Title: CAT 2018 Quant Slot 1 - Indian Game Probability Problem
Problem Statement (English):
In a traditional Indian game, two players, A and B, take turns rolling a fair six-sided die. The first player to roll a 6 wins the game. If Player A starts, what is the probability that Player A wins the game?
Solution:
Define the Probability:
Let ( P ) be the probability that Player A wins the game.
First Turn Analysis:
Player A has a ( \frac{1}{6} ) chance to win on their first roll.
If Player A fails (probability ( \frac{5}{6} )), Player B gets a chance to roll.
Second Turn Analysis:
Player B now has a ( \frac{1}{6} ) chance to win.
If Player B also fails (probability ( \frac{5}{6} )), the turn returns to Player A.
Recursive Probability Setup:
After both players fail their first attempts, the scenario repeats. Thus:
[
P = \frac{1}{6} + \left(\frac{5}{6}\right)\left(\frac{5}{6}\right)P
]
Here, the term ( \frac{5}{6} \times \frac{5}{6} ) represents both players failing their first rolls, leading back to the original probability ( P ).
Solve for ( P ):
[
P = \frac{1}{6} + \frac{25}{36}P \
P - \frac{25}{36}P = \frac{1}{6} \
\frac{11}{36}P = \frac{1}{6} \
P = \frac{1}{6} \times \frac{36}{11} = \frac{6}{11}
]
Answer:
The probability that Player A wins the game is ( \boxed{\dfrac{6}{11}} ).
Explanation:
This problem leverages geometric probability and recursion. Since each turn is independent and the game resets after both players fail, the recursive equation captures all possible winning scenarios for Player A. The final probability ( \frac{6}{11} ) reflects the higher chance of Player A (the first mover) winning due to the turn-based structure.
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