I'm assuming the paths must go right and down only.
You can count the paths by using the inclusion-exclusion principle. Without considering the points that must be avoided, there are $\binom{6+6}{6}$ paths. Now subtract the $\binom{4+1}{1}\binom{2+5}{5}$ paths that visit the first bad point, the $\binom{2+4}{4}\binom{4+2}{2}$ paths that visit the second bad point, and the $\binom{4+5}{5}\binom{2+1}{1}$ paths that visit the third bad point. Then add back in the paths that visit two bad points. No paths visit all three bad points.
\begin{align}
&\binom{12}{6}
-\left(\binom{5}{1}\binom{7}{5} +\binom{6}{4}\binom{6}{2}+\binom{9}{5}\binom{3}{1}\right)
+\left(\binom{5}{1}\binom{4}{4}\binom{3}{1}+\binom{6}{4}\binom{3}{1}\binom{3}{1}\right)\\
&=924-(5\cdot 21+15\cdot 15+ 126\cdot 3)+(5\cdot 1\cdot 3+15\cdot 3\cdot 3)\\
&=924-(105+225+378)+(15+105)\\
&=924-708+120\\
&=
{\color{red}{366}}
\end{align}
Here's an alternative solution that uses a Pascal-type recursion. Let $p(i,j)$ be the number of good paths that start from $A=(0,0)$ and reach point $(i,j)$. We want to compute $p(6,6)$.
By conditioning on the last step into $(i,j)$, we find that $p(i,j)$ satisfies the following recursion:
$$
p(i,j)=
\begin{cases}
0 &\text{if $(i,j)$ is a bad point}\\
1 &\text{if $i=0$ or $j=0$}\\
p(i-1,j)+p(i,j-1) &\text{otherwise}
\end{cases}
$$
The resulting values of $p(i,j)$ are:
\begin{matrix}
i\backslash j &0 &1 &2 &3 &4 &5 &6 \\
0 &1 &1 &1 &1 &1 &1 &1 \\
1 &1 &2 &3 &4 &0 &1 &2 \\
2 &1 &3 &6 &10 &10 &11 &13 \\
3 &1 &4 &10 &20 &30 &41 &54 \\
4 &1 &5 &0 &20 &50 &91 &145 \\
5 &1 &6 &6 &26 &0 &91 &236 \\
6 &1 &7 &13 &39 &39 &130 &{\color{red}{366}}
\end{matrix}
So $p(6,6)=366$.
(A) Some more solutions with uniform distributions
It is easy to come by more stable tournaments on $2m+1$ vertices with a uniform distribution of $\frac1{2m+1}$ on each vertex. Any regular tournament will do! To see this, note that if we start with $v_1 = v_2 = \dots = v_{2m+1}$ in one round, then in the next round we will have
$$
v_i \gets v_i^2 + \sum_{j \in N^+(i)} 2v_i v_j = \frac1{(2m+1)^2} + m \cdot \frac{2}{(2m+1)^2} = \frac1{2m+1}
$$
where $N^+(i)$ denotes the out-neighborhood of vertex $i$, so by assumption $|N^+(i)|=m$ for all $i$.
The "basic odd tournament" is one way to construct a regular tournament on $2m+1$ vertices, but for large $m$ there are many: see OEIS sequence A096368 for their quickly growing count. The first new one appears with $7$ vertices:
To see that this is not isomorphic to the basic odd tournament, note that if you take any vertex here and look at the three vertices it points to, those three vertices go in a cycle. In the basic odd tournament, if you take any vertex and look at the three vertices it points to, they will not.
(B) Solutions with more unusual distributions
But the tournament above is not very interesting. With $7$ vertices, we also get the first tournament that's stable for its own mysterious and special reasons:
This cannot be obtained from a smaller regular tournament by the vertex blow-up operation, because of the denominator of $11$.
We can get a larger variety of stable tournaments by a sort of inverse of the vertex blow-up. Do the following:
- Take a known stable tournament $T$ on $2a+1$ vertices.
- Add $b$ more vertices, all with arcs to every vertex of $T$.
- Add $b$ more vertices, all with arcs from every vertex of $T$.
- Add arcs between every pair of the new vertices arbitrarily in such a way that in the end, every vertex has indegree $a+b$ and outdegree $a+b$.
(Here, $b$ is arbitrary, but should be at least $2a+1$ for step $4$ to be possible; then we can apply Landau's theorem to the vertices outside $T$ to make sure they have the desired outdegrees. I'm omitting the details.)
The result is a regular tournament on $2a+2b+1$ vertices, thus assigning $v_i = \frac1{2a+2b+1}$ for all $i$ will work.
Now, replace tournament $T$ by a single vertex $t$, getting a tournament on $2b+1$ vertices. In this tournament, setting $v_t = \frac{2a+1}{2a+2b+1}$ (combining the weights of all the vertices of $T$) and keeping all other vertices at $\frac1{2a+2b+1}$ is still stable! (It is crucial that every vertex outside $T$ has the same behavior against every vertex of $T$, for this to work - but with that condition, the equations practically don't change.)
To get a larger variety of distributions, do this with multiple "planted" tournaments $T_1, T_2, \dots$ instead of the single $T$. For example, the $7$-vertex tournament above with distribution $(1/11, 1/11, 1/11, 1/11, 1/11, 3/11, 3/11)$ can be obtained from an $11$-vertex regular tournament in which we planted two $3$-vertex stable tournaments $T_1$ and $T_2$, then shrank $T_1$ and $T_2$ to single vertices.
(C) There are no solutions with an even number of vertices
All of the solutions above give an odd number of vertices; in particular, there are no regular tournaments with a even number of vertices. (With $2m$ vertices, every vertex would have indegree and outdegree $m - \frac12$, which is nonsense.) In fact, there are also no stable tournaments with an even number of vertices!
To prove this, first take any stable tournament, and write its stable distribution $(v_1, v_2, \dots, v_n)$ in the form $v_i = \frac{p_i}{q_i} \cdot 2^{r_i}$, where $p_i, q_i$ are odd numbers and $r_i$ is an integer (not necessarily positive).
Then, replace the $i^{\text{th}}$ vertex by a stable tournament on $\frac{p_i}{q_i} \cdot q_1 q_2 \cdots q_n$ vertices (an odd integer). The value $v_i$ is split evenly among these vertices, so each of them has value $\frac{2^{r_i}}{q_1 q_2 \cdots q_n}$. If we do this for every $i$, then in the end, every vertex will have the same value, up to a factor which is a power of $2$: there is some common denominator $Q$ such that all vertices have values from the set $\{\frac1Q, \frac2Q, \frac4Q, \frac8Q, \dots\}$. This should still be a stable tournament (with a ginormous number of vertices).
In the ginormous tournament, take another look at the identity
$$
v_i = v_i^2 + \sum_{j \in N^+(i)} 2v_i v_j.
$$
First consider a vertex $i$ such that $v_i = 1/Q$; since we take $Q$ to be the least common denominator, some such vertex must exist. Multiplying through by $Q^2$ on both sides, we get
$$
Q = 1 + \sum_{j \in N^+(i)} 2Q v_j
$$
which shows that $Q$ must be odd.
Now suppose that there is any vertex $i$ such that $v_i = 2^k/Q$ for $k>0$. Take the above identity such a vertex and multiply through by $Q/v_i = Q^2/2^k$ on both sides; we get
$$
Q = 2^k + \sum_{j \in N^+(i)} 2Q v_j.
$$
Here, the LHS is odd, but the RHS is even: contradiction! So in fact all vertices must have $v_i = 1/Q$. This is only possible if the tournament is a regular tournament, which means that it has an odd number of vertices.
But to get there, we did a number of vertex blow-up operations which replaced one vertex by an odd number. This does not affect parity. So we must have started with an odd number of vertices: a stable tournament with an even number of vertices does not exist.
Best Answer
For the sake of simplicity, I'll approximate everything with continuous functions, even though the actual game only lets you have an integer number of factories.
Let $f(t)$ be the number of civilian factories Germany has after $t$ days, assuming that production hasn't switched to military yet. The player gains $5f(t)$ industry points per day, which can be spent on new civs at a cost of 10800 points each. That's $f'(t) = \frac{5f(t)}{10800} = \frac{f(t)}{2160}$ new factories per day. Solving this differential equation with the initial condition $f(0) = 23$ gives $f(t) = 23e^{t/2160}$.
Let $S$ be the number of days between the start of the game and the day that production is switched from civilian to military factories. Note that the time interval between the start of the game (1936-01-01) and the start of the war (1939-09-01) is 1339 days.
After the switch to military factory production, the player gains $5 \times 23e^{S/2160} = 115e^{S/2160}$ industry points per day, over a period of $1339-S$ days, for a total of $115e^{S/2160}(1339-S)$ points spendable on mils.
And it turns out that this value is optimized...at $S=0$. You already have all the civilian factories you need, so go ahead and focus on military from the very beginning.