TITLE: Can $V$ only have well-orderings definable with respect to a parameter?
QUESTION [6 upvotes]: In this answer, Professor Hamkins gives a proof that for models $M$ of ZF, $M$ being a model of $\text{ZFC} + V = \text{HOD}$ is equivalent to there being a definable well-ordering of the universe:
https://mathoverflow.net/a/180734
His argument easily extends to an equivalence of these properties to $M$ having a well-ordering of the universe definable with respect to an ordinal parameter. So, if there is well-ordering of the universe with respect to some parameter $p,$ but there is not a well-ordering of $V$ definable without parameters, then necessarily $p \not \in \text{OD}$. Is this situation possible? My intuition is that it shouldn't be possible, since I don't think a non-ordinal parameter should be able to define something so fundamental when an ordinal cannot do the same.

REPLY [3 votes]: I just noticed this question, which I find quite interesting. 
I thought I'd mention the following related result, which some readers may find interesting:
Theorem. The following are equivalent.

The universe is HOD of a set: $\exists b\ (V=\text{HOD}(b))$.
The axiom V=HOD is forceable.
Somewhere in the generic multiverse, the universe is HOD of a set.
Somewhere in the generic multiverse, the axiom V=HOD holds. 

The proof is contained in my blog post, Being HOD-of-a-set is invariant throughout the generic multiverse. 
In particular, it follows that the axiom V=HOD is a switch, in models for which $V=\text{HOD}(b)$, since it can be forced on and then off again as much as you like. If $V=\text{HOD}$ holds, then you can do the forcing in Asaf's answer, adding a Cohen real, and $V\neq\text{HOD}$ in the extension $V[c]$, but then you can force $V=\text{HOD}$ again in a further forcing extension. And furthermore, whenever $V=\text{HOD}(b)$, then you can force $V=\text{HOD}$, and the assertion $\exists b\ V=\text{HOD}(b)$ is invariant by forcing.