It's well-known that AD$\_\mathbb{R}$ fails in $L(\mathbb{R})$, provably in ZFC. This is because:


* [AD$^{L(\mathbb{R})}$ implies DC$^{L(\mathbb{R})}$](https://www.jstor.org/stable/2274099).
* Over ZF+DC, AD + "Every set of reals has a scale" is equivalent to AD$\_\mathbb{R}$. *This is due to Woodin; see Theorem 32.23 in Kanamori's book.*
* But in $L(\mathbb{R})$, [properly $(\Pi^2\_1)^{L(\mathbb{R})}$ sets of reals do not have scales](https://link.springer.com/chapter/10.1007/BFb0071699).


I'm curious how much of AD$\_\mathbb{R}$ is consistent (relative to large cardinals) with ZF + $V=L(\mathbb{R})$ + AD. Specifically, it's clear that the limit should be around $\Sigma^2\_1$, but I'm not sure exactly where it lies (I can't recall precisely the level-by-level connection between scales and strategies, and I can't find it in Moschovakis); it's not even clear to me that *projective* real determinacy is consistent with this theory. Additionally, I'm interested in the large cardinals beyond a proper class of Woodins required to get (roughly) this amount of real determinacy in $L(\mathbb{R})$.



How much real determinacy can live in $L(\mathbb{R})$?
I think the following may be relevant to your question:




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First, several bullet points in your question are nontrivial results. However, there is another way to see $\mathsf{AD}\_\mathbb{R}$ does not holds in $L(\mathbb{R})$. 


Using a two step game, you can show $\mathsf{AD}\_\mathbb{R}$ can prove uniformization for all sets of reals. Consider the relation $R(x,y)$ if and only if $y$ is not ordinal definable from $x$. Assuming that $L(\mathbb{R}) \models \mathsf{AD}\_\mathbb{R}$, $R$ would have a uniformization. Every set of reals in $L(\mathbb{R})$ is ordinal definable from some real. Use this real to get a contradiction by diagonalization. 


Also note that if you put the discrete topology on $\mathbb{R}$, a two point game is a closed game. Hence $\mathsf{ZF + AD}$ can not prove the Gale-Stewart theorem (closed determinacy) for games on $\mathbb{R}$. Of course, this is because there is no well-ordering of $\mathbb{R}$. 




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"The Extent of Scales in $L(\mathbb{R})$" Corollary 6 shows that $\mathsf{DET}\_\mathbb{R}(\mathbf{\Pi}\_1^1)$ can not hold in $L(\mathbb{R})$. Steel shows this by proving that $G^\mathbb{R}\Pi\_1^1$ (the game quantifier class) is $\Sigma\_1(L(\mathbb{R}, \{\mathbb{R}\})$. By $\mathsf{DET}\_\mathbb{R}(\mathbf{\Pi}\_1^1)$, $G^\mathbb{R} \Sigma\_1^1$ is the dual of $G^\mathbb{R}\Pi\_1^1$. Hence $G^\mathbb{R}\Sigma\_1^1$ is $\Pi\_1(L(\mathbb{R}, \{\mathbb{R}\})$. But you can directly show that $G^\mathbb{R}\Sigma\_1^1$ is contained in $\Sigma\_1(L(\mathbb{R}),\{\mathbb{R}\})$. Contradiction.


($\Sigma\_1(L(\mathbb{R},\{\mathbb{R}\})$ is also $(\Sigma^2\_1)^{L(\mathbb{R}}$. Using Wadges lemma and the not-ordinal-definable relation $R$ from above, Steels showed that if $\mathsf{AD}^{L(\mathbb{R})}$ holds, then a set has a scale in $L(\mathbb{R})$ if and only if it belongs to $(\mathbf{\Sigma}^2\_1)^{L(\mathbb{R})}$. )


The same paper of Steel also notes that assuming $\mathsf{AD}$, $\mathsf{DET}\_\mathbb{R}(\mathbf{\Pi}\_1^1)$ is equivalent to the existence of $\mathbb{R}^\sharp$. (At least in the paper, Steel states it is open whether the equivalence can be proved in just $\mathsf{ZF} + \mathsf{DC}$.) Assuming $\mathsf{AD}$, $\mathsf{DET}\_\mathbb{R}(\mathbf{\Pi}\_1^1)$ has stronger consistency strength than $\mathsf{AD}$.


