TITLE: Probabilities in a riddle involving axiom of choice
QUESTION [25 upvotes]: 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.

REPLY [3 votes]: The counter-intuitiveness of the axiom of choice is clouding the real issue here ( I think). If I have a secret number and you try to guess a larger one, what are your odds of success? is there a way to make them better (if you have $99$ friends). You have only finitely many ways to fail and infinitely many to succeed, but we can't really assign a probability. That is also behind this puzzle
Going back to the given problem, suppose that each box has a guard. At a certain time each guard sees the contents of every other box then guesses the contents of their own. Under AoC with an agreed set of equivalence class representatives and the familiar protocol we know that all but finitely many will be correct. That is strange but we are used to this consequence of the axiom of choice. If you have full knowledge of all this and are required to become the guard for a box of your choice (after looking into as many other boxes as you wish)  can we say that you are certain to make a correct guess? Not really.
The situation in the question above seems clearer to me if we stipulate ahead of time that all but finitely many boxes contain $0$  (no assumption , other than finiteness on which the exceptions are).  If there are $100$ potential guards who are friends, can they be sure that each can pick a box and guess $0$ (what else can they do?) with at least $99$ correct? yes, as before split the boxes into $100$ subsequences, person $k$ looks at all the boxes in every subsequence other than his own then chooses a box further than the last non-zero box in every other subsequence.
If you are one member of this set of $100$ friends are your odds of being correct (when you guess $0$) improved $100$ fold, or at all? Not really.
Essentially $100$ people are assigned hidden integers and must name an integer greater than their assigned integer (so only finitely ways to fail and infinitely many to succeed!). If each person can see all the other integers then only one will fail. 
In other words, given AoC, we may assume a known set of coset representatives such that every real sequence is is (uniquely) the termwise sum of a coset representative and a sequence with only finitely many non-zero terms. Once we accept that (which is weird but familiar) there is no loss in saying that the chosen sequence is in a certain coset (so why not the coset of the zero sequence)?