The question is about a modification of the following riddle (you can think about it before reading the answer if you like riddles, but that's not the point of my question):


**The Riddle:**
We assume there is an infinite sequence of boxes, numbered $0,1,2,\dots$. Each box contains a real number. No hypothesis is made on how the real numbers are chosen.
You are a team of 100 mathematicians, and the challenge is the following: each mathematician can open as many boxes as he wants, even infinitely many, but then he has to guess the content of a box he has not opened. Then all boxes are closed, and the next mathematician can play. There is no communication between mathematicians after the game has started, but they can agree on a strategy beforehand.


You have to devise a strategy such that at most one mathematician fails. Axiom of choice is allowed.


**The Anwser:**
If $\vec u=(u\_n)\_{n\in\mathbb N}$ and $\vec v=(v\_n)\_{n\in\mathbb N}$ are sequences of real numbers, we say that $\vec u\approx \vec v$ if there is $M$ such that for all $n\geq M$, $u\_n=v\_n$. Then $\approx$ is an equivalence relation, and we can use the axiom of choice to choose one representant by equivalence class. The strategy is the following: mathematicians are numbered from $0$ to $99$, and the sequence of boxes $(u\_n)\_{n\in\mathbb N}$ is split into $100$ sequences of the form $\vec u\_i=(u\_{100n+i})\_{n\in\mathbb N}$ with $0\leq i\leq 99$. Mathematician number $i$ will look at all sequences $\vec u\_j$ with $j\neq i$, and for each sequence, it will compute the index $M\_j$ from which the sequence matches the representant of its $\approx$-class. He then takes $M$ to be the maximum of the $M\_j+1$ and looks at the sequence $\vec u\_i$ starting at this $M$. He can deduce the $\approx$-class of the sequence $\vec u\_i$, and guesses that $u\_{M-1}$ matches the representant. At most one mathematician will be wrong: the one who has the number $i$ with $M\_i$ maximal.


**The Modification:**
I would find the riddle even more puzzling if instead of 100 mathematicians, there was just one, who has to open the boxes he wants and then guess the content of a closed box.
He can choose randomly a number $i$ between $0$ and $99$, and play the role of mathematician number $i$. In fact, he can first choose any bound $N$ instead of $100$, and then play the game, with only probability $1/N$ to be wrong.
In this context, does it make sense to say "guess the content of a box with arbitrarily high probability"? I think it is ok, because the only probability measure we need is uniform probability on $\{0,1,\dots,N-1\}$, but other people argue it's not ok, because we would need to define a measure on sequences, and moreover axiom of choice messes everything up.



Probabilities in a riddle involving axiom of choice
The probabilistic reasoning depends on a conglomerability assumption, namely that given a fixed sequence $\vec u$, the probability of guessing correctly is $(n-1)/n$, then for a randomly selected sequence, the probability of guessing correctly is $(n-1)/n$. But we have no reason to think the event of guessing correctly is measurable with respect to the probability measure induced by the random choice of sequence and index $i$, and we have no reason to think that the conglomerability assumption is appropriate.


A quick way to see that the conglomerability assumption is going to be dubious is to consider the analogy of the Brown-Freiling argument against the Continuum Hypothesis (see [here](http://www.mdpi.com/2073-8994/3/3/636) for a discussion). Assume CH. Let $\prec$ be a well-order of $[0,1]$. Suppose $X$ and $Y$ are i.i.d. uniformly distributed on $[0,1]$. Consider the question of which variable is bigger. Fix a value $y\in [0,1]$ for $Y$. Then $P(X\preceq y)=0$, since there are only countably many points $\prec$-prior to $y$. By a conglomerability assumption, we could then conclude that $P(X\le Y)=0$, which would be absurd as the same reasoning would also show that $P(Y\le X)=0$. The argument fallaciously assumes conglomerability. We are neither justified in concluding that $P(X\le Y)=0$, nor that $\{ X \le Y \}$ is measurable (though for each fixed $y$, $\{ X \le y \}$ is measurable). And indeed it's not measurable: for were it measurable, we could use Fubini to conclude that it has null probability. Note that one can repeat the argument without CH but instead using an extension of Lebesgue measure that assigns null probability to every subset of cardinality $<\mathfrak c$, so clearly there is no refutation of CH here.


Here's another analogy: By the Hausdorff Paradox, decompose $S^2-D$ for a countable $D$ into disjoint $A\_1,A\_2,A\_3$, where $A\_1,A\_2,A\_3,A\_1\cup A\_2$ are all isometrically equivalent. Suppose a point $X$ is uniformly chosen in $S^2$. (I will ignore $D$ from now on, since it has zero measure. If you like, you can assume that the uniform choice is done so $X$ cannot lie in $D$.) Let $i$ be a random index in $\{1,2,3\}$, independent of $X$. How likely is it that the $X \in A\_i$? It's very tempting to say that it's got to be $1/3$. Any fixed value of $X$ in $S^2$ is in one of the three subsets, after all, and so if we choose a random index $i$, surely we have probability $1/3$ that it'll be in $A\_i$. Of course this easily leads to paradox. But we are not in fact entitled to assume that $J = \{ \omega : X(\omega) \in A\_{i(\omega)} \}$ is measurable. (In fact, it's easy to see that it's not measurable.)


Let's go back to the riddle. Suppose $\vec u$ is chosen randomly. The most natural option is that it is a nontrivial i.i.d. sequence $(u\_k)$, independent of the random index $i$ which is uniformly distributed over $[100]=\{0,...,99\}$. In general, $M\_j$ will be nonmeasurable (one can prove this in at least some cases). We likewise have no reason to think that $M$ is measurable. But without measurability, we can't make sense of talk of the probability that the guess will be correct.


Here's an amusing thing that may help see how measurability enters into these things. Consider a single sequence of infinitely many independent fair coin flips. Our state space is $\Omega=\{0,1\}^{\mathbb N}$, corresponding to an infinite sequence $(X\_i)\_{i=0}^\infty$ of i.i.d.r.v.s with $P(X\_i=1)=P(X\_i=0)=1/2$. Start with $P$ being the completion of the natural product measure on $\Omega$.


Can you guess the first coin flip on the basis of all the others? You might think: "Of course not! No matter what function from the values of flips $X\_1,X\_2,...$ to $\{0,1\}$ is chosen, the probability that the value of the function equals $X\_0$ is going to be $1/2$."


That's a fine argument assuming the function is measurable. But what if it's not? Here is a strategy: Check if $X\_1,X\_2,...$ fit with the relevant representative. If so, then guess according to the representative. If not, then guess $\pi$. (Yes, I realize that $\pi\notin\{0,1\}$.) Intuitively this seems a really dumb strategy. After all, we're surely unlikely to luck out and get $X\_1,X\_2,...$ to fit with the representative, and even if they do, the chance that $X\_0$ will match it, given the rest of the sequence, seems to be only $1/2$.


But if you choose shift-invariant representatives (i.e., $r([\tau \vec u])=\tau r([\vec u])$ when $(\tau\vec u)\_n = \omega$ is a left sequence shift--by Zorn, such a choice is possible), then the outer $P$-measure of the set of representatives is equal to $1$. So there is an extension $P'$ of $P$ such that $P'$-almost surely the dumb strategy works. Just let $P'$ be an extension on which the set of representatives has measure $1$ and note that the dumb strategy works on the set of representatives. 


