Two friends are playing a game where they break of pieces from a rectangular 10 x 15 candy bar. The game continues until either player gets a 1 x 1 piece. In the first move player A breaks the bar along a line. The second move is that player B chooses one of the two pieces and breaks it along a line (creating 3 total pieces). Third move is that player A can choose any of the three pieces (not just the latest two), and breaks it along a line. The game continues until either player breaks a piece leaving a 1 x 1 piece.
The question is if either of the players have a winning strategy if i) the loser is the player that creates a 1 x 1 piece ii) the winner is the player that creates a 1 x 1 piece
My reasoning is that after move $n$ there will be $n + 1$ pieces. Assuming optimal play there will be only 1 x 2 pieces left when either player is forced into losing in (i). That's exactly 75 pieces, so that's after move 74 which player B makes. That means that player A loses?
I don't really have a good idea of coming up with a strategy for (ii) though, any help of clue would be very much helpful!