# Nim related number game

Consider the following game between two players A and B. Given are $2n$ numbers $a_1\dots a_{2n}$. The game consists of two phases:

1) Player A arranges the numbers in some suitable order.

2) In each turn, the player takes either the first or the last remaining number. Player B takes first. The player which larger sum of taken numbers wins.

Draws are excluded by assumption.

My conjecture is: regardless how player A arranges the numbers in the first phase, player B can always force a win.

For $n=1$ this is trivial. The reasoning for $n=2$ is as follows: Player A arranges the numbers to be $a_1\dots a_4$. By assumption (no draw possible) one of $a_1+a_3$ and $a_2+a_4$ is larger than the other. If $a_1+a_3>a_2+a_4$ player B takes $a_1$ and is guaranteed to be able to take $a_3$ in the next turn, thus B wins.

Is this game known under some particular name? Do you have any ideas how to prove the conjecture that B can force a win?

• The numbers to take are arbitrary, not just numbers $1$ to $2n$. Also A can arrange the numbers that B is not guaranteed to take an even number first. – daw Nov 5 '17 at 15:16