In a generalized "prisoners/hats/boxes"-style logic puzzle, suppose we have the following:
- a set $W$ (of "possible world configurations")
- a set $S$ (of "possible individual strategies")
- a relation $C\subseteq W\times S$ (representing "individual success criteria")
- a solution $(s_1,\dots,s_n)\in S^n$ with the property that $\forall w\in W, |\{j\in\{1,\dots,n\}:(w,s_j)\notin C\}|\le1$.
That is, the solution gives $n$ individual strategies that are "correlated" so that at most one will fail.
Intuitively, then, if I choose $j$ uniformly at random from $\{1,\dots,n\}$ and follow strategy $s_j$, then my probability of failure is at most $1/n$. Is this true?
Surprisingly, the answers on this MathOverflow post (which is this exact question for a specific puzzle) argue that the answer is no. They seem to assume that in order to answer the question, we need a probability measure on $W\times S$; then $C$ is the event of interest and needs to be measurable, which is where problems arise.
But do we really need to bring $W$ into the probability space? We've made no assumptions of randomness about $W$. The probabilistic strategy is just a probability measure $\Bbb P$ on $S$ (assigning probability mass $\frac1n$ to each of $s_1,\dots,s_n$ and $0$ to all other strategies), and for each $w\in W$, the event $C_w=\{s\in S:(w,s)\in C\}$ is a $\Bbb P$-measurable subset of $S$. We have
$$\forall w\in W,\Bbb P(C_w)\ge\frac{n-1}n.$$
Does this not justify the informal conclusion that the probabilistic strategy guarantees success probability $\frac{n-1}n$?