Have to finish couple of things these days, so I'm half missing.
Looking at that, considering the optimality of B and the number of quantities, I reckon it is fairly easy to build a geometric progression?
You can easily build a symmetry so that B doesn't have a choice, and then every consecutive turn A does not have a choice.
So, if you start with
1/2, next you have
1/4... hence, general term
Sn=1−2−n and the limit of that is 1.
Therefore, taking into account the final turn, all values below 2 can be obtained.
For part two, I'm getting some strange series. Would you reckon you get a lower value for
x than 2?
If so, I can try to find something about the third part, but I have to finish few pages of a coursework before that tonight.
And to EAT - always forget that.