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An evil wizard plays the following game with two dwarfs $A$ and $B$: he thinks of a function $f:\mathbb{R}\to\mathbb{R}$ (which is not required to have any regularity properties, such as measurability, ..) and asks $A$ and $B$ to guess it.

$A$ and $B$ play in two separate moments.
$A$ begins and he is allowed to ask the values of $f$ on some subset $S_1\subset\mathbb{R}$.
Then he can ask the values of $f$ on some $S_2$, and so on.
He must guarantee that he will only ask about a finite number of subsets, say $N$, and that $$ \bigcup_{i=1}^N S_i\subsetneq\mathbb{R}. $$ Note that $N$ is not fixed; $A$ must only guarantee that eventually he will stop posing questions and that, at that moment, there are some values of $f$ which the wizard has not yet revealed him.
When he stops, he has to guess the remaining values of the function.

Then it's $B$'s turn, and everything proceeds exactly in the same manner.

$A$ and $B$ can not communicate with each other (except before the game begins, to decide the strategy), so one can also think of $A$ and $B$ playing at the same time but in separate rooms.

$A$ and $B$ are both freed by the wizard if at least one of them guesses the correct function, otherwise he kills both of them.

Is there a winning strategy for the two dwarfs?

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    $\begingroup$ There are no restrictions on the $S_i$, right? So why would the dwarf want to take $N$ different subsets and not just the union? $\endgroup$ – Jolien Apr 11 '15 at 18:36
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    $\begingroup$ @Math1000: That’s not permitted. $S_1$ must be a proper subset of $\Bbb R$. $\endgroup$ – Brian M. Scott Apr 11 '15 at 18:48
  • $\begingroup$ Sorry, missed that part. $\endgroup$ – Math1000 Apr 11 '15 at 19:01
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    $\begingroup$ Funny that you happen to be setting a bounty for this -- I just came across the question today and started to think about it :-) $\endgroup$ – joriki Jul 22 '15 at 21:05
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    $\begingroup$ Assuming the Axiom of Choice, they can reduce the problem to this question math.stackexchange.com/questions/371184/predicting-real-numbers/… by both picking $S_1 = \Bbb R \setminus \Bbb N$ and discarding the info the wizard tells them. $\endgroup$ – mercio Jul 23 '15 at 14:22
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Assuming the Axiom of Choice, they can reduce the problem to this question by both picking $S_1= \Bbb R \setminus \Bbb N$ and discarding the info the wizard tells them.

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