What Can We Say Complexity of Solving the Weighted Pool Ball Problem with a Variable Scale?
The asymptotically optimal algorithm is to split in 2 groups at each step.  In other words, the ability to split into more than 2 groups does not allow us to find an asymptotically more efficient algorithm.

Suppose we split into $k$ at each step.  Then we get an algorithm whose running time is given by the following recurrence relation:

$$T(n) = T(n/k) + O(k).$$

We do $\lg(n)/\lg(k)$ recursive calls to reach the base case, so the solution to this recurrence relation is

$$T(n) = O(k \lg(n)/\lg(k)).$$

Now the function $f(k) = k / \lg(k)$ attains its minimum at $k=2$; for larger $k$, it is strictly increasing.  Therefore, the optimal choice of $k$, if we split by the same ratio at each level and disregard constant factors, is $k=2$.

Splitting by a different amount at each level doesn't help.  The above argument applies to all $n$, so it applies no matter the number of balls -- i.e., no matter what level we split at.



Here is another way to think about it.  Suppose we split into $k_1,k_2,k_3,\dots,k_m$ groups at each level ($k_1$ groups in the first call, $k_2$ in the second recursive call, and so on).  Let's count the number of bits of information we've gained.

How many bits of information do we get after applying the scale when we've split into $k$ groups?  $\lg k$ bits of information, since there were $k$ possibilities.  So, we get $b_i$ bits of information at the $i$th call, where $b_i= \lg k_i$.

Now we need a total of $\lg n$ bits of information to uniquely identify the ball.  Therefore, we need to select $b_1,\dots,b_m$ so that

$$b_1 + b_2 + \dots + b_m \ge \lg n.$$

What is the running time for any particular choice of $b_1,\dots,b_m$?  Well, in the $i$th call, we split into $k_i = 2^{b_i}$ groups, and the time to use the scale is $O(k_i)$, i.e., $O(2^{b_i})$.  Therefore, the total running time for a particular choice of $b_1,\dots,b_m$ is

$$O(2^{b_1} + 2^{b_2} + \dots + 2^{b_m}).$$

Now we want to find the optimal choice of $b_1,\dots,b_m$.  This requires us to minimize $2^{b_1} + + \dots + 2^{b_m}$ subject to the constraint that $b_1 + \dots + b_m \ge \lg n$.  One can compute that the minimum is attained for $b_1= \dots = b_m = 1$.



At the end of your post, you propose an algorithm that increments the number of splits at each step.  You didn't compute the asymptotic running time of your algorithm or compare it to binary search, but if carry through that analysis, it turns out that your algorithm is asymptotically slower than binary search.  Therefore, your proposed algorithm is worse than straightforward binary search.  It does not contradict any of the above analysis.

Details:   By Stirling's approximation, $\Gamma^{-1}(n) = O(\lg n)$ (roughly, ignoring log-log factors), so your proposed algorithm's running time is something like $O((\lg n)^2)$ (roughly, ignoring log-log factors), i.e., worse than binary search.