A, B, C in order roll a dice on the condition that one who gets a six, first wins then which of the following are true

Options

A. A chances `36/91`
B. B chances `5/6.(36)/(91)`
C. C chances `(5/6)^(2)(36)/(91)`
D. All three has same chances

Answer

To solve the problem of determining the probabilities of A, B, and C winning when they roll a die in order, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem:**
- A, B, and C roll a die in that order.
- The first person to roll a six wins.
- We need to find the probabilities of A, B, and C winning.

2. **Defining Probabilities:**
- The probability of rolling a six (winning) is \( P(6) = \frac{1}{6} \).
- The probability of not rolling a six (losing) is \( P(\text{not } 6) = \frac{5}{6} \).

3. **Calculating Probability of A Winning:**
- A can win in the first round by rolling a six.
- If A does not roll a six, B and C will also roll.
- The probability that A wins on the first roll is:
\[ P(A \text{ wins in first round}) = \frac{1}{6} \]
- If A does not win, the probability that both B and C do not win is:
\[ P(A \text{ does not win}) = \frac{5}{6} \times \frac{5}{6} = \frac{25}{36} \]
- If all three do not win, the situation resets, and A has another chance to roll.
- The total probability of A winning can be expressed as an infinite series:
\[ P(A \text{ wins}) = \frac{1}{6} + \left(\frac{5}{6}\right)^3 P(A \text{ wins}) \]
- Let \( P(A \text{ wins}) = p_A \). Then:
\[ p_A = \frac{1}{6} + \left(\frac{25}{36}\right) p_A \]
- Rearranging gives:
\[ p_A - \frac{25}{36} p_A = \frac{1}{6} \]
\[ \left(1 - \frac{25}{36}\right) p_A = \frac{1}{6} \]
\[ \frac{11}{36} p_A = \frac{1}{6} \]
\[ p_A = \frac{1/6}{11/36} = \frac{6}{11} \]

4. **Calculating Probability of B Winning:**
- B can only win if A does not roll a six:
\[ P(B \text{ wins}) = P(A \text{ does not win}) \times P(B \text{ wins in first round}) \]
- The probability that A does not win is \( \frac{5}{6} \).
- B can win on the first roll with probability \( \frac{1}{6} \) or if they both fail, allowing A another chance:
\[ P(B \text{ wins}) = \frac{5}{6} \cdot \frac{1}{6} + \left(\frac{5}{6}\right)^3 P(B \text{ wins}) \]
- Let \( p_B = P(B \text{ wins}) \):
\[ p_B = \frac{5}{36} + \left(\frac{25}{36}\right) p_B \]
- Rearranging gives:
\[ p_B - \frac{25}{36} p_B = \frac{5}{36} \]
\[ \frac{11}{36} p_B = \frac{5}{36} \]
\[ p_B = \frac{5/36}{11/36} = \frac{5}{11} \]

5. **Calculating Probability of C Winning:**
- C can only win if both A and B do not roll a six:
\[ P(C \text{ wins}) = P(A \text{ does not win}) \times P(B \text{ does not win}) \times P(C \text{ wins in first round}) \]
- This can be calculated similarly:
\[ P(C \text{ wins}) = \left(\frac{5}{6}\right)^2 \cdot \frac{1}{6} + \left(\frac{5}{6}\right)^3 P(C \text{ wins}) \]
- Let \( p_C = P(C \text{ wins}) \):
\[ p_C = \frac{25}{216} + \left(\frac{125}{216}\right) p_C \]
- Rearranging gives:
\[ p_C - \frac{125}{216} p_C = \frac{25}{216} \]
\[ \frac{91}{216} p_C = \frac{25}{216} \]
\[ p_C = \frac{25/216}{91/216} = \frac{25}{91} \]

### Final Probabilities:
- \( P(A \text{ wins}) = \frac{6}{11} \)
- \( P(B \text{ wins}) = \frac{5}{11} \)
- \( P(C \text{ wins}) = \frac{25}{91} \)