Here is a variation of a Nim game : consider a full set of cuisenaire rods - 10 rods of all integer lengths between 1 and 10. Set a number N between 1 and 54. Player 1 choose one of the 10 rods and place it between the two players. Player 2 choose one of the remaining 9 rods and put it following the first one. Player 1 choose one of the remaining 8 rods and place it following the second one, and so on. The loser of the game is the player that place a rod such that the sum of the lengths is greater than N.
Quite strangely, determining whether player 1 wins or loses may be very easy or quite difficult with respect to N. Do you know anything about this game? I can't say that I invented it as I am sure that someone else has already studied it but I don't know where to search. Thanks by advance for your comments :)
Clearly the first player wins any game with $N \lt 11$. The second player wins $11, 12$ ($12$ by playing $10$ if the first player starts with $1$). First wins $13,14$ by playing $1$.
First wins $54$. Second wins $53$ by playing $1$ and $52,51$ by playing $1,2,3$
In between it gets harder.