Is there an elementary proof of a better result for the finite guessing-box puzzle?

Question

The infinitary guessing-box puzzle is amazing — see here. In the basic form, the Guessing-box Hall has infinitely many wooden boxes, each containing a real number, and there are 100 mathematicians who will enter the hall one at a time. They can open up as many boxes as they want, except they must leave at least one box unopened and predict the contents. This is recorded and the mathematician exits to the rear garden while the room is reset for the next mathematician. The amazing fact is that there is a strategy by which 99 out of the 100 will be correct. Eric Naslund proved that even with $\omega$ many mathematicians, we can still ensure that all but at most one are correct.

There are various generalizations. For example, it didn't matter at all that it was real numbers—any set of possible box contents will do. You can require them each to make infinitely many predictions, not just one, and still all of the mathematicians except at most one will be fully correct. You can have continuum many mathematicians and continuum many boxes, and every mathematician makes predictions about continuum many boxes, and they will all be fully correct, except one mathematician at most with wrong predictions. It works for any infinite cardinal $\kappa$, but you need at least many boxes as mathematicians.

My question is rather about the finite case, where of course none of that works at all.

I have a proof that for any finite number of mathematicians and any finite number of boxes, then for any strategy of guessing, there will be infinitely many instances where everybody is wrong. For example, even if we consider just natural numbers inside the boxes, by Ramsey's theorem there will be an infinite homogeneous set such that any box configuration with an increasing selection from that set will lead ultimately to the same pattern of boxes chosen for prediction and the same pattern of winners/losers. But that pattern must therefore be all losers, since we can make any mathematician a loser by changing the contents of their prediction box, while remaining inside the homogeneous set.

OK, fine. But I'm not happy with the argument, since it seems to me that we should have a much stronger result—shouldn't they almost always all be guessing wrongly? I am unsure which concept of "almost all" is relevant here, however, since the strategies might not be measurable or Borel. But also, I think there should be a much more elementary proof. Using Ramsey's theorem seems overly precious. It strikes me as intuitively obvious that they cannot be reliably correct to the slightest degree, but I struggle to find a simple proof showing this.

Question. In the guessing-box puzzle with finitely many mathematicians and finitely many boxes, each containing a real number, is there an elementary argument showing that any given strategy is almost always all wrong?

And what is the relevant sense of "almost always"?

Perhaps there is a proof by induction on the number of mathematicians, since if $n$ mathematicians are almost always all wrong, and we add another mathematician, it is easy to make them wrong in any instance by changing the content of their box to anything other than their prediction. But what is the concept of almost-always that will allow this argument to go through?

(Meanwhile, I guess the Ramsey argument applies in the case of a finite set of possible box contents, showing that if the set is large enough, there will be instances where everybody is wrong. And for this, Ramsey seems somewhat more relevant.)

Answer 1

First, an elementary argument that for any fixed $n,$ with $n$ mathematicians and $n$ boxes, and a fixed strategy, there are uncountably many instances where everyone is wrong. Fix $r \in [0,1).$ We'll show there is a tuple of natural numbers $(k_0, \ldots, k_{n-1})$ such that the box assignment $\langle k_i + r \rangle_{i<n}$ causes everyone to be wrong. If a tuple in $(n+1)^n$ is chosen uniformly at random, the expected number of correct guesses $E$ is bounded below by the probability at least one player is correct, and bounded above by $\frac{n}{n+1}$ since each player has probability at most $1/(n+1)$ of being correct. Thus, with positive probability, the tuple is such that everyone is wrong.

As for what sense it's "almost always the case" that everyone is wrong, restrict the values of the boxes to the unit interval (which is a more natural domain for a probability measure) and consider for one player the set of box assignments for which they succeed (notice that the ideal generated by such sets contains those sets $X$ such that a finite team of players can be guaranteed to have at least one success if the box assignment is in $X$). Let's further assume the player uses a "simple" strategy, by which I mean they pre-commit to which box they'll guess from (the success set of a more complicated strategy will be covered by the success sets of $n$ simple strategies). At this point, we're just looking at graphs of functions $f: [0,1]^{n-1} \rightarrow [0, 1]$ where the codomain is identified with one of the $n$ axes.

Consistently, function graphs can be rather substantial. The continuum hypothesis is famously equivalent to the plane being covered by countably many function graphs, so even for $n=2,$ it need not be the case that the possible success sets of a player generate a non-trivial $\sigma$-ideal, so a formalization of the negligibility of their chance of success could only involve finitely additive measures.

Each function graph has uncountably many disjoint translations in the axis of its codomain. If $\mu$ is a total finitely additive probability measure on $[0,1)^n$ which is invariant under the translation action by the $n$-torus, then each player succeeds on a $\mu$-negligible set. Such $\mu$ exists since this group action is abelian (and thus amenable).