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Probability

Three players...

Question

Three players $A, B$ and $C$ alternatively throw a die in that order, the first player to throw a $6$ being deemed the winner. $A$ 's die is fair where as $B$ and $C$ throw die with probabilities $p_{1}$ and $p_{2}$ respectively of throwing a $6$

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Solution

Let $q_{1} = 1 - p_{1}$ and $q_{2} = 1 - p_{2}$ .
A) $A$ wins the game at the $1 s t, 4 t h, 7 t h, . . .$ trials.
$P (A w i n s) = \frac{1}{6} + (\frac{5}{6}) q_{1} q_{2} (\frac{1}{6}) + {(\frac{5}{6})}^{2} q_{1}^{2} q_{2}^{2} (\frac{1}{6}) + . . .$
$= \frac{1 / 6}{1 - (5 q_{1} q_{2} / 6)}$
$= \frac{1}{6 - 5 q_{1} q_{2}}$
$= \frac{1}{3}$
B) $C$ wins at $3 r d, 6 t h, 9 t h . . . .$ trials
$P (C w i n s) = \frac{5}{6} q_{1} p_{2} + (\frac{5}{6} q_{1} q_{2}) (\frac{5}{6} q_{1} p_{2}) + {(\frac{5}{6} q_{1} q_{2})}_{1}^{2} (\frac{5}{6} q_{1} p_{2}) + . . .$
$= \frac{5 q_{1} p_{2} / 6}{1 - (5 q_{1} q_{2} / 6)}$
$= \frac{5 q_{1} p_{2}}{6 - 5 q_{1} q_{2}}$
$= \frac{1}{3}$
C) $B$ wins at $2 n d, 5 t h, 8 t h$ trials
$P (B w i n s) = \frac{5}{6} p_{1} + (\frac{5}{6} q_{1} q_{2}) (\frac{5}{6} p_{1}) + {(\frac{5}{6} q_{1} q_{2})}_{1}^{2} (\frac{5}{6} p_{1}) + . . . .$
$= \frac{5 p_{1} / 6}{1 - (5 q_{1} q_{2} / 6)}$
Therefore
$P (A w i n s) = P (B w i n s)$
$\Rightarrow \frac{1 / 6}{1 - (5 q_{1} q_{2} / 6)} = \frac{5 p_{1} / 6}{1 - (5 q_{1} q_{2} / 6)}$
$\Rightarrow p_{1} = \frac{1}{5}$
D) The game is equiprobable to all the three players when
$P (A w i n s) = P (B w i n s) = P (C w i n s)$
$\Rightarrow \frac{1 / 6}{1 - (5 q_{1} q_{2} / 6)} = \frac{5 p_{1} / 6}{1 - (5 q_{1} q_{2} / 6)} = \frac{5 q_{1} p_{2} / 6}{1 - (5 q_{1} q_{2} / 6)}$
$\Rightarrow p_{1} = \frac{1}{5}, p_{2} = \frac{1}{4}$

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