{\frac{1}{6}}{1 - \left(\frac{5}{6}\right)^3}$ Let's simplify this: $P_A = \frac{\frac{1}{6}}{1 - \frac{125}{216}} = \frac{\frac{1}{6}}{\frac{91}{216}} = \frac{216}{6 \times 91} = \frac{36}{91} = \frac{4}{10.11} \approx 0.396$

2. Probability that B exits first:

   • B can only exit first if A doesn't roll a six. The probability that A does not roll a six is $q = \frac{5}{6}$.
   • Then B rolls a six: $p = \frac{1}{6}$.
   • If B does not roll a six, the game continues in a similar pattern as with A.

   Thus, the probability $P_B$ can be derived similarly: $P_B = \frac{5}{6} \cdot \frac{1}{6} + \left(\frac{5}{6}\right)^4 \cdot \frac{1}{6} + \left(\frac{5}{6}\right)^7 \cdot \frac{1}{6} + \dots$ This is again a geometric series, and after simplifying: $P_B = \frac{\frac{5}{6} \cdot \frac{1}{6}}{1 - \left(\frac{5}{6}\right)^3} = \frac{\frac{5}{36}}{\frac{91}{216}} = \frac{180}{216 \times 3} = \frac{4}{15.33} \approx 0.261$

3. Probability that C exits first:

   • C can only exit first if neither A nor B rolls a six. The probability that neither A nor B rolls a six is $\left(\frac{5}{6}\right)^2$.
   • Then C rolls a six: $p = \frac{1}{6}$.

   Therefore, $P_C$ is: $P_C = \left(\frac{5}{6}\right)^2 \cdot \frac{1}{6} + \left(\frac{5}{6}\right)^5 \cdot \frac{1}{6} + \left(\frac{5}{6}\right)^8 \cdot \frac{1}{6} + \dots$ Simplifying: [ P_C = \