Probability that a die game ends on an even throw (Linear Solution) I've researched solutions for this problem seeing as I couldn't seem to solve it on my own, and there's one in this very forum that albeit looking the most simple, I can't fully understand.
Here is the proposed solution by
André Nicolas (https://math.stackexchange.com/users/6312/andr%c3%a9-nicolas), Probability of rolling the first 6 on an even throw?, URL (version: 2014-05-04): https://math.stackexchange.com/q/780464
The problem I myself am solving says the game ends when you roll the die and get a '1', which for this effect I consider to be the same, since the probability of rolling a 1 is $ \frac{1}{6} $
So, on a first glance it seems clear enough, we're calculating the probability of the game ending on an odd throw $-P(a)-$ so if it ends in the first throw naturally there'll be no doubt it is an odd throw, hence:
$$
\frac{1}{6} . 1
$$
Now, the second term. I assume $ \frac{5}{6}$  is the sum of the probabilities of 2, 3, 4, 5 and 6 of being rolled. If any of these numbers are rolled, the game continues. To fulfill what I want to calculate $-P(a)-$, I need these throws that'll not end the game to be even so that adding the final throw of that $1$ that'll eventually be rolled and end the game gives me and odd number of throws. 
Basically:
$ odd\; throws= 1 \;throw + even\;throws $
This reasoning seems solid enough until I start to question that $ (1-a)$. It's the complement of what we're calculating, so it's the probability the game will end on an even throw. The problem? To me it seems that this probability includes the final, winner roll (1), the same we're considering here separately in the first term. Being multiplied by $\frac{5}{6}$ is supposed to take away this winner roll(1) ?
Things get more confusing if I try to consider the probability of the game ending on an even throw. Again, the first term seems to be a no-brainer:
$$
\frac{1}{6} . 0 
$$
If the game ends on the first throw, it is an odd throw and so we don't consider it. But when you add the second term, what am I making of this first one?
$$
\frac{5}{6} . P(a) ... so\;in\;the\;end= 0 + \frac{5}{6} . P(a)
$$
Also if I were to use the basic probability notation... '+' would mean ∪ (or). So what, either the game ends or it doesn't? That can't be right.
As you can see I'm very confused. I'd greatly appreciate an explanation as to why the first linear equation works (because I know it does, the results I get in the end are correct: $ P(a) =$ $\frac{6}{11}$... so $P( $a')= 1 -$\frac{6}{11}$ = $\frac{5}{11}$ probability that the game ends on an even throw).
 A: Assuming you are asking "you are throwing a die repeatedly.  What is the probability that the first $1$ occurs on an even numbered throw?"
I don't understand your confusion regarding the old solution, but perhaps a different solution will clarify matters.
Since we don't care about the outcome of the throw unless it is a $1$, let us just call $X$ any roll other than $1$.  Then the winning paths are of the form $X^i1$ where $i$ is an odd number.  Since the probability of throwing $X$ is $\frac 56$ the probability of that path is $\left( \frac 56 \right)^i\times \frac 16$ so the answer is $$\frac 16\times \sum_{n=0}^{\infty}\left( \frac 56 \right)^{2n+1}=\frac 5{11}$$
Worth noting:  I'd say the other solution is better than this one, as this method depends on being able to sum the infinite series conveniently.  Of course, the other method depends critically on proving that the game terminates which, while intuitively clear, requires some effort.
To phrase the old method in (possibly) different terms:  Let $p$ be the probability you want.  Let $q$ be the probability that the first $1$ occurs on an odd numbered toss.  Since the game must end, we have $p+q=1$ or, equivalently, $q=1-p$.  Note:  to see the game must end, note that the probability that it has not ended in $N$ tosses is $\left( \frac 56 \right)^N$ and then remark that this goes to $0$ as $N$ goes to $\infty$).  Now, in order to win you have to throw $X$ the first time (probability $\frac 56$) and then win on an odd numbered toss (probability 1-p).  Thus we have $$p=\frac 56\times (1-p)\implies 6p=5-5p\implies p=\frac 5{11}$$
