Background/Definitions. Let $\alpha,\beta$ ordinal numbers. The Hessenberg sum $\alpha \# \beta$ is defined recursively as the smallest ordinal which is $>\alpha' \# \beta$ and $> \alpha \# \beta'$ for all $\alpha' < \alpha, \beta' < \beta$. One can show that $\alpha \# \beta$ is then also the mex of $\{\alpha' \# \beta : \alpha' < \alpha\} \cup \{\alpha \# \beta' : \beta' < \beta\} \cup \{\alpha' \# \beta' : \alpha' < \alpha, \beta' < \beta\}$ (the proof is not hard, but does somebody know a reference?). For $\beta < \omega$ we have $\alpha \# \beta = \alpha + \beta$. If $G$ is an impartial combinatorial game, we denote by $\alpha(G)$ its nimber, which is defined recursively as the mex of the $\alpha(G')$ with $G' \in G$. If $G,H$ are impartial combinatorial games, we denote by $G \vee H$ their selective compound game.
Question. How can we compute $\alpha(G \vee H)$ in terms of $G$ and $H$?
If $G,H$ are Nim games with only one pile, we have $\alpha(G \vee H) = \alpha(G) \# \alpha(H)$ (this is precisely the mex-description of the Hessenberg sum). In general, one can prove $\alpha(G \vee H) \geq \alpha(G) \# \alpha(H)$. We have equality when $G$ satisfies $\alpha(G') < \alpha(G)$ for all $G' \in G$, and likewise for $H$. But unfortunately, inequality seems to be quite typical. For example, if $G=\star 1$ and $H=\{\star 0,\star 2\}$ we have $\alpha(G)=\alpha(H)=1$, hence $\alpha(G) \# \alpha(H)=2$, but $G \vee H = \{\star 2,\star 1 ,\star 1 \vee \star 2,\star 0\}$, hence $\alpha(G \vee H)=4$. This shows, in particular, that selective compounds do not preserve equivalence of games (isn't this ugly?), or in other words that $\alpha(G \vee H)$ does not only depend on $\alpha(G)$ and $\alpha(H)$.
Any references to the literature are welcome, because I doubt that anything of this is new. I have already looked at the well-known books by Conway etc.