What's the probability that as I roll dice I'll see a sum of $7$ on them before I see a sum of $8$? This question is from DEGROOT's PROBABILITY and STATISTICS.
Problem
Suppose that two dice are to be rolled repeatedly and the
sum $T$ of the two numbers is to be observed for each roll. We shall determine the
probability $p$ that the value $T =7$ will be observed before the value $T =8$ is observed.
Solution
The desired probability $p$ could be calculated directly as follows: We could assume that the sample space $S$ contains all sequences of outcomes that terminate as
soon as either the sum $T = 7$ or the sum $T = 8$ is obtained. Then we could find the
sum of the probabilities of all the sequences that terminate when the value $T = 7$ is
obtained.
However,there is a simpler approach in this example. We can consider the simple
experiment in which two dice are rolled. If we repeat the experiment until either the
sum $T = 7$ or the sum $T = 8$ is obtained, the effect is to restrict the outcome of the
experiment to one of these two values. Hence, the problem can be restated as follows:
Given that the outcome of the experiment is either $T = 7$ or $T = 8$, determine the
probability $p$ that the outcome is actually $T = 7$.
If we let $A$ be the event that $T = 7$ and let $B$ be the event that the value of $T$ is
either $7$ or $8$, then $A ∩ B = A$ and
$$
p = Pr(A|B) = \frac{Pr(A ∩ B)}{Pr(B)}
=\frac{Pr(A)}{Pr(B)}
$$
But $Pr(A) = 6/36$ and
$Pr(B) = (6/36) + (5/36) = 11/36$. Hence, $p = 6/11$.
Now, my doubts are

*

*Why does the author say 

We could assume that the sample space $S$ contains all sequences of outcomes that terminate as
  soon as either the sum $T = 7$ or the sum $T = 8$ is obtained. Then we could find the
  sum of the probabilities of all the sequences that terminate when the value $T = 7$ is
  obtained.
  
*How can we go from lengthy sequences of outcomes that terminate as
  soon as either the sum $T = 7$ or the sum $T = 8$ is obtained to just the outcome of the experiment for which either $T = 7$ or $T = 8$ ?
  

 A: Question 1


*

*Why does the author say 

We could assume that the sample space $S$ contains all sequences of outcomes that terminate as
  soon as either the sum $T = 7$ or the sum $T = 8$ is obtained. Then we could find the
  sum of the probabilities of all the sequences that terminate when the value $T = 7$ is
  obtained.

Answer
Sample space $S$ has $m \rightarrow \infty$ sequences of length $n \rightarrow \infty$ that end in either $7$ or $8$. Out of these sequences we're interested in summing up the probabilities of all the series that end in a $7$. The probability of a sequence of precisely $n$ throws ending in a $7$ is:
$$
P_n(7) = \left(\frac{25}{36}\right)^{n-1} \cdot \frac{6}{36}
$$
However, since $n$ can take any value up to infinity, the overall probability of ending a sequence of any length in a $7$ is the sum of the prob. of ending a seq. after one throw plus the prob. of ending a sequence after two throws, and so on. This is the geometric series:
$$
\Phi_7 = P_1(7) + P_2(7) + P_3(7) + ... + P_n(7)
$$
which, as $n \rightarrow \infty$, sums up to (basic geometric sum formula)
$$
\Phi_7 = \lim_{n \rightarrow \infty} \frac{\frac{6}{36}\left(1-\left(\frac{25}{36}\right)^n\right)}{1-\frac{25}{36}} = \lim_{n \rightarrow \infty} \frac{6}{11}\left(1-\left(\frac{25}{36}\right)^n\right) = \frac{6}{11}
$$
This is the probability of ending a sequence of throws in a $7$ without ever hitting $8$. It's the answer you're looking for using the first, "more complicated" method.
Question 2


*

*How can we go from lengthy sequences of outcomes that terminate as
soon as either the sum $T = 7$ or the sum $T = 8$ is obtained to just the outcome of the experiment for which either $T = 7$ or $T = 8$ ?


Answer
This will become clear if we rephrase the first method a little bit. Sample space $S$ has $m \rightarrow \infty$ sequences of length $n \rightarrow \infty$ which end in either a $7$ or an $8$. The probability of you running a sequence of length $n$ which ends in $7$ is the probability
$$
P_n(7)|(P_n(7) \cup P_n(8)) = \frac{P_n(7) \cap (P_n(7)\cup P_n(8))}{P_n(7)\cup P_n(8)} = \frac{P_n(7)}{P_n(7) \cup P_n(8)}
$$
$$
\frac{P_n(7)}{P_n(7) \cup P_n(8)} = \frac{\left(\frac{25}{36}\right)^{n-1} \cdot \frac{6}{36}}{\left(\frac{25}{36}\right)^{n-1} \cdot \frac{6}{36} + \left(\frac{25}{36}\right)^{n-1} \cdot \frac{5}{36}} = \frac{6}{11}
$$
This is a lot of LaTeX for not a very impressive statement but it is useful because we can now use it to prove by induction the jump from a sequence of length $n$ to a sequence of length $1$. If we run the same formula for $n-1$ we get
$$
P_{n-1}(7)|(P_{n-1}(7) \cup P_{n-1}(8)) = \frac{P_{n-1}(7)}{P_{n-1}(7) \cup P_{n-1}(8)}
$$
where 
$$
\frac{P_{n-1}(7)}{P_{n-1}(7) \cup P_{n-1}(8)} = \frac{\left(\frac{25}{36}\right)^{n-2} \cdot \frac{6}{36}}{\left(\frac{25}{36}\right)^{n-2} \cdot \frac{6}{36} + \left(\frac{25}{36}\right)^{n-2} \cdot \frac{5}{36}} = \frac{6}{11}
$$
But this means that
$$
\frac{P_n(7)}{P_n(7) \cup P_n(8)} = \frac{P_{n-1}(7)}{P_{n-1}(7) \cup P_{n-1}(8)}
$$
and it follows, by induction, that
$$
\frac{P_n(7)}{P_n(7) \cup P_n(8)} = \frac{P_{1}(7)}{P_{1}(7) \cup P_{1}(8)}
$$
Therefore, whatever value $n$ takes, the probability of a sequence of that length ending in $7$ given that it ends in either $7$ or $8$ is equal to the probability of a sequence of length $1$ ending in $7$ given it is a part of $S$.
A: *

*Is a definition of what the probability of an event $A$ is e.g. 
$$
  P(A) = \frac{\text{number of ways $A$ can happen}}{\text{total number of things that can happen}}
$$ Like when you try to figure out what is the probability that you roll a two given that you know you rolled an even number is $1/3$ since there is just one way that a two can come up, and 3 ways that an even number could have come up. 
So, using this definition, to get $P(\text{we roll a 7 before we roll an 8})$ is out of all the sequences of two dice rolls that end in either 7 or 8, how many of them have a 7 occur before 8. 
The author notes that such a strategy for determining the probability is a bit too complex and continues to describe a simpler strategy

*This can be justified with noting that 
$$P(\text{you roll 7 before 8}| \text{you rolled either a 7 or an 8}) = P(\text{you roll 7 before 8})$$ that is the event of rolling a 7 before an 8 is independent of the event of rolling either a 7 or an 8 (i.e. If I told you the result one one die was either 7 or 8, it will not impact the probability that a 7 happens before an 8)
