Probability of rolling the first 6 on an even throw? I came across this question and I wasn't sure how to approach it. The question says basically roll a die, what's the probability you get the first 6 on an even throw? I.e., the 2nd, 4th, 6, 8th, etc. die throw. Part of my issue is that obviously you never know when it may come. Here's my guess:
The probability of you getting a 6 on your $n$th  throw is:
$$\left(\frac{5}{6}\right)^{n-1}\cdot\frac{1}{6}$$
Now we want to solve when $n=2k$ where $k$ is a natural number which gives us:
$$\left(\frac{5}{6}\right)^{2k-1}\cdot\frac{1}{6}=\left(\frac{5}{6}\right)^{2k}\cdot\left(\frac{5}{6}\right)^{-1}\cdot\frac{1}{6}$$
The obvious guess is $\frac12$, since it seems it would be equally likely to fall on an even vs. odd, though odd should have an edge since the number of odd rolls are always greater than or equal to the even throws, but unsure.
$$=\left(\frac{5}{6}\right)^{2k}\cdot\frac{1}{5}=\left(\frac{25}{36}\right)^{k}\cdot\frac{1}{5}$$
This point I have no idea how to proceed, have I been doing this right so far? What's the next step?
 A: It is intuitively clear, and not hard to show, that the game terminates with probability $1$. Let $a$ be the probability that the game ends on an odd throw. Then our desired probability is $1-a$.
On the first toss, with probability $\frac{1}{6}$ the game ends immediately, and hence on an odd throw. With probability $\frac{5}{6}$, the game does not end immediately, and therefore the probability the game ends on an odd throw is $1-a$. Thus
$$a=\frac{1}{6}\cdot 1+\frac{5}{6}\cdot(1-a).$$
Solve the linear equation for $a$. We get $a=\frac{6}{11}$, so the probability the game ends on an even throw is $\frac{5}{11}$.    
A: Your final expression:
$$a_k = \frac{1}{5} \cdot\left( \frac{25}{36} \right)^k $$
This correctly gives the probability that your first 6 will occur on the $2k$th roll. So, $a_1$ is the probability that your first 6 is on roll 2; $a_2$ is the probability that your first 6 is on roll 4; etc. 
In order to find the probability $p$ that your first 6 is on any even roll, we just need to add up all the $a_k$s:
$$p = \sum_{k=1}^\infty a_k = \sum_{k=1}^\infty \left[ \frac{1}{5} \cdot \left( \frac{25}{36} \right)^k \right]$$
The formula for the sum of an infinite geometric series with a ratio $r < 1$ (of the form $a_1, a_1 r^1, a_1 r^2, \dots$) is well-known:
$$\sum_{k=1}^\infty a_1 r^k = \frac{a_1}{1-r}$$
Comparing the form of an infinite geometric series to the formula we've come up with for $a_k$, we see that $a_1 = (1/5)(25/36)$ and the ratio is $r = (25/36)$. Hence:
$$p = \frac{a_1}{1-r} = \frac{25/180}{11/36} = \frac{5}{11}$$
This is close to $1/2$, but not exactly equal to it. 
One way to intuitively understand why this is is to think of this as a "game" between two players who take turns rolling the die. In each pair of turns, the first player to roll automatically has an advantage, since if they roll a 6, they win, whereas the second player only gets a chance to roll if the first player hasn't already won. This is why it's less likely that you get the first 6 on an even roll than on an odd roll - the even rolls correspond to the "second player", who is at a disadvantage.
