87

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?

Jared
  • 30,335
  • 10
  • 55
  • 131
  • I do not understand this game. If each mathematician can open countably many boxes, here is a winning strategy: everyone open all the boxes; then they vacuously win. Who chooses the unopened box for the guessing? Is this uniform in any way? – Asaf Karagila Apr 24 '13 at 06:28
  • @Asaf: I think that the statement precludes the possibility of opening every box: it is implicitly required that there be an unopened box in his room. As I read it, each mathematician guesses the number in some unopened box in his room and any uniformity is solely the result of their arrangements during the strategy discussion. – Brian M. Scott Apr 24 '13 at 11:03
  • If the numbers are different, there are uncountable numbers possible for the lone unopened box in each room. No possible guess in sight, unless there is some relation between the numbers. I.e., $b_i \to b_0$ or some such. – vonbrand Apr 24 '13 at 15:24
  • 1
    Surely a single mathematician can guess the contents of a box with probability $0$, regardless of how (countably) many boxes he has opened. In your description guesses are independent, so surely the probability to win is $100 \cdot 0^{99}\cdot 1 = 0$. – Karolis Juodelė Apr 24 '13 at 15:28
  • 5
    As I said, at first a solution seems impossible, but it also seems impossible that two unit spheres can be assembled from the pieces of one unit sphere. Strange results are possible with the axiom of choice. Also, to address vonbrand's comment, there is not necessarily any relation between the numbers, and it is only required that *at least*, and not *exactly*, one box be left unopened. – Jared Apr 24 '13 at 15:30
  • 1
    Does each mathematician have to guess the number and the box or is it sufficient to say $r$ is in one of those unopened boxes? – John Douma Apr 24 '13 at 15:44
  • 1
    A couple questions. (1) Do the mathematicians guess their numbers simultaneously, or one after another? (2) Assuming (as is very likely) the latter in (1), do they determine the order of the guessing, or some malevolent entity? (3) Again assuming that latter in (1), do the mathematicians know the guesses of previous mathematicians? – user642796 Apr 24 '13 at 16:28
  • The requirement that at most one of the guesses is wrong makes this problem a lot more "constructive" than Banach-Tarski sort of argument. Had you said infinitely many mathematicians, and finitely many mistakes... **maybe** it was possible. – Asaf Karagila Apr 24 '13 at 16:58
  • 1
    @Asaf Karagila: It should possible for that case as it should be equivalent to the infinite prisoners and hats problem if the mathematicians label themselves 1-100 and open all boxes that are one more than their label. I think you are invoking some uncountable axiom of choice on the equivalence classes in that case. For reference: http://en.wikipedia.org/wiki/Prisoners_and_hats_puzzle#Countably_Infinite-Hat_Variant_without_Hearing – Alex R. Apr 24 '13 at 17:05
  • 3
    @Alex, but the difference is that the prisoners puzzle allow for an arbitrarily large, but finite, failure. Reducing this failure to a single failure seems... unlikely. – Asaf Karagila Apr 24 '13 at 17:08
  • @Arthur They certainly can't guess in front of each other! Otherwise they can communicate an answer trivially. If you mean "are they allowed to hear whether the others were successful" then this doesn't affect whether or not there is a positive probability of winning. – not all wrong Apr 24 '13 at 17:34
  • 12
    If I fill the boxes with undefinable numbers, then no mathematician can guess the number in any box, since there is no way he can name it :p – Nick Matteo Apr 24 '13 at 18:43
  • 2
    @ArthurFischer: The mathematicians don't interact with each other until all have made their prediction, so we may as well assume they all guess simultaneously. – Jared Apr 24 '13 at 20:11
  • 1
    @Karolis: but the mathematicians are playing a deterministic strategy. I don't think a probabilistic argument holds water. – Qiaochu Yuan Apr 24 '13 at 20:37
  • Usually, when a solution depends upon the axiom of choice, we cannot explicitly write the solution. We can only say it exists. Are we to give the explicit solution or is it sufficient to show that a winning strategy exists? – John Douma Apr 25 '13 at 01:31
  • Anyway, here's a thought. The mathematicians can open boxes adaptively; that is, they can first open a box, and then based on what they've seen so far, choose another box to open accordingly. In particular they do not need to decide all of the boxes they will open ahead of time (but only the first box they will open). – Qiaochu Yuan Apr 25 '13 at 01:32
  • @Jared Is there a point at which you intend to post your solution? (Yes, I realize you want to provide enough time for us to find our own solutions.) – Karl Kroningfeld Apr 25 '13 at 01:42
  • 3
    Since a couple of comments have mentioned the "probability" of correct guesses, I'd like to second (and perhaps amplify) Qiaochu Yuan's comment: There is nothing probabilistic going on in the problem. Furthermore, I don't see anything the mathematicians can gain by randomizing their guesses. – Andreas Blass Apr 25 '13 at 02:15
  • Is the diabolical mastermind who planned this riddle (diabolical because in my mind he kills the mathematicians if they lose) prepare the boxes before or after the mathematicians have devised their strategy? Is he allowed to know what strategy they will use? – Asaf Karagila Apr 25 '13 at 07:46
  • @QiaochuYuan, AndreasBlass, the probabilistic argument is natural because he wants the probability of being right to be $99 \over 100$ (well, he wants something even stronger). Surely you agree that one mathematician would guess correctly with probability 0. Do you want to argue that the guesses are dependent? – Karolis Juodelė Apr 25 '13 at 11:03
  • 1
    @AsafKaragila: if it's a winning strategy, it doesn't matter. – Nick Matteo Apr 25 '13 at 12:28
  • 5
    @KarolisJuodelė No, he does not ask for some probability to be $\frac{99}{100}$. He asks for a strategy that ensures that 99 of the 100 guesses are correct. – Andreas Blass Apr 25 '13 at 17:12
  • Great problem. Does anyone have information as to the originator? And I am told that a strategy here leads to a nonmeasurable set, which means that the Axiom of Choice (or something close to it) is essential. I see that it is mentioned with no specifics in The Mathematical Intelligencer, Jim Henle's column, 36:4, Winter 2014. But he says there that he does not know who the originator is. – stan wagon Nov 20 '19 at 16:22

2 Answers2

41

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.

Jonathan Hebert
  • 4,988
  • 1
  • 17
  • 35
mercio
  • 48,894
  • 2
  • 77
  • 128
  • This is more or less the solution I know, exposited very nicely! I like the idea to re-label the boxes. The solution I know has mathematician $m$ open all boxes not congruent to $m\operatorname{mod}100$, which is just a specific re-labeling in your solution. – Jared Apr 25 '13 at 16:18
  • The representative before opening $(m, x)$ for large $x$ can (will) be different from the representative after. Hence you actually have two different values of $x(m')$ and of $y(m)$. After opening the $(m, x)$ boxes, the old values might as well be random numbers. While they can all be recalculated, what guarantee is there, that the box $(m, y_2(m))$ was not opened already? – Karolis Juodelė Apr 25 '13 at 19:10
  • @KarolisJuodelė: I don't understand. The entire sequence for $m'$ was already opened for each $m' \neq m$, so the values of $x(m')$ (and hence $y(m)$) won't change. – Nick Matteo Apr 25 '13 at 21:14
  • @Kundor, for simplicity, say $M = \{m_1, m_2\}$ and $n \in N$ is relabled so that even numbers belong to $m_2$. Say after $m_2$ opens boxes for $m_1$, he sees $4?2?0?0?0?0?0\dots$ (where $?$ is an unopened box). He picks the representative $0000000000000\dots$ and thus $y_1(m_2) = 3$. He then opens his own boxes and sees $4?2?0?0101010\dots$, but now the equivalence class has changed and the new representative is $654321010101\dots$ and $y_2(m_2) = 4$ but the box $(m_2, 4)$ is already open. – Karolis Juodelė Apr 26 '13 at 05:13
  • @KarolisJuodelė: Your example is helpful in understanding where lies the misunderstanding. When the boxes for $m'$ are opened, we view those boxes as an entire real sequence themselves, not as a part of the original real sequence. In a sense, we are partitioning the original sequence into $|M|$ subsequences, so that the representative of each subsequence does not change when opening boxes from other subsequences. – Jared Apr 26 '13 at 05:47
  • What is "the sequence $(m',x)$"? – JiK Aug 11 '15 at 12:17
  • @JiK it's the sequence of the real numbers in the boxes labelled $(m',0), (m',1), (m',2), (m',3), \ldots$ – mercio Aug 11 '15 at 13:20
  • 1
    Correct me if I'm wrong, but there's nothing special about real numbers here. You can have a fixed collection of objects, say ordinals below $\aleph_7$, and say 100 mathematicians see a countable sequence of boxes, each containing an ordinal below $\aleph_7$. Choose equivalence class representatives of the countable sequences of these ordinals, etc. – mbsq Mar 15 '17 at 17:28
  • @mbsq you are correct: as long as the mathematicians know a fixed set (of arbitrary cardinality) that each box will contain an element of, the strategy holds. – StevenClontz Jul 31 '18 at 17:07
22

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.

StevenClontz
  • 874
  • 6
  • 14
  • Nice argument but the paragraph before the conclusion confuses me. Shouldn't it read "for each mathematician #m, all boxes x(m) and onward match the representative"? I am also not sure about the three uses of the x(1) box that follow in the next sentence; wasn't the mathematician #m supposed to guess the y(m)-1 box corresponding to his representative? – Luke Skywalker Aug 14 '13 at 18:46
  • 1
    Ah! I was forgetting that in the case x(1) is the greatest, then y(m)-1 is actually x(1). I leave the comment as a remark just in case someone else bumps into that confusion. For a moment it looked like everyone was guessing on the first mathematician sequence and not on their own. – Luke Skywalker Aug 14 '13 at 19:00
  • 1
    +1 for clearer explanation, e.g. your phrase "sequences of real numbers" with examples, and your clearer explanation of the $x$ and $y$ functions. – Rosie F Jul 20 '18 at 08:13
  • Hey, this same strategy works when there are a finite number of numbers, say for example 1000 boxes in each room, so each room has a 100 sequences. But it doesn't seem intuitive that it works for a finite number of boxes. Did I do wrong anywhere? – Mahathi Vempati Jul 30 '18 at 15:35
  • 1
    @Tinkidinki There's a couple issues with only having finitely many boxes. For one, all the finite sequences match on all but finitely many terms, so everything belongs to the same equivalence class. Another is that we cannot define $x(n)$ to be the first box that matches the equivalent class representative, since that box may not exist (all boxes may be different). – StevenClontz Jul 31 '18 at 17:06