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Consider $B$ a finite-type integral quasi-projective scheme over $\mathbb{Z}$ such that $B(\mathbb{Z})$ infinite. (If you like, take $B$ to be the affine line). Let $X \to B$ be a generically smooth proper family of curves of genus $g > 1$. Assume the $b \in B(\mathbb{Z})$ for which $X_b(\mathbb{Q}) = \emptyset$ are Zariski-dense in $B$. Must then the proportion of members $X_b$, over $b \in B(\mathbb{Z})$, for which $X_b(\mathbb{Q}) \neq \emptyset$, be equal to $0$? (Variant: the same question with $B(\mathbb{Z})$ replaced by $B(\mathbb{Q})$.)

An explicit variant (although not quite a special case as it stands). Let $f \in \mathbb{Z}[x]$ be irreducible of degree $> 4$. Do the integers $N$ for which $f(x) = Ny^2$ has a rational solution, have density zero? Variant: let $N$ range over the primes, or over the squarefrees.

In the above setup, we may include this example by assuming more generally that $X_{\mathbb{Q}} \to B_{\mathbb{Q}}$ admits exactly $m$ sections defined over $\mathbb{Q}$, and ask whether the density of $b \in B(\mathbb{Z})$ with $|X_b(\mathbb{Q})| > m$ is zero. Alternatively, drop the properness and generic smoothness assumptions, including instead the assumption that the smooth projective model of the generic fibre has genus $> 1$.

A comment. This, of course, is easily seen to fail for families of genus zero. Various heuristics, and some standard conjectures, imply that it should also fail for genus one families. Is there an example in genus one for which it can be proved that a positive proportion, yet not almost all of the members have rational points -- or is this also out of reach?

Added later. Indeed, relaxing the assumption on the genus, are there any possibilities for the proportion other than $0$, $1/2$, and $1$?

Consider $B$ a finite-type integral quasi-projective scheme over $\mathbb{Z}$ such that $B(\mathbb{Z})$ infinite. (If you like, take $B$ to be the affine line). Let $X \to B$ be a generically smooth proper family of curves of genus $g > 1$. Assume the $b \in B(\mathbb{Z})$ for which $X_b(\mathbb{Q}) = \emptyset$ are Zariski-dense in $B$. Must then the proportion of members $X_b$, over $b \in B(\mathbb{Z})$, for which $X_b(\mathbb{Q}) \neq \emptyset$, be equal to $0$? (Variant: the same question with $B(\mathbb{Z})$ replaced by $B(\mathbb{Q})$.)

An explicit variant (although not quite a special case as it stands). Let $f \in \mathbb{Z}[x]$ be irreducible of degree $> 4$. Do the integers $N$ for which $f(x) = Ny^2$ has a rational solution, have density zero? Variant: let $N$ range over the primes, or over the squarefrees.

In the above setup, we may include this example by assuming more generally that $X_{\mathbb{Q}} \to B_{\mathbb{Q}}$ admits exactly $m$ sections defined over $\mathbb{Q}$, and ask whether the density of $b \in B(\mathbb{Z})$ with $|X_b(\mathbb{Q})| > m$ is zero. Alternatively, drop the properness and generic smoothness assumptions, including instead the assumption that the smooth projective model of the generic fibre has genus $> 1$.

Consider $B$ a finite-type integral quasi-projective scheme over $\mathbb{Z}$ such that $B(\mathbb{Z})$ infinite. (If you like, take $B$ to be the affine line). Let $X \to B$ be a generically smooth proper family of curves of genus $g > 1$. Assume the $b \in B(\mathbb{Z})$ for which $X_b(\mathbb{Q}) = \emptyset$ are Zariski-dense in $B$. Must then the proportion of members $X_b$, over $b \in B(\mathbb{Z})$, for which $X_b(\mathbb{Q}) \neq \emptyset$, be equal to $0$? (Variant: the same question with $B(\mathbb{Z})$ replaced by $B(\mathbb{Q})$.)

An explicit variant (although not quite a special case as it stands). Let $f \in \mathbb{Z}[x]$ be irreducible of degree $> 4$. Do the integers $N$ for which $f(x) = Ny^2$ has a rational solution, have density zero? Variant: let $N$ range over the primes, or over the squarefrees.

In the above setup, we may include this example by assuming more generally that $X_{\mathbb{Q}} \to B_{\mathbb{Q}}$ admits exactly $m$ sections defined over $\mathbb{Q}$, and ask whether the density of $b \in B(\mathbb{Z})$ with $|X_b(\mathbb{Q})| > m$ is zero.

A comment. This, of course, is easily seen to fail for families of genus zero. Various heuristics, and some standard conjectures, imply that it should also fail for genus one families. Is there an example in genus one for which it can be proved that a positive proportion, yet not almost all of the members have rational points -- or is this also out of reach?

Consider $B$ a finite-type integral quasi-projective scheme over $\mathbb{Z}$ such that $B(\mathbb{Z})$ infinite. (If you like, take $B$ to be the affine line). Let $X \to B$ be a generically smooth proper family of curves of genus $g > 1$. Assume the $b \in B(\mathbb{Z})$ for which $X_b(\mathbb{Q}) = \emptyset$ are Zariski-dense in $B$. Must then the proportion of members $X_b$, over $b \in B(\mathbb{Z})$, for which $X_b(\mathbb{Q}) \neq \emptyset$, be equal to $0$? (Variant: the same question with $B(\mathbb{Z})$ replaced by $B(\mathbb{Q})$.)

An explicit variant (although not quite a special case as it stands). Let $f \in \mathbb{Z}[x]$ be irreducible of degree $> 4$. Do the integers $N$ for which $f(x) = Ny^2$ has a rational solution, have density zero? Variant: let $N$ range over the primes, or over the squarefrees.

In the above setup, we may include this example by assuming more generally that $X_{\mathbb{Q}} \to B_{\mathbb{Q}}$ admits exactly $m$ sections defined over $\mathbb{Q}$, and ask whether the density of $b \in B(\mathbb{Z})$ with $|X_b(\mathbb{Q})| > m$ is zero. Alternatively, drop the properness and generic smoothness assumptions, including instead the assumption that the smooth projective model of the generic fibre has genus $> 1$.

A comment. This, of course, is easily seen to fail for families of genus zero. Various heuristics, and some standard conjectures, imply that it should also fail for genus one families. Is there an example in genus one for which it can be proved that a positive proportion, yet not almost all of the members have rational points -- or is this also out of reach?

Added later. Indeed, relaxing the assumption on the genus, are there any possibilities for the proportion other than $0$, $1/2$, and $1$?

Consider $B$ a finite-type integral quasi-projective scheme over $\mathbb{Z}$ such that $B(\mathbb{Z})$ infinite. (If you like, take $B$ to be the affine line). Let $X \to B$ be a generically smooth proper family of curves of genus $g > 1$. Assume the $b \in B(\mathbb{Z})$ for which $X_b(\mathbb{Q}) = \emptyset$ are Zariski-dense in $B$. Must then the proportion of members $X_b$, over $b \in B(\mathbb{Z})$, for which $X_b(\mathbb{Q}) \neq \emptyset$, be equal to $0$? (Variant: the same question with $B(\mathbb{Z})$ replaced by $B(\mathbb{Q})$.)

An explicit variant (although not quite a special case as it stands). Let $f \in \mathbb{Z}[x]$ be irreducible of degree $> 4$. Do the integers $N$ for which $f(x) = Ny^2$ has a rational solution, have density zero? Variant: let $N$ range over the primes, or over the squarefrees.

In the above setup, we may include this example by assuming more generally that $X_{\mathbb{Q}} \to B_{\mathbb{Q}}$ admits exactly $m$ sections defined over $\mathbb{Q}$, and ask whether the density of $b \in B(\mathbb{Z})$ with $|X_b(\mathbb{Q})| > m$ is zero.

A comment. This, of course, is easily seen to fail for families of genus zero. For example, if $\deg{f} = 2$ above, then for half the primes $N$ (those that split in the quadratic field $\mathbb{Q}[x]/(f(x))$) the equation $f(x) = Ny^2$ has a solution. Various heuristics, and some standard conjectures, imply that it should also fail for genus one families. I would be interested in seeing an example in genus one for which it can be proved that a positive proportion (yet not almost all, for the Zariski topology) of the members have rational points -- or is this out of reach?

Consider $B$ a finite-type integral quasi-projective scheme over $\mathbb{Z}$ such that $B(\mathbb{Z})$ infinite. (If you like, take $B$ to be the affine line). Let $X \to B$ be a generically smooth proper family of curves of genus $g > 1$. Assume the $b \in B(\mathbb{Z})$ for which $X_b(\mathbb{Q}) = \emptyset$ are Zariski-dense in $B$. Must then the proportion of members $X_b$, over $b \in B(\mathbb{Z})$, for which $X_b(\mathbb{Q}) \neq \emptyset$, be equal to $0$? (Variant: the same question with $B(\mathbb{Z})$ replaced by $B(\mathbb{Q})$.)

An explicit variant (although not quite a special case as it stands). Let $f \in \mathbb{Z}[x]$ be irreducible of degree $> 4$. Do the integers $N$ for which $f(x) = Ny^2$ has a rational solution, have density zero? Variant: let $N$ range over the primes, or over the squarefrees.

In the above setup, we may include this example by assuming more generally that $X_{\mathbb{Q}} \to B_{\mathbb{Q}}$ admits exactly $m$ sections defined over $\mathbb{Q}$, and ask whether the density of $b \in B(\mathbb{Z})$ with $|X_b(\mathbb{Q})| > m$ is zero.

A comment. This, of course, is easily seen to fail for families of genus zero. Various heuristics, and some standard conjectures, imply that it should also fail for genus one families. Is there an example in genus one for which it can be proved that a positive proportion, yet not almost all of the members have rational points -- or is this also out of reach?