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This question already has an answer here:

The scenario is like this:

There's a room, inside the room there are 100 cards with numbers [1,100] on them (No duplications), the cards are ordered randomly.

Two persons stand outside the room, They can decide on a tactic before the following:

  • One person enters the room, he can see the order, he can make one switch between places of two numbers. (e.g: (3,1,2) ---switch 3 and 1---> (1,3,2)).
  • After the switch (Or not switching) the person flips the cards and exits the room. (Other than the switch, the person does not change the order)
  • The second person enters the room, A number between [1,100] is called out of the void, he has 50 guesses to get that number.

A few notes:

  • The cards are in a perfect line, you can't order them vertically or something like that.
  • The cards are identical, There are no particular features for specific cards or something like that.
  • You can't leave anything in the room (items / human parts / ...)
  • The two persons have no idea about the number that will be called.
  • A guess = Flips a particular card.
  • This is not the same as the prisoners, Since the first person doesn't know about the number the second person is supposed to find.
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marked as duplicate by Ivo Beckers, boboquack, elias, Glorfindel, Rubio Apr 10 '17 at 16:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Isn't this the same as this one? puzzling.stackexchange.com/q/16/19989 $\endgroup$ – Marius Apr 10 '17 at 7:35
  • $\begingroup$ @Marius No, because that one has 100 people while this one has only 2. $\endgroup$ – Rand al'Thor Apr 10 '17 at 8:34
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    $\begingroup$ @randal'thor But the case with 2 people is a subcase in that question. $\endgroup$ – boboquack Apr 10 '17 at 8:41
  • $\begingroup$ You explain a scenario, but you put no actual question. What are we, as solvers, to do? $\endgroup$ – LeppyR64 Apr 10 '17 at 15:41
  • $\begingroup$ Possible duplicate of “How to beat Count Dracula”. $\endgroup$ – Peregrine Rook Apr 10 '17 at 16:34
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This is the same as 100 Prisoners' Names in Boxes

The algorithm is... Mentally number the cards 1 to 100 from left to right for example. turn the card with the (mental) number that you are supposed to find. If you found it, you are done.
If not, turn the cart with the (mental) number same as on the card you just turned.
The first person that goes into the room needs to make sure that there are no cycles more than 50 cards.
And this can be done easily since it can be max 1 cycle that is 51 or more. You can just break that cycle by changing 2 cards.
If you make sure there are no 51+ cards cycle you are always sure that you start in the right cycle and get to the card you need in 50 guesses or fewer.

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  • $\begingroup$ But the first user doesn't know about the number. $\endgroup$ – StationaryTraveller Apr 10 '17 at 8:38
  • $\begingroup$ @StationaryTraveller. He does not need to know the number. He just needs to make sure that there are no cycles more than 50. Then he does not care what the number is. Because he will be sure that the second person will start on the right cycle and find the card in the same number of guesses as the length of the cycle. $\endgroup$ – Marius Apr 10 '17 at 8:47

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