Let $S_n = W_1 + ... + W_n$, where all $W_j$ are independent and identically distributed. Let $P(W_j=1)= 0.75$ and $P(W_j=2) = 0.25$. We wish to compute $P(S_n = 5)$ for all possible $n$ and find the probability that the sequence $\{ S_n \}_n$ never reaches the value "5".
Looking at this initially, it feels like the desired probability must be zero. Since it's clear that $n \le S_n \le 2n$, as soon as we reach $n=5$, the probability that $\{ S_n \}_n$ never passes the value of 5 must be $0$. Or am I missing something?
To get the $P(S_n = 5)$, I found what should be the generating function of $S_n$:
$g_{S_n}(s) = (0.75s + 0.25s^2)^n$
from $E(s^{S_n})= (E(s^W))^n$
Then applied the relation: $$P(S_n = 5) = \frac{{g_{S_n}}^{(k)}(0)}{k!} = \frac{{(0.75s + 0.25s^2)^n}^{(k)}(0)}{k!}$$
This gave me a big expansion:
$$ \frac{(n-4)(n-3)(n-2)(n-1)n(0.75^5)+5(n-3)(n-2)(n-1)n(0.75)^3 +2.8125(n-2)(n-1)n}{5!} $$
Does this seem reasonable? I'm pessimistic about this result, since the above gives a probability bigger than one when we let $n$ be five. Thank you for your ideas/input!