A and B throw a die alternatively till one of them gets 3 or 6 and wins the game. If B starts the game, then the probability of winning the game by A is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2/5
In one throw, getting 3 or 6 is a success:
\( \displaystyle p = \frac{2}{6} = \frac{1}{3},\quad q = 1 - p = \frac{2}{3} \)
B throws first, so A throws 2nd, 4th, 6th, …. A wins on his first throw if B fails and A succeeds; on his second throw if B, A, B fail and A succeeds; and so on:
\( \displaystyle P(\text{A wins}) = qp + q^3p + q^5p + \cdots \)
This is an infinite geometric series with first term qp and common ratio \(q^2\):
\( \displaystyle P(\text{A wins}) = \frac{qp}{1 - q^2} \)
\( \displaystyle = \frac{\frac{2}{3} \cdot \frac{1}{3}}{1 - \frac{4}{9}} = \frac{2/9}{5/9} = \frac{2}{5} \)
Why the other options are incorrect
Option (2) 3/5 — Incorrect. This is the probability that B, who starts, wins: \(1 - \frac{2}{5} = \frac{3}{5}\).
Option (3) 6/11 — Incorrect. With success probability \(\frac{1}{6}\) (only one winning face) the first thrower wins with probability \(\frac{6}{11}\); here a success is 3 or 6, so \(p = \frac{1}{3}\).
Option (4) 5/11 — Incorrect. This is the second thrower’s chance when \(p = \frac{1}{6}\); with \(p = \frac{1}{3}\) A’s chance is \(\frac{2}{5}\).