First one to roll an even number wins the game 
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 will win the game?

As far as I know, for the first roll Bill has a chance of $\frac{1}{2}$, then followed by George with a less than $\frac{1}{2}$ chance and Ross has a lesser than the second roll from George. Why is the answer $\frac{4}{7}$? 
 A: Assume, the probability , that Bill wins , is $p$.
If Bill does not win with the first throw, then George has probability $p$
to win.
If both Bill and George fail, then the probability that Ross wins, is again $p$.
So, Bill has probability $p$, George has probability $\frac{p}{2}$ and George probability $\frac{p}{4}$ to win. So, we have $p+\frac{p}{2}+\frac{p}{4}=\frac{7p}{4}=1$. 
Therefore $p=\frac{4}{7}$.
A: You need to account for all outcomes favorable for Bill to win: eitger ob the first toss, your 0.5 OR on 4th, i.e. all 3 previous tosses are odd, OR on 7th and so on. Can you take it from here?
A: Call the winning probabilities $b,g,r$ respectively. Then clearly $b + g + r = 1$.  Also, call the probabilities that they each win in the first round $b_1, g_1, r_1$. Then $b_1 = \frac{1}{2}, g_1 = \frac{1}{4}$ and $r_1 = \frac{1}{8}$. So we have no decision in the first round with chance $\frac{1}{8}$.
Now, note that $b = b_1 + \frac{1}{8}b = \frac{1}{2} + \frac{1}{8}b$, because $b$ wins with chance $b_1$ in the first round and if gets a second round (which happens with chance $\frac{1}{8}$) he has chance $b$ again, as if he started anew. 
This gives us $b = \frac{8}{7} \cdot \frac{1}{2} = \frac{4}{7}$.
Similarly $g = g_1 + \frac{7}{8}p = \frac{1}{4} + \frac{7}{8}p$, so $g = \frac{8}{7}\cdot \frac{1}{4} = \frac{2}{7}$.
So $r = \frac{1}{7}$, by summing to 1, or doing a similar calculation.
