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.

$100$ rooms each contain countably many boxes labeled with the natural numbers. Inside of each box is a real number. For any natural number $n$, all $100$ boxes labeled $n$ (one in each room) contain the same real number. In other words, the $100$ rooms are identical with respect to the boxes and real numbers.

Knowing the rooms are identical, $100$ 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.

$99$ out of $100$ mathematicians must correctly guess their real number for them to (collectively) win the game.

What is a winning strategy?

Solution 1:

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.

Solution 2:

Found this via Reddit. Here's my writeup of the solution.

The strategy involves the axiom of choice like so: the mathematicians group sequences of real numbers such that two sequences are in the same group if and only if they agree on all but the first few terms.

For example, let $\pi_i$ denote the $i$-th digit of $\pi$ (i.e. $\pi_0=3,\pi_1 = 1,\pi_2=4,\dots$).
Then the sequences $(\gamma,e,\sqrt 2,2^{4/3},\pi_1,\pi_2,\pi_3,\dots)$ and $(\ln(2),\gamma,-7.8,\pi_0,\pi_1,\pi_2,\pi_3,...) $ are in the same group, because they both are equal on all but the first 4 elements.

The Axiom of Choice is required to choose an arbitrary representative from each group. For example, I can choose $(1.49,3,-\cos(4),\pi_0,\pi_1,\pi_2,...)$ to represent the group I described above, but since there's infinitely many groups and I only have a finite amount of space to describe my strategy, I must appeal to the Axiom of Choice to produce a "choice function" that tells me which representative should be chosen from each of the groups.

Now I'll describe the plan. Let's say I'm mathematician #1. I'm going to open every box except 1, 101, 201, 301, and so on. Meanwhile mathematician #2 will open every box except 2, 102, 202, 302, etc., and in general mathematician #$n$ will open all boxes except those labeled $n,100+n,200+n,300+n,...$.

Back to me. I know what's inside the boxes that my buddy in room 2 didn't open. Let's suppose the numbers are:

• 2 -> 1739218.33
• 102 -> sqrt(5)-sqrt(2)
• 202 -> Arctan(37.238)
• 302 -> 382
• 402 -> -832.019
• 502 -> 4
• 602 -> $\pi_3$
• 702 -> $\pi_4$
• 802 -> $\pi_5$
• ...

Okay, I know the group that falls in. (Coincidentally, it's the group I talked about above.) I'm going to write a note that it started matching the representative from box $602$ onward. Let's write that note like this: "x(2)=6". I'll repeat that process for #3, noting "x(3)=5" for box $503$, and for #4, perhaps I note that "x(4)=7" for box $704$, and so on.

What I've done is defined $x(m)$ for $m\in\{2,\dots,100\}$ to be the first box that disagrees with the associated representative given by the Axiom of Choice. However, I don't know the value of $x(1)$ since I haven't opened boxes $1,101,201,\dots$ yet.

What I can do though is let $y(1) = \max(x(2),...,x(100))+1$ be larger than all the observed numbers. It just so happened that $x(4)=7$ was the biggest, so $y(1)=8$.

It's finally time to open most of the remaining boxes. I'll open all the boxes in my sequence $1, 101, 201, \dots$, starting with the box given by $y(1)=8$: box $801$. (Since $y(1)$ has to be at least $1$, this strategy always leaves at least box $001$ closed.) Let's see what I got (let $e_i$ denote the $i$-th digit of $e$):

• 1 -> ???
• 101 -> ???
• ...
• 701 -> ???
• 801 -> 7pi+sqrt(3)
• 901 -> $e_{9}$
• 1001 -> $e_{10}$
• 1101 -> $e_{11}$
• ...

Seeing those last digits, I know enough to figure out which group it belongs to: the group with representative $(\sqrt 2,e_1,e_2,e_3,...)$.

I now know enough to make my guess. I'm going to use the representative, and pick the box given by $y(1)-1=7$, which is the maximum value of $\{x(2),\dots,x(100)\}$ (which we said was $x(4)$ in this example). In this case, the seventh entry is $e_7=1$ (note that we're zero-indexing so that the first entry goes with box $001$). So I'll guess that box $701$ contains $1$.

Of course, mathematician #$n$ will do exactly the same thing by considering the values of $\{x(1),\dots,x(n-1),x(n+1),\dots,x(100)\}$, computing $y(n)$, and so on. Now, I need to prove to you that these strategies work.

To do this, I just need to prove that if I'm wrong, then everyone else is right! (99/100 ain't bad, according to the rules.)

Okay, I'm wrong, so what happened? Obviously, the box given by $y(1)-1$ didn't match the Axiom of Choice's representative. That means that $x(1)>y(1)-1$, since $x(1)$ is the number for which every other mathematician knew the boxes from that point forward matched the Axiom of Choice's representative.

With this information, I realize something. $y(1)-1\geq x(n)$ for every other number n, since $y(1)$ is defined to be the maximum of the $x(n)$ plus one! So here's what I now know:

$x(n)\leq y(1)-1<x(1)$

This is great news. Everyone else defined $y(n)$ knowing $x(1)$, and $x(1)$ has just been shown to be bigger than the other $x(n)$. So:

$x(n)\leq y(1)-1<x(1)<x(1)+1=y(n)$

So, for each mathematician #$n$, all boxes $x(1)$ and onward match the representative given by the Axiom of Choice. Every mathematician #$n$ opened $x(1)+1$ onward, and guessed the choice function's $x(1)$ entry for the $x(1)$ box, which has to match up!

Thus, if I'm wrong, everyone else has to be right. And that beats the game.