For $a=1$ and $c>2$ it is always possible for both players to force a draw.

Doesn't that follow from Gelfond–Schneider theorem? ($x=x^{\frac{x+1}{x}}$ is transcendental for irrational algebraic $x$ which is impossible)

The correct way to formulate it would be "What is the probability that both are girls" instead of "What is the probability that the other one is a girl" (but once formulated correctly it suddenly becomes less counterintuitive)

The problem with the current formulation is that, if you only know that there is at least one girl, "the other one" is not well-defined because you could have 2 possibilities for the child that is a girl and no way to distinguish them.

Note that if $\sigma$ is an inner automorphism, then this is equivalent to $G$ having exponent $p$.

Yes, it actually be improved to $2^{2^n}$ using Zsigmondy's theorm

I think that Ben's point is that the fibers are not all the same so it's not a fibration, over t=0 it's two lines crossing while everywhere else it's an hyperbola

Do you have examples of such extensions where $A$ is not of the form $B[x_1,\cdots, x_n]$?

As $|\lambda|=1$ doesn't $\lambda^*=\lambda^{-1}$?

Cartier duality is one of my favorite dualities! Even though the group multiplication of the commutative algebraic group and the multiplication in its coordinate ring look at the first glance (from the usual definition of group schemes) totally different things, they turn out to be perfectly dual to each other if you unravel the axioms in commutative diagrams, which is what allows Cartier duality. The duality is best seen by noting that commutative algebraic groups are basically equivalent to Hopf algebras, which is an obviously self-dual structure.

More generally, if we consider the graph whose vertices are the irreducible components and two components are connected by an edge iff they intersect, then two points of the spectrum are path connected iff their irreducible components are in the same connected component of the graph.

As we can pass through the generic point, there exist a path between any two closed points of the same irreducible component. Moreover if two irreducible component share a point we can use it to pass from one to the other. Hence, at least if the number of irreducible components is finite (for example if R is noetherian), such a path exist iff the two closed points are in the same connected component.

You used $n$ both as the size of $\Omega$ and the size of the basis, but these should be different letters right?