Probability that a 40% shooter is better than a 30% shooter given 20 shots? I'm working through the Regression and Other Stories (ROS) textbook and it's not clear to me how Gelman etal. would like a reader to answer question 4.3 (which I summarize below):

Two basketball shooters. Shooter A has an accuracy of 30% while shooter B has an accuracy of 40%. They each take 20 shots. What is the probability that shooter B makes more shots?

Now, I'm aware of the following question and it all makes sense. What confuses me however, is that at no point in the textbook (unless I missed it) does Gelman etal. demonstrate that approach. Therefore, I wonder if there is another way to answer the question using a "Gelman" method?
So far I have the following R code:
N = 20
p_a = 0.3  # shooter A has a 30% accuracy
se_a = sqrt(p_a * (1 - p_a) / N)
p_b = 0.4  # shooter B has a 40% accuracy
se_b = sqrt(p_b * (1 - p_b) / N)

# The probability that shooter B makes more shots than B with mean and s.e. of: 
p_diff = p_b - p_a
se_diff = sqrt(se_a**2 + se_b**2)

My intuition tells me that the difference in their ability is described by a (Gaussian?) distribution with mean p_diff (0.1) and standard error of se_diff (0.15). Given that p_diff = p_b - p_a, positive values of p_diff account for instances when B "beats" A while negative values are the opposite. See below:

Therefore, the probability of B beating A is the area under the graph to the right of the red line. Which can be calculated in r with:
# p_diff is 0.1
# se_diff is 0.15
p_b_beats_a = 1 - pnorm(0, p_diff, se_diff)

Which gives the incorrect value of ~0.75 (see link above for correct value).
What step/intuition have I missed? Given that at no point in ROS has Gelman etal. mentioned continuity correction (as the above link does), I'm assuming there is some other "Gelman" method/intuition that I've missed ...
Alternatively, how does one go from the method as described in the link to the "Gelman" approach?
I'd appreciate both the maths/code but also an intuitive explanation.

Since the referenced thread only includes a direct approach in a comment, here goes.
You can directly evaluate the probability of the discrete events when shooter A makes more shots than shooter B and sum them together rather than approximating a normal distribution. If $A \sim Bin(20, 0.3)$ and $B \sim Bin(20, 0.4)$ then you are looking for $P(A < B)$, which is the same as $\Sigma_{n=0}^{20} P(A < n, B = n)$
We can probably assume the shots are independent, so
$$
P(A<B) = \Sigma_{n=0}^{20} P(A \le n-1)P(B = n)$:
$$
p <- 0
for (n in 0:20) {
  p <- p + dbinom(n, 20, 0.4) * pbinom(n-1, 20, 0.3)
}

If you want to approximate the Binomial distribution by a Normal one and $A \sim Bin(n,p)$, you need $np$ to be large: it might be large enough here, but it, but given that you have to do the whole continuity correction thing, it might be more hassle than its worth!
A quick search shows evaluating the Binomial distribution comes up in ROS in Ex. 3.5, but there doesn't appear to be a solid section on probability theory.
I hope this demonstrates another approach. I made the mistake of getting A and B the wrong way around initially, hopefully I haven't missed any less than/equals signs!
