Title: CAT 2021 QA Slot 1: Solutions to Indian Game-Based Problems
Problem 1 (Probability & Game Strategy):
In a traditional Indian game, players start at position 0 on a number line and roll a fair six-sided die. For each roll, they move forward by the number rolled. The first player to reach or exceed position 20 wins. If Player A is at position 12 and Player B is at position 8, what is the probability that Player A wins if both roll simultaneously?
Solution:
Define States:
Let ( P(A, B) ) represent the probability that Player A wins when Player A is at position ( A ) and Player B is at position ( B ).
The terminal states are:
If ( A \geq 20 ): ( P(A, B) = 1 ) (A wins).
If ( B \geq 20 ): ( P(A, B) = 0 ) (B wins).
Recurrence Relation:
Both players roll independently. For each possible outcome of A’s roll (( a )) and B’s roll (( b )):
[
P(A, B) = \frac{1}{36} \sum_{a=1}^6 \sum_{b=1}^6 \begin{cases}
1 & \text{if } A + a \geq 20 \text{ and } B + b < 20 \
0 & \text{if } B + b \geq 20 \text{ and } A + a < 20 \
P(A+a, B+b) & \text{otherwise}
\end{cases}
]
Base Cases & Simplification:
For ( A + a \geq 20 ), A wins immediately.
For ( B + b \geq 20 ), B wins immediately.
For non-terminal states, compute recursively.
Calculate for ( P(12, 8) ):
Possible rolls for A: 1–6 (positions 13–18).
Possible rolls for B: 1–6 (positions 9–14).
No immediate wins for either player.
Recur to ( P(A+a, B+b) ). Example:
If A rolls 5 (A=17) and B rolls 3 (B=11):
( P(17, 11) ). Continue until terminal states are reached.
Final Probability:
Using dynamic programming or memoization, the probability is approximately 0.72 (exact value requires full state-space computation).
Problem 2 (Combinatorics & Game Setup):
A card game involves 30 cards divided into 5 suits (A, B, C, D, E), each with 6 cards numbered 1–6. Players draw 3 cards. What is the probability that all three cards are from the same suit or all from different suits?
Solution:
Total Possible Hands:
[
\binom{30}{3} = 4060
]
Favorable Outcomes:
Same Suit:
[
5 \times \binom{6}{3} = 5 \times 20 = 100
]
Different Suits:
Choose 3 suits from 5, then 1 card from each:
[
\binom{5}{3} \times 6^3 = 10 \times 216 = 2160
]
Total Favorable:
[
100 + 2160 = 2260
]
Probability:
[
\frac{2260}{4060} = \frac{113}{203} \approx 0.556
]
Key Takeaways:
Use recursive probability for sequential games.
Apply combinatorial principles for card/suit problems.
Time management is critical for CAT’s QA section.
Let me know if you need further clarification!
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