In two players “Pop It!” game players take turns to choose a row and pop as many bubbles as they want in that row. The player who pops the last bubble loses the game. What is the optimal strategy for this game?
Assume that all bubbles in all rows are popped, except one row. If that row contains more than one unpopped bubbles then the player who plays in this turn wins: pop all the bubbles except one.
Pop-it board is (usually) symmetrical. Each row has a twin, another row with the same number of bubbles. Therefore we can separate the board into two symmetrical boards G1, G2. In this case, the second player has an optimal play, it is ensured that the last move is by the opponent.
The first player makes a move in say G1, the second player makes a mirror move in G2. Except that if the first player leaves a row with one bubble then the second player will make sure that there is an odd number of rows with exactly one bubble.
Consider the case that all rows have one bubble each. Players are forced to alternate popping one bubble until the game ends. So if there is an odd number of rows then the first player takes the last.
A board could have an odd number of rows. In this scenario, the first player clears one of the rows completely. This move makes the remaining board symmetrical. By repeating our previous arguments, we can see that the first player will win the game.