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.

All credit must go to @Geoffrey Liddell and @whuber for their help. I post the below for completeness.
As pointed out by @Geoffrey Liddell, I was confusing standard deviations and standard errors. The solution I was looking for (as guided by @whuber) looks something like:
N = 20
p_a = 0.3  # Shooter A has a 30% shooting accuracy
mu_a = N * p_a
std_a = sqrt(N * p_a * (1 - p_a))
p_b = 0.4  # Shooter B has a 40% shooting accuracy
mu_b = N * p_b
std_b = sqrt(N * p_b * (1 - p_b))

# determine coefficients of the Gaussian that describes the difference between
# B and A: 
mu_diff = mu_b - mu_a
std_diff = sqrt(std_a**2 + std_b**2)

x = seq(-20, 20, 0.001)
y = dnorm(x, mu_diff, std_diff)
threshold = 0.5  # Positive threshold because of "B - A"
{
  plot(x, y, 'l', )
  abline(v=threshold, col='red')
  text(-10, 0.06, 'A wins')
  text(10, 0.06, 'B wins')
}

p_b_beats_a = 1 - pnorm(threshold, mu_diff, std_diff)



Where p_b_beats_a is ~0.69 which is the correct answer.
Thanks again to @Geoffrey Liddell and @whuber.
ADDENDUM: In conjunction with confusing standard deviations vs standard errors, I think another issue that was confusing me was working in "proportion space" vs "actual value space".
For example, originally I had p_diff = p_b - p_a. That meant I had to convert my threshold of 0.5 in "actual value space" to a value in "proportion space" (which I still didn't know how to do - assuming it's even possible?). However, by working with actual values i.e mu_diff = mu_b - mu_a, the threshold of 0.5 is now intuitive and the maths follows.
