Every day 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. He calls a day “bad” if he wins less than $50 \%$ of the games in a day. Note that a day is a “bad day” with a probability of $\frac{1}{2}$. Approximate the probability that within $64$ days, he experienced more than $29$ “bad days”.
My approach so far :
Let $S_n = X_1+X_2+...+X_n$ where $X_i = 1_{\text{ ith day is "bad"}}$.
So, we need $P[S_{64} > 29]$. We can use the CLT to approximate this probability.
Here $E[X_i]=\frac{1}{2}$ and $\sigma = \frac{1}{2}$, so the probability is equivalent to $P[\frac{S_{64}-64 \times \frac{1}{2}}{4} > \frac{29}{4}] = 1-\Phi(\frac{29}{4})$ but that is very very close to $0$.
Am I going wrong somewhere with this?
What you wrote for the mean and standard deviation of a single day's outcome is correct; however, your application of the central limit theorem is incorrect.
The CLT states $$\sqrt{n} (\bar X_n - \mu) \overset{d}{\to} \operatorname{Normal}(0, \sigma^2),$$ where $\bar X_n$ is the sample mean $\frac{1}{n} \sum X_i$, and $\mu$, $\sigma$ are the mean and standard deviation of the $X_i$. But since $S_n = n \bar X_n$, that is to say, $S_n$ is the sample total, we have
$$\sqrt{64} \left(\frac{S_{64}}{64} - \frac{1}{2}\right)$$ is approximately normal with mean $0$ and variance $1/4$; equivalently, $$\frac{S_{64}}{4} - 8$$ is approximately standard normal. We then compute using continuity correction $$\Pr\left[S_{64} > 29\right] \approx \Pr\left[\frac{S_{64}}{4} - 8 > \frac{29.5}{4} - 8 \right] \approx \Pr[Z > -0.625] = 1 - \Phi(-0.625) \approx 0.734014.$$
The exact probability is binomial with $n = 64$ and $p = 1/2$:
$$\Pr[S_{64} > 29] = 2^{-64} \sum_{x=30}^{64} \binom{64}{x} = \frac{13537062193921960835}{18446744073709551616} \approx 0.733846,$$ which shows that the normal approximation with continuity correction performs fairly well.
As a side note, $\Pr[X_i = 1] = 1/2$ is correct. Specifically, the probability of a "bad day" is $$\Pr[X_i = 1] = \frac{1}{2^9} \sum_{x=0}^4 \binom{9}{x} = \frac{1}{2}.$$