Question

A and B throw a die alternatively till one of them gets a number more than 4 and wins the game. Then the probability of winning the game by B, if A starts first:

• A. \(\frac{2}{5}\)
• B. \(\frac{3}{5}\)
• C. \(\frac{1}{5}\)
• D. \(\frac{4}{5}\)

Answer 1

The Correct Option is A

Let's determine the probability of player B winning when A and B throw a die alternately, and A starts first. To win, a player needs to roll a number greater than 4 (i.e., 5 or 6). The probability of rolling a number greater than 4 with a fair die is:

\(P(\text{win}) = \frac{2}{6} = \frac{1}{3}\)

Conversely, the probability of not winning (rolling 1, 2, 3, or 4) is:

\(P(\text{not win}) = \frac{4}{6} = \frac{2}{3}\)

Let \(P_B\) be the probability that B wins the game. There are two scenarios where B can win:

1. A does not win on the first roll, allowing B to have a chance to roll. The probability of this happening and then B winning is:

\(P(\text{A doesn't win first roll}) \times P(\text{B wins}) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}\)

2. A does not win on the first roll, B does not win on his roll, and the game resets to the initial state where it is B's turn to win again. Thus, the probability of B winning after this sequence is \(P_B\). The probability of this scenario is:

\(P(\text{A doesn't win first roll}) \times P(\text{B doesn't win}) \times P_B = \frac{2}{3} \times \frac{2}{3} \times P_B\)

Since both scenarios are mutually exclusive, we can sum up these scenarios to get the total probability of B winning:

\(P_B = \frac{2}{9} + \frac{4}{9}P_B\)

Solving for \(P_B\):

\(P_B - \frac{4}{9}P_B = \frac{2}{9}\)

\(\frac{5}{9}P_B = \frac{2}{9}\)

\(P_B = \frac{2}{9} \times \frac{9}{5}\)

\(P_B = \frac{2}{5}\)

Thus, the probability of B winning the game is \(\frac{2}{5}\).