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

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

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.