Let $\kappa$ and $\lambda$ be infinite cardinals. Construct a random function $\phi:\kappa\times\lambda\rightarrow\{0,1\}$ by flipping a fair coin for each element of the domain. What is the probability, $P(\kappa,\lambda)$, that there exists $\alpha<\kappa$ such that $\forall x \, \phi(\alpha,x)=0$. Intuitively, it seems like there should be a cardinal $\eta_\lambda$ (for each $\lambda$) such that: $P(\kappa,\lambda)=0$ if $\kappa<\eta_\lambda$ and $P(\kappa,\lambda)=1$ if $\kappa\geq\eta_\lambda$. (EDIT 2:) However, based on the comments, it sounds like this intuitive question cannot be formalized in a reasonable way. Where should I look to get a better understanding of why this is the case? Are there specific theorems that would be useful for understanding this?
EDIT 1: There has been debate in the comments about whether this question is meaningful without specifying a measure on $^{\kappa\times\lambda}2$. If $X=[0,\frac{1}{2})$, then $[0,1]$ is a disjoint union of $X$ and a translate of $X$. Therefore, $\mu(X)=\frac{1}{2}$ for any translation invariant probability measure, $\mu$, on $[0,1]$ (for which $X$ is measurable). So, we can argue that $X$ has measure $\frac{1}{2}$ for any probability measure on $[0,1]$ having certain natural properties. In the setting of binary functions on $\kappa\times\lambda$, there are similar "symmetries" akin to translation invariance. For example, one would expect $\mu(X)=\mu(\{1-\phi:\phi\in X\}$. So, what I'm interested in is whether one can make an argument about the measure of $Y=\{\phi\in^{\kappa\times\lambda}2:\exists\alpha<\kappa\forall \beta<\lambda(\phi(\alpha,\beta)=0)\}$ that is based on conditions we would want any "uniform" probability measure to satisfy.
Special case $P(\aleph_0,\aleph_0) = 0$.
Let $\kappa = \lambda = \aleph_0$. All subsets of $\kappa \times \lambda$ are Borel sets. For each $\alpha \in \kappa$, [assuming indpendence, which is not stated] $$ \mathbb P \big(\{\forall x \in \lambda, \phi(\alpha,x)=0\}\big) = \mathbb P \left(\bigcap_{x \in \lambda} \{\phi(\alpha,x)=0\}\right) = \prod_{x\in\lambda} \mathbb P\big(\{\phi(\alpha,x)=0\}\big) = \prod_{x \in \lambda}\frac{1}{2} = 0. $$ so that $$ P(\kappa,\lambda) = \mathbb P\big(\{\exists \alpha \in \kappa, \forall x \in \lambda, \phi(\alpha,x)=0\}\big) =\mathbb P\left(\bigcup_{\alpha \in \kappa} \{\forall x \in \lambda, \phi(\alpha,x)=0\}\right) = 0 $$ a countable union of sets of measure zero.
What if $\lambda$ is uncountable?
Then for each $\alpha\in \kappa$, the event $$ \bigcap_{x \in \lambda} \{\phi(\alpha,x)=0\} \subseteq \bigcap_{x \in \lambda_0} \{\phi(\alpha,x)=0\} $$ for some countably infinite subset $\lambda_0 \subseteq \lambda$, so it is a subsetof a Borel set of measure zero. Let's assume our probability measure is complete. Then we conclude $$ \mathbb P\left(\bigcap_{x \in \lambda} \{\phi(\alpha,x)=0\}\right) = 0 $$ anyway.
what if $\kappa$ is uncountable?
Then we get $$ P(\kappa,\lambda) = \mathbb P\big(\{\exists \alpha \in \kappa, \forall x \in \lambda, \phi(\alpha,x)=0\}\big) =\mathbb P\left(\bigcup_{\alpha \in \kappa} \{\forall x \in \lambda, \phi(\alpha,x)=0\}\right) $$ an uncountable union of (possibly non-Borel) sets of measure zero. Even if it is measurable (which we do not know), we do not know that it still has measure zero.