The following simple observation leads to a solution of the task:

Since the game board initially contains an even number of elements arranged in a sequence, when the first player moves one end of the sequence is in odd and the other end is in even position. Therefore the first player can always select an element either from an odd or even position.

The algorithm preprocesses the contents of the initial board before starting the game to compute the values ~OddSum~ and ~EvenSum~, the sum of the elements in odd positions and the sum of the elements in even positions, respectively. If ~OddSum \ge EvenSum~ then the first player always selects from an odd position and forces the second player to select from even position. The case ~OddSum < EvenSum~ treated similarly.