Dice roll game probability Three people, X, Y, and Z, in order, roll an ordinary die. The first one to roll an even number wins. The game continues until someone rolls an even number. Find the probability that X will win.
X rolls on the first, fourth, seventh, tenth, etc. turn. I figure the probability that the first time someone rolls an even number on a turn is $1 \over 2^n$. For example, on the first turn, the probability of rolling the first even number on the first turn is $ 1\over 2$. The probability of rolling the first even number on the fourth turn is (3 odds x 1 even) or ($1 \over 2^3$)($1 \over 2^1$)=$1 \over 2^4$. Seventh turn, (6 odds x 1 even) or ($1 \over 2^6$)($1 \over 2^1$)=$1 \over 2^7$. Basically, the probability that X will win is
= $1\over2$+$1\over2^4$+$1\over2^7$+$1\over2^{10}$+...
This is where I got stuck. According to the book, the answer is 4/7. How?
 A: It will probability that X be even in the first roll, i.e., $\frac{1}{2}$.
Then probability that X, Y and Z will not be even and X, in the 4 roll, will be, i.e. $(\frac{1}{2})^4$.
As you said the general probability will be
$$\sum_{k=0}^{\infty}\frac{1}{2^{1+3k}}=\frac{1}{2}\sum_{k\ge 0}\frac{1}{8^k}$$
This is a geometric series where $r=\frac{1}{8}<1$ so it converges. You can see here how to continue.
A: X== $1\over2$+$1\over2^4$+$1\over2^7$+$1\over2^{10}$+...
So
8X = 4+$1\over2$+$1\over2^4$+$1\over2^7$+$1\over2^{10}$+...
And subtracting X from 8X we have 7X=4
No sigmas needed. 
A: You can see the answer without any equations.  Note that X is twice as likely to win as Y who in turn is twice as likely to win as Z.  This is because to win on any round, X requires an even (probability 1/2), Y requires an odd followed by an even (probability 1/4), and Z requires 2 odds followed by an even (probability 1/8).  Their winning probabilities are in the ratio of 4:2:1,  so their probabilities are 4/7, 2/7, 1/7.
Note that this depends on the fact that that the probabilities sum to 1 because the probability that someone will win is 1.  If the maximum number of rounds were limited, this would not be the case. In reality, the maximum number of rounds will always be limited.  The players won't live forever.  There will be a limit on the amount of time they are willing to spend playing this game.  So there will always be a tiny probability that no one will ever roll an even.  That probability can easily be made negligible by allowing enough rounds.  But even with an upper limit on the number of rounds, the player's probabilities will still be in the ratio of 4:2:1.  They will sum to 1 minus the probability that the game produces no winner.
A: An interesting way to solve it is using recursion.
Let's say $X$'s probability of winning the game is $p$.
The probability of $X$ winning on the first throw is $\frac{1}{2}$, the probability of the game continuing and $X$ throwing again is $\frac{1}{2^{3}}$. Once it's $X$'s turn again, you can think of it as if the game were starting again from scratch.
Therefore we propose the following:
$$p = \frac{1}{2} + \frac{1}{8}p$$
$$p = \frac{4}{7}$$
