A plays $9$ games with his computer. In each game, he will win, draw or lose with probabilities $\frac{1}{2}$ , $\frac{1}{3}$ and $\frac{1}{6}$ respectively.A day is “great” if he wins all of the $9$ games. Approximate the probability that within $128$ days, he experienced exactly three “great days”.
My approach
See that $P[\text{great day}] = (\frac{1}{2})^9$.
Let $S_n = \sum_{i=1}^{n} X_i$ where $X_i = 1_{\text{ith day is great}}$.
We need to approximate $P[S_{128}=3]$. For using the CLT, we regard $3$ as the interval $(2.5,3.5)$ and thus calculate $P(|S_{128}-3| \le \frac{1}{2}) = \Phi(6.506)- \Phi(4.504)$ which is almost $0$.
I cannot better this estimate at all! Is there any other way out?
The exact probability is simply $$\Pr[X = 3] = \binom{128}{3} (1/512)^3 (1 - 1/512)^{128-3} = \frac{(128)(127)(126)}{6} \cdot \frac{511^{125}}{512^{128}} \approx 0.00199201.$$ If you don't have a computer to evaluate this expression, then we would employ a Poisson approximation, not a normal approximation, because the probability of a "great" day is $p = 1/512$ which is quite small; the expected number of great days in $n = 128$ days is only $\lambda = np = \frac{1}{4}$.
Recall that if $X \sim \operatorname{Binomial}(n, p)$ with $p$ sufficiently small and $n$ sufficiently large, then we may approximate $X$ with $$N \sim \operatorname{Poisson}(\lambda = np),$$ and $$\Pr[N = x] = e^{-\lambda} \frac{\lambda^x}{x!}.$$ In your case, $\lambda = 1/4$, so $$\Pr[X = 3] \approx e^{-1/4} \frac{(1/4)^3}{3!} \approx 0.00202813.$$
I don't understand why you would think of using a normal approximation here. The CLT does not apply because $p$ is so small; thus you would need a much larger $n$ in order for the sample total to be well-modeled by a normal distribution.