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: On the first round, Bill has a probability of winning of $0.5$. 
Knowing that Bill did not win or lose on the first round, he has a probability of winning on the second round of $0.5^4$, This is because if he did not win or lose in the first round, the probability of that happening is $0.5^3$, and now the probability of his winning again is $0.5$.
Now, on the third round, his probability of winning is is $0.5^7$, by the same logic.
And so on.
Now we add up the probability he has of winning for every round, and we get a geometric series with reason $q=0.5^3$. His probability of winning is therefore:
$p = 0.5 + 0.5^4 + 0.5^7 + \cdots = 0.5 \times \frac{1}{1-0.5^3}$
A: 
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 $(.5)^3$

You are almost there.   Call a group of rolls by the three players a round.
Let $P_B$ be the probability that Bill wins the game.   Bill wins the game (starting from the first round) if either he wins the first round, or if nobody wins the first round and Bill wins the game starting from the second round.   Use your evaluation to build a recursive formula:
$$P_B= 0.5+0.5^3P_B$$
This is because the (conditional)probability that Bill wins the game when given nobody wins the first round, is equal to the probability that Bill wins the game — because they basically just start over again from the beginning.
Solve to find: $P_B\phantom{= 4/7}$
Likewise evaluate the probabilities that George wins the game and that Ross wins the game.   Comment on whether/why going first conveys any advantage and verify that the sum equals one.

 $P_G =0.5 P_B$ and $P_R=0.5^2P_B$ because of some reason.

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. 
A: Good start! Rather, he wins on the first roll with probability $0.5,$ on the fourth roll with probability $(0.5)^4,$ on the seventh roll with probability $(0.5)^7,$ and so on. These are the only rolls on which he can win, and if he doesn't win on one of those rolls, then he loses with probability $1$ (that is, there is effectively no chance that the game will go on forever). Thus, since he can't win on multiple turns, then the probability of winning is $$\begin{eqnarray}\sum_{n=0}^\infty\left(\frac12\right)^{3n+1}&=&\sum_{n=0}^\infty\frac12\cdot\left(\frac12\right)^{3n}\\&=&\frac12\cdot\sum_{n=0}^\infty\left(\frac12\right)^{3n}\\&=&\frac12\cdot\sum_{n=0}^\infty\left(\left(\frac12\right)^3\right)^n\\&=&\frac12\cdot\sum_{n=0}^\infty\left(\frac18\right)^n,\end{eqnarray}$$ and using the formula for the sum of a convergent geometric series, we have that the probability is $$\frac12\cdot\frac{1}{1-\frac18}=\frac12\cdot\frac1{\frac78}=\frac12\cdot\frac87=\frac47.$$
