If you represent the numbers in the multiplication game to the base 18, it becomes clearer which will win and lose. My worksheet now shows numbers followed by the number-base-18 in parentheses, e.g. 18(10).

The winning and losing ranges for the first player are as follows

wins [2(2) - 9(9)]; loses [10(a) - 18(10)]  (Note I'm using 'a' for 10 to the base 18.)

wins [19(11) - 162(90]); loses [163(91) - 324(100)]

wins [325(101) - 2916(900)]; loses [2917(901) - 5832(1000)]

In general:

wins: [18^n+1 - 18^n*9]; loses: [18^n*9+1 - 18^(n+1)] or

wins: [10 ... 01 - 90...0]; loses: [90 ... 1 - 100 ... 0]

This works because a player with a number in a winning range (18^n+1 - 18^n*9) can always leave the next player with a number in a losing range (18^n*9+1 - 18^(n+1)). Similarly a player with a number in a losing range (18^n*9+1 - 18^(n+1)) must leave the next player with a number in a winning range  (18^(n+1)+1 - 18^(n+1)*9).

So, an algorithm to determine which player wins is as follows.

if (target<= 18) { if (target <= 9) player 1 wins else player 2 wins}

else {

val firstTwoDigits = the first two digits of target when represented to the base 18

if (firstTwoDigits are in the range 11 - 90)  player 1 wins

else /* firstTwoDigits are either in the range 91-99 or are 10 */ player 2 wins

}

This is essentially Daniel's algorithm.