2 people are playing a game with coins. First, 10 coins are put in a circular form. Then, the players take turns removing 1 coin or 2 coins that are together at each turn. The player who takes the last coin wins. Who has a winning strategy here if both players play optimally? How about if there are 11 coins. Is it possible to generalize for more than 11 coins?
Claim 1: If there are two contiguous groups of coins left of equal size, then the player to move loses.
Proof: If the player makes a certain move in one group, the other player can mimick the same move in the other group. At some point the first player will eliminate one of the groups, at which point the second player can also remove the other group, winning the game.
Claim 2: If there is one (non-cyclic) chain left of any length, then the player to move always wins.
Proof: If there is an even number of coins in the chain, you remove the middle $2$ coins, and Claim 1 then proves the result. Similarly, if there is an odd number of coins, you remove the middle coin, and use Claim 1 to finish the proof.
Claim 3: If there is one cyclic chain left of any length $\geq 3$, then the player to move always loses.
Proof: Removing any number of coins from anywhere in the cycle always leaves a non-cyclic chain of a certain length, allowing the next player to win by Claim 2.
So I think the final answer is that the player to move always loses, unless there are only $1$ or $2$ coins in a "circle".