# Predicting Real Numbers

Here is an astounding riddle that at first seems impossible to solve. I’m certain the axiom of choice is required in any solution, and I have an outline of one possible solution, but would like to see how others might think about it.

$$100100$$ rooms each contain countably many boxes labeled with the natural numbers. Inside of each box is a real number. For any natural number $$nn$$, all $$100100$$ boxes labeled $$nn$$ (one in each room) contain the same real number. In other words, the $$100100$$ rooms are identical with respect to the boxes and real numbers.

Knowing the rooms are identical, $$100100$$ mathematicians play a game. After a time for discussing strategy, the mathematicians will simultaneously be sent to different rooms, not to communicate with one another again. While in the rooms, each mathematician may open up boxes (perhaps countably many) to see the real numbers contained within. Then each mathematician must guess the real number that is contained in a particular unopened box of his choosing. Notice this requires that each leaves at least one box unopened.

$$9999$$ out of $$100100$$ mathematicians must correctly guess their real number for them to (collectively) win the game.

What is a winning strategy?

Before entering, the mathematicians agree on a choice of representatives for real sequences when two sequence are equivalent if they are equal past some index ; and a re-labeling of $\Bbb N$ into $M \times \Bbb N$ where $M$ is the set of mathematicians.
Once a mathematician $m$ is in the room, he opens every box not labeled $(m,x)$ for $x \in \Bbb N$, and for $m' \neq m$ he carefully notes the greatest index $x(m')$ (which is independent of $m$) where the sequence $(m',x)$ has a different value from that of its corresponding representative, and $x(m') = -1$ if it is the representative.
Then, $m$ computes $y(m) = \max_{m' \neq m} x(m') +1$, and opens every box labeled $(m,x)$ for $x > y(m)$. He finds the representative of that sequence, and guesses what’s inside box $(m,y(m))$ according to that representative. He has the risk of guessing wrong if $y(m) \le x(m)$ (he is the only one not knowing the value of $x(m)$).
If there is an $m$ such that $x(m') < x(m)$ for every $m' \neq m$, then $m$ will be the only mathematician that can answer wrongly (for the others, $y(m') > x(m) > x(m')$). If there are several $m$ whose $x(m)$ tie for greatest, then they will all answer correctly.