Question

In a hockey match, both teams A and B scored the 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

Hint: To solve this problem, we try to find the probability for A to win. With this information, we can decide if the judgement is fair or not. We will further require the formula of infinite geometric progression. This is given by –
$a+ar+a{{r}^{2}}+...=\dfrac{a}{1-r}$
Here, a is the first term and r is the common ratio.

Complete step-by-step solution -
To solve the problem of probability, we define the required events associated with the winning of A and B. Thus, we have,
A: event that A wins (that is gets six)
B: event that B wins (that is gets six)
Now, probability of event A [p(A) = $\dfrac{1}{6}$]. This is because there is only desirable outcome (that is 6) out of all possible outcomes (1, 2, 3, 4, 5, 6).
Now, we are given that A and B toss the coins alternately starting from A. Thus, to find the desired probability for A to win, we have,
Favourable probability = $p(A)+\left( p(\overline{A})p(\overline{A})p(A) \right)+\left( p(\overline{A})p(\overline{A})p(\overline{A})p(\overline{A})p(A) \right)+...$
Here, $p(\overline{A})$ is the complement (that is 1 – p(A)) of event A. Now, we get this formula of p(A), since, the first term represents that A got 6 in the first turn and thus won. However, if A doesn’t win in the first term, then he can win in the third term (since turns are alternate). For this, to happen, in addition of A losing (which has probability of $\dfrac{5}{6}$, since there are 5 possible outcomes possible apart from 6), B also has to lose (which also has probability of $\dfrac{5}{6}$). However, now in the third term A wins. Similarly, in the next addition term, it is the probability that A wins in the fifth term and this series repeats infinitely. Thus, to find the answer, we have,
Favourable probability = \[\dfrac{1}{6}+\left( \dfrac{5}{6} \right)\left( \dfrac{5}{6} \right)\left( \dfrac{1}{6} \right)+\left( \dfrac{5}{6} \right)\left( \dfrac{5}{6} \right)\left( \dfrac{5}{6} \right)\left( \dfrac{5}{6} \right)\left( \dfrac{1}{6} \right)+...\]
Favourable probability = \[\dfrac{1}{6}+\left( \dfrac{1}{6} \right){{\left( \dfrac{5}{6} \right)}^{2}}+\left( \dfrac{1}{6} \right){{\left( \dfrac{5}{6} \right)}^{4}}+...\]
Since, this is a geometric progression with the first term as $\dfrac{1}{6}$ (as a) and the common ratio as ${{\left( \dfrac{5}{6} \right)}^{2}}$ (as r), thus, we have,
$a+ar+a{{r}^{2}}+...=\dfrac{a}{1-r}$
= $\dfrac{\dfrac{1}{6}}{1-{{\left( \dfrac{5}{6} \right)}^{2}}}$
= $\dfrac{6}{11}$
For B to win, the required probability would be $1-\dfrac{6}{11}=\dfrac{5}{11}$

Since, this probability is greater than 0.5, hence, A has higher chances of winning. Thus, the decision made by the referee was unfair.

Note: In general, although non-intuitive, the person beginning the rolling of the die has a slightly higher chance to get six first. This result would also have been true if the decision would have been taken on the basis of coin toss rather than rolling a die. In that case, the probability for A to win would be $\dfrac{2}{3}$. (this can be similarly calculated as was done for the case of the die)