BxMO Team Selection Test

We will show that the capacity of the two boxes does not matter, as long as the total capacity is $n$ (and at least 1 for each box). Jesse can always win this game, and can do that by first playing the power $2^0=1$ of two, and then in each following turn the next power of two that is smaller or greater. That means: if he played the numbers $$2^{-i},2^{-(i-1)},\dots ,2^{-1},2^0,2^1,\dots ,2^{j-1},2^j$$ at a certain moment, he will play either $2^{-(i+1)}$ or $2^{j+1}$ in his next turn. By playing cleverly, Jesse can make sure that the greatest power of two among the stones played so far is always contained in the black box. We will prove this by induction. In his first move, he puts the stone with value $2^0$ in the black box and the claim is true; this is the base case of the induction. When it is his turn again, and Tjeerd moved the greatest power of two so far, which according to the induction hypothesis was contained in the black box, to the white box, then the black box actually has a free space, and Jesse can put a new greater power of two in there, and the claim is true. If Tjeerd moved some other stone or did nothing, then the greatest power of two so far is still in the black box, and Jesse can play a smaller power of two; it does not matter where he puts it. Also in this case, the claim is true. This proves the induction step, and the claim is proved.

Therefore, after playing the last stone, the greatest power of two is in the black box. It is greater than the sum of all smaller powers of two played ($2^j>2^j-2^{-i}=2^{j-1}+2^{j-2}+\cdots +2^{-(i-1)}+2^{-i}$), hence it is certainly greater than the sum of the powers of two in the white box. Therefore, the total value inside the black box is greater than the total value in the white box. $\square$