Slice

On every turn, exactly 1 extra rectangle is created. In the beginning, there is only 1 rectangle, and we want to end with $m \cdot n$ rectangles. If $m \cdot n$ is even, the the first player is the winner, but if $m \cdot n$ is odd, then player 2 wins (we can see this by testing the first few products). Therefore we want to find the total number of ways such that $m \cdot n$ is even. $m \cdot n$ is odd if and only if $m$ and $n$ are odd themselves. There are then $(10)(10)=100$ odd products. Since there are a total of $(20)(20)=400$ products, $400-100=300$ products are even, and thus this is the desired answer.

Denote P as Player 1 and N as player 2 In order for P to win, there must be an odd number of moves available. When there is a 1x1 grid, there is no way for P to move so N wins, making it a N position. When there is a 1 x ${Z}$ , where ${Z}$ is an integer 1 ≤ ${Z}$ ≤ 20, there will be ${Z}$ - 1 possible moves, half of them being P positions and the other half being N positions. If there is a Z1 x Z2 grid, where either Z1 is odd and Z2 is even, there will be a total of cuts of Z2xZ1 + (Z2-1), thus, an odd number of cuts would be possible thus they would be P Positions. If Z1 and Z2 are both even, there would be a (Z1xZ2)+(Z2-1) number of cuts, still resulting in an odd number of cuts, thus they would all be P positions. However, if Z1 and Z2 both are odd, (Z1xZ2)+(Z2-1) would be an even number, and thus, be a N position. We know that the four scenarios is Z1 and Z2 are both odd, Z1 is odd while Z2 is even, Z1 is even while Z2 is odd, and Z1 and Z2 are both even. There would be 3/4 chance of there being a P position, and since there is 400 possible results, we can multiply 400 by 3/4 and have a resultant of 300 possible P positions, or games which guarantee Player 1 to win.

Player one will win when the number of cuts required to finish the game is odd.

Let's look at a 1\times n board. The number of cuts required to finish the game is n-1. Thus in this case, player one will win half of the time, or in 10 cases.

Let's look at a 2\times n board. We can cut the board along the n-length, in which case it will take 2n more moves to finish the game since we have two 1\times n boards. If we cut the board along the shorter side, we split it into two different boards with each having a side of 2. The number of cuts for each of those boards have the same parity, so the sum of cuts must be even.

Therefore, the number of cuts for a 2\times n board is an odd number. Thus player one wins in all 20 2\times n scenarios.

In a 3\times 1 board, player one loses. In a 3\times 2 board, player one can cut it into two 3\times one rectangles, or can cut it into a 2\times 1 and a 2\times 2. The number of cuts here is 1 + 4, so player one wins. Similar to our 1\times n board case, player one wins half the time, in 10 cases.

By virtue of a pattern, player one wins 10 + 20 + 10 + 20 + ... + 10 + 20 times, or 10(10 + 20) = 300 times.

Studiyng the simplest boards we gain the winning strategy for the first player. Board 2x2 $\Rightarrow$ Player 1 wins - Players sliced the board 3 times

Board 2x3 $\Rightarrow$ Player 1 wins - Players sliced the board 5 times

Board 2x4 $\Rightarrow$ Player 1 wins - Players sliced the board 7 times

We understand that the number of cuts made is m*n-1

Board 3x3 $\Rightarrow$ Player 1 loses - Players sliced the board 8 times

Board 3x4 $\Rightarrow$ Player 1 wins - Players sliced the board 11 times

Board 3x5 $\Rightarrow$ Player 1 loses - Players sliced the board 14 times

We understand that if the number of cuts is odd, player 1 wins (we can try with other boards and it happens every time).

We have to count the boards with m n-1 = 2k-1 $\Rightarrow$ m n = 2k

If n=2k-1 (10 possibilities) $\Rightarrow$ m=2h (10 possibilities) = 10*10=100

If n=2k (10 possibilities) $\Rightarrow$ Every m $\Rightarrow$ 10*20=200

The total boards which make player 1 win are 300

Via an inductive process, it is clear that for an n-by-m rectangle there mn-1 cuts are needed. If mn-1 is odd then player 1 wins. And of the 400 combinations there are 300 for which mn-1 is odd.

The game ends when the starting rectangle has been cut into $mn$ squares of size $1 \times 1$ . After the $k$ th turn, there will be $k+1$ rectangles. Thus, the game will always end after exactly $mn - 1$ turns. The first player will always win (no matter how they play!) if $mn$ is even and the second player will always win if $mn$ is odd. If $mn$ is odd, then $m$ and $n$ are both odd. Of the $400$ possible starting board sizes, $100$ of them have odd size. Thus, the first player will win for $400 - 100 = 300$ of the possible starting board sizes.