TITLE: Is it possible to formulate the axiom of choice as the existence of a survival strategy?
QUESTION [19 upvotes]: Consider the following situation:

There is an infinite set $G$ of giraffes.
  A lion comes and announces a set $C$ of all possible colours and an infinite cardinal $\kappa$.
  The hungry lion tells the giraffes that when she comes back, the following happens:
  The giraffes may no longer speak to each other.
  The lion gives each giraffe a scarf of some colour in the set $C$.
  Every giraffe sees all the scarves (including their own) but may be mistaken of the colour of strictly less than $\kappa$ scarves.
  The giraffes do not know which scarves they see correctly, only that there is a strict upper bound on the number of false colours.
  Then each giraffe must guess the colour of their own scarf.
  The lion eats all the giraffes that guess wrong.
  Can the giraffes agree on a survival strategy before the lion returns so that strictly less than $\kappa$ giraffes are eaten?

In the typical formulation of this problem $|G|=\kappa=\aleph_0$ and the giraffes have prior knowledge of which scarves they see.
Let us ignore the practical issues related to this generalized savanna for a moment.
This is what the existence of a survival strategy means in more mathematical terms:
Take any infinite set $G$, any nonempty set $C$ and any infinite cardinal $\kappa$.
A survival strategy is a function $S:G\times C^G\to C$ (strategy) with the following property.
Take any any function (colouring) $f\in C^G$ and any $o\in (C^G)^G$ (the giraffe $g\in G$ observes the colour pattern $o(g)$) so that $\left\lvert\{h\in G;o(g)(h)\neq f(h)\}\right\rvert<\kappa$ for all $g\in G$.
Let $a\in C^G$ (the answers given by the giraffes) be defined by $a(g)=S(g,o(g))$.
This $a$ always satisfies $\left\lvert\{g\in G;a(g)\neq f(g)\}\right\rvert<\kappa$.
Note that to make sense of this existence we need not be able to compare all cardinals; it suffices that $|A|<\kappa$ is well defined.
(I am not a set theorist myself. If this formulation seems inappropriate or does not seem to correspond to the story, let me know. If you know how to add clarifying details to this question, feel free to edit.)
Claim:
The giraffes have a survival strategy in ZFC.
Proof:
In the set $C^G$ of all possible scarf colourings define an equivalence relation $\sim$ by
$$
a\sim b
\iff
\left\lvert\{g\in G;a(g)\neq b(g)\}\right\rvert<\kappa.
$$
Let $q:C^G\to C^G/{\sim}$ be the quotient map and $r$ a right inverse for it.
The giraffes can agree on the map $r$ based on the information given by the lion.
Suppose the scarf colour pattern chosen by the lion is $f\in C^G$.
All giraffes can see the equivalence class $q(f)$ but cannot be sure about $f$ itself.
The giraffe $g\in G$ the guesses $r(q(f))(g)\in C$, so that the answers form the colour pattern $r(q(f))$.
Since $r(q(f))\sim f$, the number of mistaken giraffes is strictly less than $\kappa$.
In terms of the more mathematical formulation, the strategy is $S(g,v)=r(q(v))(g)$.
$\square$
The problem:
The proof relies heavily on the axiom of choice (the existence of $r$).
But is it equivalent to it?
That is, if we assume that the giraffes have a survival strategy for any $G$, $C$ and $\kappa$, can we deduce the axiom of choice?
My vague intuition suggests an affirmative answer, but I do not have any real arguments to support it.
I have not seen such claims in lists of equivalent forms of the axiom of choice.
One issue is that the strategy is somewhat indeterministic in the sense that it does not give any control over the identities of the surviving giraffes.
There are of course variations of the problem:
Do we really need the strategy for all $G$, $C$ and $\kappa$, or is it enough let let one or two of them vary (perhaps $|G|=\kappa$ and $C=\{0,1\}$)?
What if the giraffes have prior knowledge of which scarves they don't see correctly (e.g. their own)?
Note:
One purpose of the animal story is to give simple and reasonably intuitive names for the objects.
But my main purpose is that if the result indeed is equivalent with the axiom of choice, this could be used to demonstrate what the axiom of choice is about to non-mathematical (or choice-ignorant mathematical) audience.
It is of course interesting as such that the existence of a survival strategy follows from AC, but it would be much more interesting if the existence had abstract consequences, AC itself or something else.

REPLY [5 votes]: The question makes sense when $\kappa=\aleph_0$ since the statement involves no choice. But according to the interesting book by Hardin and Taylor, "The mathematics of coordinated inference" (2013), mentioned above in a comment by François, even the existence of a survival strategy for $\kappa=\aleph_0$ and arbitrary $G$ and $C$ is not known to be equivalent to the Axiom of Choice. They state this as "the single most prominent open question" of chapter 3 (Question 3.6.1). 
ADDENDUM
There is one reasonable reformulation of your question which avoids mentioning cardinalities (which are always tricky in a choiceless context) and which is equivalent to it in ZFC. It turns out that such a reformulation is actually a particular case of a theorem in the book that is unknown to imply AC, which would indicate that your question is also an open problem.
Fix any proper ideal $\mathcal{I}$ in the powerset of $G$. Then you can stipulate that the subset of scarves each giraffe sees wrongly is a member of $\mathcal{I}$, and ask whether there is a strategy under which only a member of $\mathcal{I}$ guesses wrong (when AC is available, the subsets of cardinality strictly less than $|G|$) would form such an ideal).
Under this reformulation, $G$ can be turned into a topological space in which the neighbourhood filter of a point consists precisely on those subsets containing the point whose complements belong to $\mathcal{I}$. For each $g \in G$ you can define an equivalence relation $\sim_g$ on $C^G$ by $f \sim_g f'$ if and only if $f$ and $f'$ agree on a deleted neighbourhood $U - \{g\}$ of $g$. Given an assignation $f$, each giraffe $g$ can detect the equivalence class of $f$ under $\sim_g$ and predict their own colour. This is precisely what Hardin and Taylor call a $\textit{near neighbourhood predictor}$. Moreover, one can easily see that all points in a given member $i \in \mathcal{I}$ are topologically isolated, and in particular any point in a nonempty subset of $i$ has a neighbourhood intersecting only finitely many points in that subset, making $i$ something which they call $\textit{weakly scattered}$. Their corollary 7.2.4 then states:
"For any space $X$ of agents and set $Y$ of colours, there exists a weakly scattered-error near neighbourhood predictor"
Once again, their list of open questions includes question 7.8.1, which asks whether the above proposition, provable in ZFC, actually implies AC over ZF.

REPLY [5 votes]: This is related to Nate Eldredge's answer that shows some amount of choice is required. In the case $G=\mathbb{N}$, $\kappa = \omega_0$ and $C=2$, the equivalence relation you define is eventual equality of 0-1 sequences and it has a special name, $E_0$.
Choose one element from each equivalence class of $E_0$, and assume that this set is Borel. In this case, there would be a Borel map $f: 2^{\mathbb{N}} \rightarrow 2^{\mathbb{N}}$ such that $xE_0y \Leftrightarrow f(x)=f(y)$. Such Borel equivalence relations are known as "smooth" (check out this MO question) and $E_0$ is known to be not smooth. (You can see page 5 of this workshop talk by S. Thomas, or Su Gao's Invariant Descriptive Set Theory Proposition 6.1.7 on page 135)
Therefore, the giraffes' strategy you define is not Borel. To see that some amount of choice is necessary to construct this strategy, or any set that intersects each $E_0$-class at a single point, you can move to a model of set theory where choice fails badly and every set of reals is Borel (for example, see here). Of course, Nate Eldredge's answers reveals more than this argument but I thought it is worth mentioning.