Let $A$ be a set, and let $\mu,\lambda$ be cardinals.
We denote by $[A]^{\mu}$ the set of subsets of $A$ having cardinality $\mu$.
We call a function $c:[A]^{\mu}\rightarrow \eta$ a coloring of $[A]^{\mu}$.
We say that $c$ has a homogeneous subset of size $\lambda$ if there exists $B\subseteq A$ with $|B|\geq\lambda$ such that $c|_{[B]^{\mu}}$ is constant.
Let $\kappa,\lambda,\mu$ and $\nu$ be cardinals, where $\mu,\eta\leq\kappa$. The notation $\kappa\rightarrow (\lambda)^{\mu}_{\eta}$ means:
For every set $A$ with $|A|\ge\kappa$, every coloring $c$ of $[A]^{\mu}$ has a homogeneous subset of size $\lambda$.
Over $\mathsf{ZFC}$ (as clearly we need some form of choice when $\kappa\geq\aleph_{0}$ - e.g. if $\kappa=\aleph_{0}$ we would need at least the weak form of choice every infinite set has a countable subset) can the words "every set" in the definition of $\kappa\rightarrow (\lambda)^{\mu}_{\eta}$ be replaced by the words "some set"?
Here is is my reasoning:
Suppose $A$ is some set with $|A|\geq\kappa$ such that every coloring of $[A]^{\mu}$ has a homogeneous subset of size $\lambda$, and suppose that $\kappa\not\!\rightarrow (\lambda)^{\mu}_{\eta}$. That is, there is some set $S$ with $|S|\ge\kappa$ for which there is a coloring $c_{S}$ of $[S]^{\mu}$ having no homogeneous subset of size $\lambda$.
If $|A|\leq|S|$, I can reach a contradiction. But I am having trouble with the case $|A|\geq|S|$. Could someone nudge me in the right direction? Or perhaps indicate a counterexample?
One easy remark: If we relax the condition $|A|\ge\kappa$ to $|A|=\kappa$ in the definition of $\kappa\rightarrow (\lambda)^{\mu}_{\eta}$, then the answer is indeed affirmative.
It can be replaced with "some set $A$ with $|A| = \kappa$": if we have set $B$ s.t. $|B| \geq \kappa$ and coloring $c$ of $[B]^\mu$, we can find subset $A' \subseteq B$ with cardinality $\kappa$, bijection $f: A' \to A$, create coloring $c'$ of $A$ defined as $c'(X) = c(f^{-1}(X))$, find homogeneous subset $Y$ for $c'$, and then $f^{-1}(Y)$ will be homogeneous subset for $c$.
It can't be replaced with just "some set $A$ with $|A| \geq \kappa$". For example, take $\kappa = 2$, $\mu = 1$, $\eta = \lambda = 2$.
Then, according to original definition, $\kappa \not \to (\lambda)_\eta^\mu$, because $A = \{p, q\}$ and coloring $c(p) = 0$, $c(q) = 1$ has no homogeneous subset of size $2$.
However, according to your variant, it does, because for set $A = \{p, q, r\}$ for any coloring we have two elements of the same color, and so they form homogeneous set of size $2$.
Intuitively, if we can't color smaller set while avoiding homogeneous subsets, we also can't color larger set. But it doesn't work in reverse: it's possible that we can color smaller set, but not larger one.