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

captains 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

probability that A gets a six be 1/6 .
e.g., P = 1/6
so, Probability that A doesn't get a six be 1 - 1/6 = 5/6 e.g., q = 1/6

we know from formula ,

probability of A's winning = P(A) + P(A'B'A) + P(A'B'A'B'A) +......
= p + pq² + pq⁴ + ..... this is in geometric progression
so, Sn = P/(1 - q²)

put values of p and q ,
so, probability of A's winning = (1/6)/(1 - 5²/6²)
= (1/6)/{(36 - 25)/36}
= 6/11

and hence, probability of B's winning = 1 - 6/11
= 5/11

Answer 2

According to the given situation:

Captain of Team A throw a dice, six comes in first attempt than Team A wins,

probability of winning team A is P(A)= 1/6

( total outcome of a dice are 1,2,3,4,5,6 = 6, favourable outcome is 6 ,which is 1 in counting)

probability of losing the game P(A')= 1-1/6 = 5/6

probability of winning of team A =

P(A)+ P(A')P(B')P(A)+ P(A')P(B')P(A')P(B')P(A)...

=
$\frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} ... \\ \\ = \frac{1}{6} + ( { \frac{5}{6} })^{2} \times \frac{1}{6} + ({ \frac{5}{6} })^{4} \times \frac{1}{6} ...$
it is a GP, first term a= 1/6

common ratio r = 25/36

sum of GP ,if r < 1

$= \frac{a}{(1 - r)} \\ \\ = \frac{1}{6}(\frac{1}{1- \frac{25}{36}} ) \\ \\ = \frac{1}{6} \times \frac{36}{11} \\ = \frac{6}{11}$
Probability of winning team B:

P(A')P(B)+P(A')P(B')P(A')P(B)+P(A')P(B')P(A')P(B')P(A')P(B)+...

$\frac{5}{6} \times \frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} + \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} ... \\ \\ \frac{5}{36} + {( \frac{5}{6} })^{2} \times \frac{5}{36} + ( { \frac{5}{6} })^{4} \times \frac{5}{36} ...$
it is a GP

with a = 5/36

r = 25/36

sum of infinite terms =
$\frac{5}{36}(\frac{1}{ 1 - \frac{25}{36} }) \\ \\ = \frac{5}{36} \times \frac{36}{11} \\ \\= \frac{5}{11}$
Yes,the decision of refree was fair,but it is time consuming.

if he chosen a coin then result might came easily