With $n = 400$ trials, the exact probability distribution for the number of heads $X$ observed is given by $X \sim {\rm Binomial}(n = 400, p = 1/2)$, assuming the coin is fair. Since calculating $\Pr[160 \le X \le 200]$ requires a computer, and $n$ is large, we can approximate the distribution of $X$ as ${\rm Normal}(\mu = np = 200, \sigma^2 = np(1-p) = 100)$. Thus $$\begin{align*} \Pr[160 \le X \le 200] &\approx \Pr[159.5 \le X \le 200.5] \\ &= \Pr\left[\frac{159.5 - 200}{\sqrt{100}} \le \frac{X - \mu}{\sigma} \le \frac{200.5 - 200}{\sqrt{100}} \right] \\ &= \Pr[-4.05 \le Z \le 0.05] \\ &= \Phi(0.05) - \Phi(-4.05) \\ &\approx 0.519913. \end{align*}$$ Note that we employed continuity correction for this calculation. The exact probability is $0.5199104479\ldots$.
A similar calculation applies for $\Pr[160 \le X \le 190]$. Using the normal approximation to the binomial, you would get an approximate value of $0.171031$. Using the exact distribution, the probability is $0.17103699497659\ldots$.
An analytic calculation shows the probability that the second player wins is $0.521491$.
For a start, consider the probability that the sum $S$ is equal to $n$ at some point, where $n \le 100$; let's call that probability $p(n)$. On the previous step, the sum must have been $n-i$ for some $i$ with $0 \le i \le n-1$, and then the player must have drawn $i$, with probability $1/100$. So
$$p(n) = \sum_{i=0}^{n-1} \frac{p(i)} {100}$$
where we define $p(0) = 1$. The solution to this recurrence is
$$p(n) = \frac{(1+1/100)^{n-1}} {100}$$
for $0 \lt n \le 100$. (This formula does not hold for $n > 100$, but we will not need values of $p(n)$ in that range.)
Now that we know how to compute $p(n)$, let's consider how the player's scores can be $x$ and $y$ for the first and second players, respectively. We might as well consider a slightly more general problem and ask how the first player's score can be $x$ when the score is the first number drawn with $S \ge G$ for some $G \le 100$. Let's say the previous number drawn was $m$, where $m \le G$, and then the next number was $x$, where $m+x > G$. The probability of this sequence of events is $p(m) / 100$. For the first player's score, we are interested only in the case $G=100$.
Suppose we then continue drawing numbers until $S \ge 200$, with the last number drawn being $y$ and the previous number being $n$, so $n+y > 200$. Since we started at $m+x$, this is just like starting from zero as in the first case, but now with a goal of $200 - (m+x)$ instead of $100$. Then the associated probability is $p(n -(m+x)) / 100$. So the overall probability of the sequence of numbers $m, m+x$, (zero or more numbers omitted), $n, n+y$ is
$$\frac{p(m) \cdot p(n-(m+x))}{100^2}$$
We are interested in the total probability of the cases where $x < y$. Taking into account the constraints on $m, x, n$ and $y$, this probability is
$$\sum_{m=1}^{100} \sum_{x=101-m}^{100} \sum_{n=m+x}^{200} \sum_{y= \max(200-n,x)+1}^{100} \frac{p(m) \cdot p(n-(m+x))}{100^2}
$$
Observing that the summand does not involve $y$, we can simplify this sum to
$$
\sum_{m=1}^{100} \sum_{x=101-m}^{100} \sum_{n=m+x}^{200} \frac{[100-\max(200-n,x)] \cdot p(m) \cdot p(n-(m+x))}{100^2}$$
which evaluates to $0.521491$.
Best Answer
I think this is a slightly more intuitive way of looking at the question:
Suppose $f (x)=k$. Once we have chosen $k$, there are $200$ possible values for $g (x)$, one of which is $k$, hence we get an answer of $\frac{1}{200}$.