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

In a tie-breaker hockey match where the winner is decided by rolling a die, the probability of the captain of team A winning by rolling a six first is 6/11, while team B's probability is 5/11.

Step-by-step explanation:

In the scenario described, we need to calculate the probability that the captain of team A wins the game by rolling a six first, as it is a game of chance involving dice. To determine if the referee's decision is fair, we'll compare the captain of team A's probabilities to the captain of team B's.

To find the probability of team A winning, we consider two scenarios: A wins on the first roll, or the game continues without a winner until A's second roll, third roll, and so on. The probability of winning on the first roll for A is 1/6 since there is one winning number out of six possible outcomes. If A does not win on the first roll, B gets a chance to roll. The probability that the game reaches A's second roll is the probability that both A and B fail their first roll, which is a (5/6) chance of not rolling a six each. Therefore, the probability of A winning on the second roll is (5/6) * (5/6) * (1/6).

This pattern continues indefinitely as a geometric series because A will have a chance to win on the third roll if both fail twice, and so on. The total probability of A winning, which can be found through the sum of an infinite geometric series, is:

P(A wins) = (1/6) + (5/6) * (5/6) * (1/6) + (5/6)^4 * (1/6) + ... = (1/6)/(1 - (5/6)^2) = 6/11.

Similarly, for B to win, it can win on its first turn or future turns, following a similar series but delayed by one term since B starts second:

P(B wins) = (5/6) * (1/6) + (5/6)^3 * (1/6) + (5/6)^5 * (1/6) + ... = (5/6) * (1/6)/(1 - (5/6)^2) = 5/11.

Since team A has a probability of 6/11 and team B has 5/11, team A has a slight advantage, meaning the referee's decision is not perfectly fair.

The comparison to the toss of a coin is significant as it produces a fair decision with equal chances of 1/2 for either outcome. This is not the case in the die-rolling scenario, where the team starting has a higher chance of winning over the team going second.