This is a wonderful puzzle, this time from a relatively lesser used source – A Moscow Math Circle, Week by week Problem Sets, by Sergey Dorichenko.
One marker is placed in the center of a 9×9 board. Anisha and Arushi take turns moving the marker to one of the adjacent squares – one sharing a side – provided that this square has never been occupied by the marker. Anisha goes first. The first player unable to move loses. Which of the players can guarantee a win?
As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.
Happy father’s day!
Alok Goyal's Puzzles Page 
The first person is likely to lose. If one cell containing the initial coin is removed, we have even number of cells left. If second person were trapped, first one has made odd or even moves and the second one has made the opposite. Either ways , o + e = o which is not equal to even cells. If first person was trapped, the number of moves is odd + odd or even + even, thereby an even number of cells used up. Therefore only the second move makes you ensure victory
Agreed, but the proof does not appear to be bullet proof, I will share the answer soon