Title: CAT 2020 Slot 3: Analyzing Game-Based Logic Problems
Content: Solutions to Game-Related Questions
Problem 1: Strategy in a Multi-Player Game
Question:
Three players (A, B, and C) take turns choosing numbers from 1 to 5. Each number can be chosen only once. The player who picks the last remaining number wins. If all numbers are chosen without a winner, the game restarts. What is the optimal strategy for Player A to ensure victory?
Solution:
Game Analysis:
Total numbers: 5 (1, 2, 3, 4, 5).
Players alternate turns.
The player who picks the 5th number wins.
Key Insight:
The game can be broken into pairs: (1,2), (3,4), and 5.
If Player A picks 5 first, they win immediately.
If Player A picks a number from a pair, Player B can mirror the choice to force Player C into the last pair.
Optimal Strategy:
Step 1: Player A picks 5 on their first turn.
Step 2: This guarantees Player A picks the last number and wins.
Answer: Player A should select 5 first to win immediately.
Problem 2: Probability in a Card Game
Question:
A deck of 20 cards contains 8 red and 12 black cards. Two cards are drawn at random without replacement. What is the probability that both cards are red, given that the first card drawn is red?
Solution:
Conditional Probability Formula:
[
P(\text{Both Red} \mid \text{First Red}) = \frac{P(\text{Both Red})}{P(\text{First Red})}
]
Calculations:
( P(\text{First Red}) = \frac{8}{20} = \frac{2}{5} ).
( P(\text{Both Red}) = \frac{8}{20} \times \frac{7}{19} = \frac{14}{95} ).
Substitute into the formula:
[
\frac{\frac{14}{95}}{\frac{2}{5}} = \frac{14}{95} \times \frac{5}{2} = \frac{14}{38} = \frac{7}{19}
]
Answer: The probability is (\boxed{\frac{7}{19}}).
Problem 3: Logic Grid Puzzle
Question:
Four friends (X, Y, Z, W) play a game where each picks a unique color (Red, Blue, Green, Yellow). The rules are:
X does not pick Red.
Y picks a color adjacent to Z’s color in the rainbow (Red, Orange, Yellow, Green, Blue, Indigo).
W picks a color opposite to X’s.
Who picks Green?
Solution:
Constraints:
Possible colors: Red (R), Blue (B), Green (G), Yellow (Y).
Rainbow adjacency: R-Y-G-B (simplified).
W’s color is opposite X’s.
Step-by-Step Deduction:
X ≠ R.
Assume X = B → W = G (opposite).
Y must pick adjacent to Z. If W = G, Z cannot be G.
Valid assignments:
X = B, W = G.
Y and Z take R and Y.
Y cannot be adjacent to Z (R and Y are adjacent).
Contradiction.
Final Assignment:
X = Y (Yellow), W = R (opposite).
Z = G (remaining), Y = B.
Y (B) is adjacent to Z (G).
Answer: Z picks Green.
Conclusion
CAT 2020 Slot 3 game-based questions tested strategic thinking, conditional probability, and logical deduction. Mastery of adjacency rules, probability formulas, and systematic elimination ensured success.
Key Takeaways:
Break complex problems into smaller constraints.
Use conditional probability for dependent events.
Visualize relationships (e.g., color opposites, rainbow order).
Let me know if you need further clarification!
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