Game to maintain distinct number of balls in glasses There are $n$ glasses, containing $n+1,n+2,\ldots,2n$ balls, respectively. Two players $A$ and $B$ play a game, alternately taking turns with $A$ going first. In each move, the player must choose some balls (perhaps all, but not empty) from one glass, so that the number of balls remain pairwise distinct. The player who cannot move loses. Who can win?
The position where the game ends must be $0,1,\ldots,n-1$, since otherwise there is a move left. To reach this position, the game must be in the position where $n-1$ of the $n$ glasses coincide with the end position. For $n=1$, $A$ wins after the first move. For $n=2$, $A$ takes off two balls from $4$ to get to $3,2$. Then whatever $B$ does, some glass contains $0$ or $1$ balls, so $A$ wins in the next turn.
 A: If n is even, A's first move is to move n balls out of the 2n-glass. Partner adjacent numbers together in such a manner (0,1), (2,3), (4,5), ..., (2n-2,2n-1). Whenever B turns an n-glass into an m-glass, A responds by turning partner(n)-glass into the partner(m)-glass.  
If n is odd, A's first move is to take all balls out of the (n+1)-glass. Then we partner numbers (1,2), (3,4), ..., (2n-1,2n) and do the same as before.  
_

To see why this works let's work thru an example where $n=6$. The initial setup is
$\;\;\;\;\;\;\;\; \_\__0 \;\;\; \_\__1 \;\;\; \_\__2 \;\;\; \_\__{3} \;\;\; \_\__{4} \;\;\; \_\__{5} \;\;\; \_\__{6} \;\;\; \text{O}_{7} \;\;\; \text{O}_{8} \;\;\; \text{O}_{9} \;\;\; \text{O}_{10} \;\;\; \text{O}_{11} \;\;\; \text{O}_{12} $
where O$_m$ represents a glass with $m$ balls. The game ends when the slots $\_\__k$ are filled for $k=0..5$. Player $A$ makes his first move by taking out $6$ balls from O$_{12}$, resulting in
$\;\;\;\;\;\;\;\; \_\__0 \;\;\; \_\__1 \;\;\; \_\__2 \;\;\; \_\__{3} \;\;\; \_\__{4} \;\;\; \_\__{5} \;\;\; \text{O}_{6} \;\;\; \text{O}_{7} \;\;\; \text{O}_{8} \;\;\; \text{O}_{9} \;\;\; \text{O}_{10} \;\;\; \text{O}_{11} $  
The key observation is that there are an even number of slots and an even number of glasses -- so we can pair them.
$\;\;\;\;\;\;\;\; (\_\__0 \;\;\; \_\__1) \;\;\; (\_\__2 \;\;\; \_\__{3}) \;\;\; (\_\__{4} \;\;\; \_\__{5}) \;\;\; (\text{O}_{6} \;\;\; \text{O}_{7}) \;\;\; (\text{O}_{8} \;\;\; \text{O}_{9}) \;\;\; (\text{O}_{10} \;\;\; \text{O}_{11}) $   
Now $A$ wants to keep the pairs together. Say player $B$ takes 7 balls out of $\text{O}_9$, forming   
$\;\;\;\;\;\;\;\; (\_\__0 \;\;\; \_\__1) \;\;\; (\text{O}_2 \;\;\; \_\__{3}) \;\;\; (\_\__{4} \;\;\; \_\__{5}) \;\;\; (\text{O}_{6} \;\;\; \text{O}_{7}) \;\;\; (\text{O}_{8} \;\;\;\_\__9) \;\;\; (\text{O}_{10} \;\;\; \text{O}_{11}) $  
Player $A$ responds by changing $\text{O}_8$ into $\text{O}_3$ (by taking out 5 balls).  
$\;\;\;\;\;\;\;\; (\_\__0 \;\;\; \_\__1) \;\;\; (\text{O}_2 \;\;\; \text{O}_{3}) \;\;\; (\_\__{4} \;\;\; \_\__{5}) \;\;\; (\text{O}_{6} \;\;\; \text{O}_{7}) \;\;\; (\_\__8 \;\;\;\_\__9) \;\;\; (\text{O}_{10} \;\;\; \text{O}_{11}) $   
This ensures that $A$ always has a valid move after $B$, so player $A$ makes the last move and thus wins.
