Q:

What logic does one use to solve this probability question?

Bill, George, and Ross, in order, roll a die. The first one to roll an even number wins and the game is ended. What is the probability that Bill win the game ?. How I approached the question: Since Bill is the first one to roll the die, he has a $.5$ probability of winning the game, if he does win then the question is solved the probability is $.5$, however if he doesn't roll an even number he has still a probability of winning, this probability would be $(.5)^3$. It seems to me that the probability of him winning is $(0.5)^n$ where $n$ is the number of die rolls. Is my logic correct, if not how is the problem answered ?.

A:

Your logic is somehow not bad. However it is not complete, because Bill can win at the first roll, the 4th roll, the 7th roll, etc etc. Let $X_n$ be the event that Bill wins at the $n$-th roll. Clearly $P(X_1)=\frac{1}{2}$ (you know this). $P(X_2)=P(X_3)=0$ because then it is not his game. $P(X_4)$ then Bill must not win one time, George must not win one time, and Ross neither, that together gives $(\frac{1}{2})^3$ and then Bill must win the fourth roll so $P(X_4)=(\frac{1}{2})^4$. In general we have:
\begin{align}
P(\text{ Bill wins }) &= P(\text{ Bill wins at the first roll, fourth roll, seventh roll,..})\\
&=\sum_{n=0}^\infty \left( \frac{1}{2}\right)^{3n+1}\\
&=\frac{1}{2}\frac{1}{1-\frac{1}{8}}\\
&=\frac{4}{7}
\end{align}
So Bill wins with probability $4/7$. You can check that the probability that George and Ross win is $2/7$ and $1/7$ respectively. 

