Question

In a hockey match, both teams $A$ and $B$ scored same number of goals up to the end of the game, so to decide the winner, the referee asked both the captains to throw a die alternately and decided that the team, whose captain gets a six first, 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 referee was fair or not.

Answer 1

Let A: team A is declared as a winner

Let B: team B is declared as a winner.

Let E: die shows six on the throw.

Clearly, $P\left(E\right)=\frac{1}{6}$

$P\left(\overline{E}\right)=1-P\left(E\right)=1-\frac{1}{6}=\frac{5}{6}$

If captain of team A starts then he may get a six in $1^{st}$ throw or $3^{rd}$ throw or $5^{th}$ throw and so on.

Therefore, $P\left(A\right)=P\left(E\right)+P\left(\right(\overline{E}\left)\right(\overline{E}\left)\right(E\left)\right)+P\left(\right(\overline{E}\left)\right(\overline{E}\left)\right(\overline{E}\left)\right(\overline{E}\left)E\right)+......$

Using sum of infinite G.P. $S_{infinite}=\frac{a}{1-r}$

$=\frac{1}{6}+{\left(\frac{5}{6}\right)}^2\frac{1}{6}+{\left(\frac{5}{6}\right)}^4\frac{1}{6}+.....=\frac{\frac{1}{6}}{1-\frac{25}{36}}$

i.e., $P\left(A\right)=\frac{6}{11}$ and $P\left(B\right)=1-P\left(A\right)=1-\frac{6}{11}=\frac{5}{11}$

The decision of referee wasn't fair since team A has more chances of being declared a winner despite the fact that both the teams had secured same number of goals.