I'm not sure how to solve this puzzle:
Alice and Bob play a simple dice game. They take turns rolling a single die, starting with Alice, until one of them wins by getting a 6. In the end it turns out that Bob won. What is the probability that he won on his first roll?
Let $B_i$ be the outcome of Bob's die on round $i$. Clearly, $\mathbb{P}(B_i = k) = 1/6$ for any value $k \in \{1,2,3,4,5,6\}$. However, the question is effectively asking
$$
\mathbb{P}(B_1 = 6 \mid \text{Bob wins}) = \;?
$$
Is the answer just $1/6$ since $B_1$ is independent of whether or not Bob wins? This seems counterintuitive to me, because it seems like conditioning on the event that Bob wins should change things. I am uncertain.
Best Answer
Guide:
You have to compute it.
$$P(B_1=6|\text{Bob wins}) = \frac{P(A_1 \ne 6, B_1 = 6)}{P(\text{Bob wins})}$$
$$P(\text{Bob wins}) = \sum_{j=1}^\infty P(\text{Bob wins at } j^{th} \text{ round})$$
Try to write down the $P(\text{Bob wins at } j \text{ round})$ and then use geometric series to compute the quantity.