TITLE: Is every element of $Mod(S_{g,1})$ a composition of right handed Dehn twists?
QUESTION [7 upvotes]: Let $S_{g,1}$ be the surface of genus $g \geq 1$ and $1$ boundary component. Let $Mod(S_{g,1})$ be the mapping class group in which we allow isotopies to rotate the action on the boundary (equivalently think of it as the mapping class group of the once-punctured surface of genus $g$).
Is every element of $Mod(S_{g,1})$ a composition of right-handed Dehn twists?
Note that this is true for $S_{g,0}$ as stated in page 124 of  A primer on Mapping Class Groups by Farb and Margalit under the name of "a strange fact".
Edit: I am going to comment ThiKu's answer to avoid further confusion. 
In A. Wand: Factorisation of Surface Diffeomorphisms and in Baker, Etnyre and Van Horn-Morris: Cabling, Contact Structures and and Mapping Class Monoids the authors, independently, provide with examples of diffeomorphisms in $Veer(\Sigma_{2,1},\partial \Sigma_{2,1})$ which are not in $Dehn^+(\Sigma_{2,1}, \partial \Sigma_{2,1})$. That is, right-veering diffeomorphisms which are not a product of right-handed Dehn twists. However, the mapping class group in which these results hold is $Mod( \Sigma_{2,1}, \partial \Sigma_{2,1})$, that is, the mapping class group of automorphisms fixing the boundary and isotopies fixing the boundary as well. This, a priori, does not yield counter-examples to my question (unless it does together with some other result that I do not know).
Observe that for all $g \geq 1$ there is a central extension 
$$1 \to \mathbb{Z} \to Mod(\Sigma_{g,1}, \partial \Sigma_{g,1}) \to Mod( \Sigma_{g,1}) \to 1 $$
which is not split in general.

REPLY [4 votes]: Yes. As pointed out by Ian Agol, I actually answered my question.
I observed that the identity is a non-empty composition of right handed Dehn twists in $Mod(S_{g,1})$. A priori this is not trivial. I was thinking about monodromies on Brieskorn-Pham singularities $(x^p+y^q)$ which are freely periodic and a composition of right-handed Dehn twists (by morsifying the singularity). One can easily see that this solves the problem in the cases that I was asking originally since all surfaces $S_{g,1}$ appear as Milnor fibers of such singularities for all $g$.
EDIT: I removed the last part of the answer since it was not true in that generality and this answers completely what I wanted.