# János Kramár comments on Concise Open Problem in Logical Uncertainty

• Ah, I think I can stymy with 2 nonconstant advisors. Namely, let and . We (setting up an adversarial ) precommit to setting if and if ; now we can assume that always chooses , since this is better for .

Now define and . Note that if we also define then is bounded; therefore if we can force or then we win.

Let’s reparametrize by writing and , so that .

Now, similarly to how worked for constant advisors, let’s look at the problem in rounds: let , and for . When determining , we can look at . Let . Let’s set to 1 if ; otherwise we’ll do something more complicated, but maintain the constraint that : this guarantees that is nondecreasing and that .

If then and we win. Otherwise, let , and consider such that .

We have . Let be a set of indices with for all , that is maximal under the constraint that ; thus we will still have . We shall set for all .

By the definition of :

For , we’ll proceed iteratively, greedily minimizing . Then:

Keeping this constraint, we can flip (or not flip) all the s for so that . Then, we have , if , and for , if .

Therefore, , so we win.