TITLE: Does Turing determinacy imply full determinacy?
QUESTION [13 upvotes]: The axiom of Turing determinacy is a weakening of the full axiom of determinacy, $AD$, in which only games with payoff sets which are $\equiv_T$-invariant are demanded to be determined.
In "Turing determinacy and the continuum hypothesis" (published in 1989), Ramez Sami writes:

"The main question so far unsettled in this particular domain can be roughly put this way: is it true that for any "reasonable" pointclass $\Gamma$ we have: Turing-Det$(\Gamma)\implies$Det$(\Gamma)$? In particular is it the case that: [over $ZF+DC$, presumably] Turing $AD$ implies $AD$?"

http://link.springer.com/article/10.1007%2FBF01622874, page 153


My question is, what is the status of this question currently? Do we know whether Turing $AD$ is strictly weaker than $AD$? The only recent work I know of around Turing determinacy is from the reverse mathematical side (http://www.math.cornell.edu/~shore/papers/pdf/TDet21.pdf); I'm not at all familiar with the set theory on the subject.
(I vaguely recall that Turing determinacy implies that every Suslin set is determined, but I can't remember where I supposedly learned this "fact.")

REPLY [5 votes]: To show this in L(R) (or any well understood determinacy model) one apparently has to run a core model induction as in section 6.2 of  https://ivv5hpp.uni-muenster.de/u/rds/core_model_induction.pdf  which explains why it is open in the abstract.