Question

In a hockey match, both teams A and B scored same number of goals upto the end of the game, so to decide the winner, the refree asked both the captains to throw a die alternately and decide that the team, whose captain gets a first six, will be declared the winner. If the captain of team A was asked to start, find their respective probabilities of winning the match and state whether the decision of the refree was fair or not.

Answer 1

\[P\left( \text{ a six} \right) = \frac{1}{6}\]
\[P\left( \text{ not a six } \right) = 1 - \frac{1}{6} = \frac{5}{6}\]
\[P\left(\text{  A wins } \right) = P\left( 6 \text{ in first throw }  \right) + P\left( 6 \text{ in third throw } \right) + . . . \]
\[ = \frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} + . . . \]
\[ = \frac{1}{6}\left[ 1 + \left( \frac{5}{6} \right)^2 + \left( \frac{5}{6} \right)^4 + . . . \right]\]
\[ = \frac{1}{6}\left[ \frac{1}{1 - \frac{25}{36}} \right] . . . \left[ {1+a+a}^2 {+a}^3 + . . . =\frac{1}{1 - a} \right]\]
\[ = \frac{1}{6} \times \frac{36}{11}\]
\[ = \frac{6}{11}\]
\[P\left( B \text{ wins }  \right) = P\left( 6 \text{ in second throw } \right) + P\left( 6\text{  in fourth throw }  \right) + . . . \]
\[ = \frac{5}{6} \times \frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} + . . . \]
\[ = \frac{5}{36}\left[ 1 + \left( \frac{5}{6} \right)^2 + \left( \frac{5}{6} \right)^4 + . . . \right]\]
\[ = \frac{5}{36}\left[ \frac{1}{1 - \frac{25}{36}} \right] . . . \left[ {1+a+a}^2 {+a}^3 + . . . =\frac{1}{1 - a} \right]\]
\[ = \frac{5}{36} \times \frac{36}{11}\]
\[ = \frac{5}{11}\]

It can be seen that the probability that team A wins is not equal to the probability that team B wins.
Thus, the decision of the referee was not fair.