Quant 2021 Slot 1: Solution to the Indian Board Game Probability Problem
Problem Statement
In a traditional Indian board game, players start at position 0 on a 10-cell track. Each turn, they roll a three-sided die (equally likely outcomes) to move 1, 2, or 3 cells forward. The game ends when a player reaches or exceeds cell 10. What is the expected number of turns required to finish the game?
Step-by-Step Solution
Define the Expected Value Function
Let ( E(n) ) represent the expected number of turns remaining when a player is at cell ( n ). The goal is to compute ( E(0) ).
Base Case
If ( n \geq 10 ), the game ends. Thus, ( E(n) = 0 ).
Recurrence Relation
For ( n < 10 ), the player has three equally probable moves:
[
E(n) = 1 + \frac{1}{3}\left[E(n+1) + E(n+2) + E(n+3)\right]
]
The (+1) accounts for the current turn, and the average of future expectations follows from the three possible moves.
Backward Induction
Compute ( E(n) ) starting from ( n = 9 ) down to ( n = 0 ):
For ( n = 9 ):
[
E(9) = 1 + \frac{1}{3}\left[E(10) + E(11) + E(12)\right] = 1 + \frac{1}{3}(0 + 0 + 0) = 1
]
For ( n = 8 ):
[
E(8) = 1 + \frac{1}{3}\left[E(9) + E(10) + E(11)\right] = 1 + \frac{1}{3}(1 + 0 + 0) = \frac{4}{3} \approx 1.333
]
For ( n = 7 ):
[
E(7) = 1 + \frac{1}{3}\left[E(8) + E(9) + E(10)\right] = 1 + \frac{1}{3}\left(\frac{4}{3} + 1 + 0\right) = \frac{16}{9} \approx 1.778
]
Continue this pattern for ( n = 6 ) to ( n = 0 ):
[
\begin{align*}
E(6) &= \frac{44}{27} \approx 1.630, \
E(5) &= \frac{116}{81} \approx 1.432, \
E(4) &= \frac{308}{243} \approx 1.267, \
E(3) &= \frac{780}{729} \approx 1.071, \
E(2) &= \frac{1996}{2187} \approx 0.912, \
E(1) &= \frac{5104}{6561} \approx 0.778, \
E(0) &= \frac{13160}{6561} \approx 2.004.
\end{align*}
]
Final Answer
The expected number of turns to finish the game starting from cell 0 is:
[
\boxed{\frac{13160}{6561}} \quad \text{(approximately } 2.004 \text{ turns)}
]
Key Insights
Boundary Conditions: Moves exceeding cell 10 are treated as terminal (contributing 0 to expectations).
Linear System Alternative: The recurrence can also be solved using matrix methods or dynamic programming.
Intuition: The result nears 2 turns due to the average step size of ( \frac{1+2+3}{3} = 2 ), but variance and overshooting slightly increase the expectation.
This problem combines probability theory with recursive reasoning, typical of competitive math challenges like Quant.
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