Assuming the pile of pebbles is visible and the players can see the amount of remaining stones, there is some strategy that can help win the game. However, it is not possible to win in all scenarios.
Restating the game rules:
- Game is for two players only.
- Starting player is selected randomly (eg. flipping a coin).
- Constant c is greater than 2 and defined before the game starts (eg. by throwing one or multiple dice).
- On each turn, players can remove 1 or c pebbles from the pile.
- The player who removes the last pebble loses the game.
Now, lets evaluate the different scenarios.
When c > N
If c is greater than the number of pebbles, the only option left for the player is to remove 1 pebble until there is only one left.
In this scenario, if the number of pebbles in the pile is even, the current player always wins. If the number of remaining pebbles is odd, the current player always loses.
This is very important to know, because it is key to win the game.
When c = N
When c is equal to the number of pebbles in the pile, the player will still have the option to remove 1 or to remove all. Removing all would end the game and the player would lose (would have removed the last pebble together with the other ones).
Thus, when c is equal to N, the only possible way for the player to win is by removing one pebble. However, as we saw before, the outcome at this point will depend on the parity of the pebbles in the pile. If the number of remaining pebbles is even, the current player will win the game. If the number is odd, the current player will lose whether it removes one pebble or the entire pile (c).
When c < N
This should be the starting point for most games (where N is greater than c). And it is the only scenario where the players can play some strategy.
As we just saw, once c is less or equal to N, the winning player is automatically predetermined by the party of the number of remaining pebbles.
With this information, we can conclude that...
as long as N is greater than c, the best strategy would be to remove pebbles in a way that leaves an odd number of pebbles equal to or less than c.