A probability question that uses the binomial expansion The question is as follow:
(i) Find the binomial expansion of $(1-x)^{-3}$ up to and including $x^{4}$.
(ii)
A player throws a 6-sided fair die at random. If he gets an even number, he loses the game and the game ends. If he gets a "1", "3"  or "5" he throws the die again. He wins the game if he gets either "3" or "5" thrice consecutively (eg. 335, 555, 353) and the game ends. Find the exact probability of him winning the game.
I have been thinking about this question for quite a while. Obviously, the author of the question wants us to solve part (ii) with the help of part (i). However, to solve part (ii), it looks more an infinite series to me (the possible combinations of winning the game).
Could anyone contribute to solve this question please? 
 A: Binomial expansion of 
$$(1 + x)^a = 1 + ax + \frac{a(a-1)}{2!} x^2 + \frac{a(a-1)(a-2)}{3!} x^3 + \frac{a(a-1)(a-2)(a-3)}{4!} x^4 + {\cal O}(x^5)$$
when $|x| < 1$.
Part 2) I have just written out the pattern of him winning in 3 steps, 4 steps, 5 steps, etc.
$$ \Pr(Win) = \frac{1}{3^3} \bigg[ 1 + \Big(\frac{1}{6}\Big) + \Big(\frac{1}{6^2} + \frac{1}{6 \cdot 3}\Big) + \Big(\frac{1}{6^3} + 2 \frac{1}{6^2\cdot3} + \frac{1}{6\cdot 3^2}\Big) + \Big(\frac{1}{6^4} + 3 \frac{1}{6^3\cdot3} + 3\frac{1}{6^2\cdot 3^2}\Big) + \Big(\frac{1}{6^5} + 4 \frac{1}{6^4\cdot3} + 6\frac{1}{6^3\cdot 3^2}\Big) + \cdots \bigg]
$$
Do you see what can be done from here?

I wrote them as series which can be summed.
$$\Pr(Win) = \frac{1}{3^3}\bigg[\sum_{n=0}^\infty 6^{-n} + \frac{1}{3}\sum_{k=0}^\infty k 6^{-k} + \frac{1}{3^2} \sum_{h=1}^\infty {h+1 \choose 2}6^{-h}  \bigg],$$
Each of which can be summed without too much difficulty.
A: This doesn't really answer your question how to make use of part (i), but since you already accepted an answer that doesn't do that either, perhaps that's not so bad, and perhaps this answer can help someone else answer the original question.
The generating function of three variables $x,y,z$ for sequences of three types of rolls with exactly one roll of type $z$ at the end and with no three consecutive rolls of type $x$ is
$$
z\left(1+x+x^2\right)\sum_{j=0}^\infty\left(y\left(1+x+x^2\right)\right)^j=z\cdot\frac{1+x+x^2}{1-y\left(1+x+x^2\right)}\;.
$$
Substituting the probabilities $x=\frac13$, $y=\frac16$, $z=\frac12$ yields the probability
$$
\frac12\cdot\frac{1+\frac13+\frac19}{1-\frac16\left(1+\frac13+\frac19\right)}=\frac{39}{41}
$$
for losing the game, and thus $\frac2{41}$ for winning the game.
The expansion of $(1-x)^{-3}$ might conceivably enter into it if $1+x+x^2=\frac{1-x^3}{1-x}$ is used.
