Title: VARC 2023 Slot 3 Answer: Game Problem Solution
Problem Statement:
In a game with four players (A, B, C, D), each round involves rolling a standard six-sided die. The player with the highest unique roll advances. If multiple players roll the same highest number, they compete in a tiebreaker round. The game continues until one player is the sole leader. Calculate the probability that Player A wins the game in the first round.
Solution:
Understanding the Rules:
Players roll a die (faces 1–6).
The highest unique roll wins.
If the highest roll is shared, a tiebreaker occurs in the next round.
We need the probability that A wins in the first round (no tiebreaker).
Key Calculations:
Probability A rolls a 6: ( P(A=6) = \frac{1}{6} ).
Probability no other player rolls a 6: ( P(\text{B,C,D} \leq 5) = \left(\frac{5}{6}\right)^3 ).
Probability A wins immediately with a 6:
[
P_{\text{win}} = \frac{1}{6} \times \left(\frac{5}{6}\right)^3 = \frac{125}{1296} \approx 0.09645.
]
Handling Ties at the Highest Roll:
If A rolls ( k ) (2–6), others must roll ≤ ( k-1 ):
[
P_{\text{win, A=k}} = \frac{1}{6} \times \left(\frac{k-1}{6}\right)^3.
]
Summing for ( k=2 ) to ( 6 ):
[
\sum_{k=2}^6 \frac{1}{6} \times \left(\frac{k-1}{6}\right)^3 = \frac{1}{6^4} \sum_{k=1}^5 k^3.
]
Sum of cubes: ( 1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 225 ).
[
P_{\text{win, non-6}} = \frac{225}{1296} \approx 0.1736.
]
Total Probability A Wins in First Round:
[
P_{\text{total}} = P_{\text{win}} + P_{\text{win, non-6}} = \frac{125 + 225}{1296} = \frac{350}{1296} \approx 0.2704 , (27.04%).
]
Conclusion:
The probability that Player A wins the game in the first round is approximately 27.04%. If a tiebreaker occurs, the probability would require recursive calculations based on subsequent rounds.
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