I'm asked to do answer the following question:
Two players are playing a game. They have a gameboard with boxes $1-100.$ When it's player 1's turn, he places a goose (this is a Dutch game) on one of the boxes. If this is box t, player $2$ is not allowed to put another goose on $t-20, t-10, t, t+10$ or $t+20.$ So, one cannot place a goose at box $11$ if there is already a goose on $1, 21$ or $31$ (91 doesn't count here). The player who cannot place a goose on one of the boxes anymore, has lost the game. After how many turns do we know who the winner is?
I think it might have to do something with the pigeonhole principle, but I'm not sure how to prove this.
Is there someone who knows how to start this?
Here is a winning strategy for player 2.
If player $1$ plays a goose on box $t$, then player $2$ plays on box $t+1$ if $t$ is odd, $t-1$ if $t$ is even.
By induction, we can show that in each pair $(2i-1,2i)$, $i=1\ldots50$, after player 2's turn either both are allowed or both are forbidden. Thus it is always possible for player 2 to follow this strategy, no matter what player 1 does.