Title: CAT 2022 Slot 3 Quant - Solution to Indian Probability Game
Problem Statement (Hypothetical Example):
In a traditional Indian game, two players take turns flipping a coin. The first player to flip heads wins. What is the probability that the second player wins?
Solution:
Define Probability:
Let ( P ) be the probability that the second player wins.
First Player’s Turn:
The first player flips heads with probability ( \frac{1}{2} ). If they win, the second player loses (( 0 )).
If the first player flips tails (( \frac{1}{2} )), the turn passes to the second player.
Second Player’s Turn:
The second player now has the same scenario:
Wins with probability ( \frac{1}{2} ).
If tails (( \frac{1}{2} )), the game restarts, and the second player’s probability becomes ( P ).
Recursive Equation:
[
P = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times P
]
Explanation:
The second player wins only if:
First player tails (( \frac{1}{2} )) + second player heads (( \frac{1}{2} )) → ( \frac{1}{4} ).
Both players tails (( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} )), then the game restarts, giving the second player another chance with probability ( P ).
Solve for ( P ):
[
P = \frac{1}{4} + \frac{1}{4}P \implies P - \frac{1}{4}P = \frac{1}{4} \implies \frac{3}{4}P = \frac{1}{4} \implies P = \frac{1}{3}
]
Answer:
The probability that the second player wins is ( \boxed{\frac{1}{3}} ).
Key Takeaways:
Recursive Probability: Use conditional probability to model turn-based games.
CAT Strategy: Simplify complex scenarios by breaking into smaller, repeating cases.
Time-Saving Tip: For geometric series like this, recognize the pattern and solve algebraically (avoids lengthy summation).
Let me know if you need further clarification or additional examples!
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