# Board game 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?

• Just to clarify, placing a goose on box $t$ forbids future placements on boxes $\{t-20, t-10, t, t+10, t+20\} \cap \{1\ldots 100\}$ (for both players)? – Robert Israel Mar 8 '17 at 19:45
• In principle we know it on turn 0, since this game is solvable. But I'm guessing what you're asking is more along the lines of "What is the maximum number of turns a game can take?". – eyeballfrog Mar 8 '17 at 19:51
• I love that 'it is a Dutch game' instantly explains why it should be a goose! Everything instantly makes sense now. – user334732 Mar 8 '17 at 21:32
• yes, @RobertIsrael, thats what I mean! – jbuser430 Mar 9 '17 at 10:14
• @eyeballfrog yes thats what I mean – jbuser430 Mar 9 '17 at 10:14

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.