I'm unsure what it means exactly to win this game with probability $p$, since the choice of the distribution is also part of the game, but we don't have a distribution for how that distribution is picked. Only part of player 1's first move, after all, involves random variables. (Presumably you want to block the observation that for any given $P$ there is a winning player 2 strategy of playing the finite support set itself.)

I had meant that $\kappa$ has the property that $\forall\beta\exists \lambda\exists j:V_\lambda\to V_\lambda$ with $\text{cp}(j)=\kappa$ and $j(\kappa)>\beta$.

Yes, that is the question I answered. Your alternative question, also suggested by user 喻 良 on the main post, is subject to what else you want true in $L_\alpha$. For example, it will be sensitive to how exactly you define the measure and how you define what it means to be measurable. You should have a specific theory, such as $\text{ZFC}^-$. But in this case, once you specify the theory, the answer will be connected with the least $\gamma$ for which $L_\gamma$ satisfies that theory. For these reasons, I don't find that version of the question as robust.

@newaccount What would you mean by projective code? I guess just the (finite) projective formula, with its parameters. In our case here, we know the definition of a projective nonmeasurable set already (and my set above is projective in $L$, with no parameters), the code of it appears right away at a small finite stage.

I think it is spelled "measurable". Although some words do keep the e, as with forceable and changeable, so I would not know what is the general rule or even if there is a coherent rule.

It does make a difference whether you assume $V=L$ or not, since if $\omega_1^L$ is countable in $V$, then every set of reals in $L$ is countable and hence measurable, and in this case there is no such $\alpha$.

It has to be bigger than $\omega_1^{CK}$ simply because it has to be uncountable. So it has to be bigger than $\omega_1$, since all sets in $L_{\omega_1^L}$ are countable in $L$.

@MikhailKatz In the forcing sense, no, since some models satisfy the ground axiom, which asserts that they are not realized as forcing extensions of any model and have no deeper grounds. Meanwhile, Benedikt Löwe and I investigated the bimodal logic, going up and down, in this paper: arxiv.org/abs/1208.5061.

According to what he says, it seems that we will have $2^\kappa=\aleph_{\kappa+1}$ for singular infinite cardinals $\kappa$ in the relevant Easton model, since such a cardinal is always a limit cardinal, and so $2^\lambda$ for regular $\lambda<\kappa$ will be $\aleph_{\lambda^+}<\aleph_\kappa$, meaning that $2^{<\kappa}=\aleph_\kappa$ and so $2^\kappa$ is $\aleph_{\kappa+1}$ as he explains.

That is what David Gao's comment is about on the main question.

Furthermore, precisely because the field is saturated, it is highly homogeneous, with numerous automorphisms. Therefore there is no canonicity to the function $x\mapsto \omega^x$. For example, there are automorphisms of No moving $\omega$ to $\sqrt{\omega}+17+\epsilon$ or moving $\pi$ to $\pi+1/\sqrt{\omega}$ and so forth. This is because real-closed fields admit elimination of quantifiers and so in each case these numbers have the same 1-type.

You haven't mentioned saturation at all in the question, but this seems to be the central feature to be discussing, since it also determines all the rest. Every saturated order of the given size supports all the rest of the structure, since one can take a saturated model of that size and the order will be saturated and hence isomorphic to the given order. The surreal construction is all about saturating the order, filling gaps at each step.