There is something lost in the translation. Let say that after $B$ chooses a box, not only player $A$ says the distance of the ball to the selected box to player $B$, but even shows it to $B$, before moving the ball to an adjacent box, of course without $B$ knowing where. So $B$ knows the two places where the ball can be, but how can he decide?

mavropnevma wrote: There is something lost in the translation. Let say that after $B$ chooses a box, not only player $A$ says the distance of the ball to the selected box to player $B$, but even shows it to $B$, before moving the ball to an adjacent box, of course without $B$ knowing where. So $B$ knows the two places where the ball can be, but how can he decide? I think $B$ can guess just after $A$ says the distance.

How can $A$ say the distance before $B$ takes a guess? Then $A$ only knows where the ball is, not where $B$ will choose, so he cannot say the distance.

It's obviously wrong for $n=3$

mavropnevma wrote: How can $A$ say the distance before $B$ takes a guess? Then $A$ only knows where the ball is, not where $B$ will choose, so he cannot say the distance. I am also unsure about the statement, it is originally not clear, however I suppose that B can guess any time, I mean at each step B can guess more than one times. Because otherwise it is clear that B can never find the ball.

crazyfehmy wrote: mavropnevma wrote: How can $A$ say the distance before $B$ takes a guess? Then $A$ only knows where the ball is, not where $B$ will choose, so he cannot say the distance. I am also unsure about the statement, it is originally not clear, however I suppose that B can guess any time, I mean at each step B can guess more than one times. Because otherwise it is clear that B can never find the ball. If this is a original problem and $B$ can guess more than one time in each step , so $B$ can ask and after $A$ answer , $B$ guess the two boxes and he can win !!!

The question can be salvaged by requiring that the ball always move in the same direction. If $B$ knows the direction, he can obviously find the ball in no more than three guesses. If $B$ does not know the direction beforehand, it may take four.

This is the official English translation of the problem statement: There are $n$ chests places on the vertices of a regular $n-$gon and a bead. Alice and Bob play a game. At the beginning Alice hides the bead in one of the chests. Each move consists of three stages: Alice if wishes can secretly move the bead from the located chest to the any of two neighboring chest if the neighboring chest is not chosen by Bob at the previous move Bob chooses one of the chests Alice tells the distance from the chosen chest the chest where the bean is located If Bob can determine the chest containing bean at the end of some move he wins. For all values of $n\geq 3$ determine the minimal number of moves necessary for Bob to guarantee winning.