Question

A die is cast independently until the first 6 appears. If the casting stops on an odd number of times, Bob wins; otherwise, Joe wins. (a) Assuming the die is fair, what is the probability that Bob wins? (b) Let \(p\) denote the probability of a 6 . Show that the game favors Bob, for all \(p\), \(0

Short answer

(a) The probability that Bob wins the game if the die is fair is \(\frac{6}{11}\). \n(b) For any probability \(p\), \(0 \frac{1}{2}\).

Step-by-step solution

Understanding the Probability of Rolling a 6

From the problem, a six-sided die is thrown, and a 'win' is when a 6 is rolled. The probability of rolling a 6 on a fair six-sided die is \(\frac{1}{6}\).

Application of Geometric Distribution - Part (a)

We need to find the probability that Bob wins, which happens when the game stops after an odd number of tries. Since the trials are independent and the probability of 'success' (rolling a 6) stays the same, this is a situation described by the geometric distribution. The sum of probabilities where the first 6 appears on an odd throw is calculated as follows: \[ \sum_{{i=0}}^{{\infty}} {\left(\frac{5}{6}\right)}^{2i} * \frac{1}{6} = \frac{1}{6} * \frac{1}{{1 - {\left(\frac{5}{6}\right)}^2}} = \frac{6}{11} \]

Application of Geometric Distribution - Part (b)

Now let's generalize the probability P to be any value between 0 and 1. The probability that Bob wins becomes:\[ \sum_{{i=0}}^{{\infty}} {(1 - p)}^{2i} * p = p * \frac{1}{{1 - {\left(1 - p\right)}^2}} = \frac{p}{{2p - p^2}} \]Since \(0 \frac{1}{2}\) for all \(0<p<1\), implying the game favors Bob.