Exercise 4.21: In a game you win \$10 with probability $\frac{1}{20}$ and lose \$1 with probability $\frac{19}{20}$. Approximate the probability that you lost less than \$100 after the first 200 games. How will this probability change after 300 games?
Attempt:
First, we show the winnings and losings together in a single variable. Define \begin{equation*} W_n = 10S_n - (n - S_n) \end{equation*} where $W_n$ denotes the winnings after $n$ games and $S_n$ defines the number of wins in $n$ games. Thus, \begin{equation*} P(W_n \geq 100) = P(11S_n - n> -100) = P\biggr(S_n > \frac{n - 100}{11}\biggr). \end{equation*} Now, we apply Central Limit Theorem in either case, with different values of $n$.
Let $n = 200$, then $S_n \sim Bin(200, \frac{1}{20})$. We thus wish for $S_n > \frac{100}{11}$. Moreover, $E[S_n] = 200 \cdot \frac{1}{20} = 10$ and Var$(S_n) = 200 \cdot \frac{1}{20} \cdot \frac{19}{20} = \frac{19}{2}.$ Thus, it follows from CLT with the continuity correction that \begin{equation*} P\biggr(\frac{\frac{100}{11} - 10}{\sqrt{\frac{19}{2}}}< \frac{S_n - 10}{\sqrt{\frac{19}{2}}}\biggr) \approx 1 - \Phi(-0.457169) \approx 0.6772. \end{equation*}
Now, the book gives a different answer for the first case of 200 games, that being of 0.5636. I wish to understand my mistake before moving on to the next case
Intuitively this also makes sense as the condition of $S_n > \frac{100}{11}$ should be near the top of the bell curve of the normal distribution, as the expected value of 10 is close to $\frac{100}{11}$. However, for the life of my I can't spot the error in my calculation.
(The other Math Stack Exchange question for this questions clarified nothing essentially for me, hence this post.)
If $X$ is the random number of wins in $n$ games, then $$X \sim \operatorname{Binomial}(n = 200, p = 0.05)$$ and the net win/loss random variable is $$W = 10X - (n-X) = 11X - n.$$ Thus $$\Pr[W > -100] = \Pr[11X - 200 > -100] = \Pr[X \ge 10].$$ This last expression is due to the fact that $X$ cannot take on fractional values. Consequently, $$\Pr[X \ge 10] = 1 - \sum_{x=0}^{9} \binom{200}{x} (0.05)^x (0.95)^{200-x} \approx 0.54529\ldots.$$ This is the exact probability: the only approximation here is in the rounding of the fraction to a decimal.
This also gives a key insight as to why your answer is incorrect: just because you are using a normal approximation with continuity correction does not mean the outcomes for $W$ that you want to include in the desired probability can be outside the sample space for $W$.
For instance, if $U \sim \operatorname{Binomial}(n = 500, p = 0.5)$, and I ask you for $\Pr[U < 225.999]$, you would first need to write $\Pr[U < 225.999] = \Pr[U \le 225]$, then you apply continuity correction to approximate it as $$\Pr\left[Z \le \frac{225.5 - 250}{5 \sqrt{5}}\right].$$ The same applies here; thus $$\Pr[W > -100] = \Pr[X \ge 10] \approx \Pr\left[\frac{X - 10}{\sqrt{9.5}} \ge \frac{9.5 - 10}{\sqrt{9.5}}\right] \approx \Pr[Z \ge -0.162221] \approx 0.564434.$$ Evidently, your text is rounding before completing the calculation, or it is using a standard normal table lookup without interpolation, since $\Pr[Z \ge -0.16] \approx 0.563559$. In any case, the approximation $$\Pr[Z \ge -0.457169] \approx 0.676225$$ deviates far too much.