Everything is over $\mathbb C$. Let $X\to \mathbb P^1$ be a fiber bundle with fiber some smooth projective variety $F$. If we have another such bundle $Y \to \mathbb P^1$, then, for which $p,q$ we have the Hodge number $h^{p,q}(X)=h^{p,q}(Y)$?
Apparently, the outer Hodge numbers (essentially $h^{p,0}$) are equal since they are birational equivalent. But can we expect more? Is it true that all the Hodge numbers of $X$ and $Y$ are equal? (I know this is true for projective bundles, i.e. $F\cong \mathbb P^n$, but I have no idea for general bundles. I guess it is not.)
Thanks!
It is true in general, the keyword is Grothendieck ring of varieties. I am not sure for a reference, but it's probably in the recent book by Chambert-Loir, Nicaise and Sebag about motivic integration.
There is a ring $\mathrm K_0(\mathrm{Var}/\Bbb C)$ defined as follows : take the free abelian group generated by algebraic varieties over $\Bbb C$. Define a product as $[X] \cdot [Y] = [X \times Y]$, and impose the relations $[X] = [U] + [Z]$ for any open set $U \subset X, Z = X \backslash U$.
Theorem : There is a unique morphism $\mathrm K_0(\mathrm{Var}/\Bbb C) \to \Bbb Z[u,v]$ that coincide with Hodge polynomials $\sum_{p,q}h^{p,q}u^pv^q$ on projective smooth varieties.
It is easy to check that if $X \to B$ is a fiber bundle in the Zariski topology with fiber $F$, then $[X] = [F] \cdot [B]$ in $K_0$. In particular, with your notation $[X] = [Y]$ in $\mathrm{K}_0(\mathrm{Var}/ \Bbb C)$ so their Hodge polynomial coincide.