From a bag containing $a_i$ candies, Alice receives $\lfloor a_i/2\rfloor$ candies whether Alice or Bob opens it.

The only difference is the extra candy when $a_i$ is odd. Alice receives that candy exactly when she opens the bag.

Therefore, her total is a fixed contribution $B=\sum_{i=1}^{n}\lfloor a_i/2\rfloor$, plus the number of odd-sized bags she opens.

This explains why choosing the largest bags is not the right goal: opening an even-sized bag gives Alice no advantage, while opening any odd-sized bag gives her exactly one extra candy.