Essentially, given a start of an inclusive integer range $s$, an end of the range $f$ such that $s \le f$, and an integer $n$ such that $s \le n \le f$, how many guesses would binary search take to find $n$?
For example, let $s = 1$, $f = 20$, and $n = 9$. Then, binary search would take $5$ guesses as follows: 10, 5, 7, 8, 9.
For another example, let $s = 1$, $f = 20$, and $n = 10$. Then, binary search would take $1$ guess as follows: $10$.
Can this be generalized to a formula with these 3 inputs?
Here is some pseudocode for reference that calculates the number of guesses for a specific combination with $g$ being the current number of guesses and $m$ being the current guess:
binary_search(s, f, n)
g = 0
while s <= f
g = g + 1
m = int((s + f) / 2)
if m == n
return g
else if m < n
s = m + 1
else
f = m - 1
Edit:
Using the information provided by ronno, here is the recurrence that we would have to solve to find the closed-form expression after performing the transformations $(s, f, n) \mapsto (0, f - s, n - s)$:
Note: Removed $s$ below as $s = 0$ after the above transformations. $$ \text{binary_search}(f, n) = \begin{cases} 1 &\text{if } m = n \\ 1 + \text{binary_search}(f - m - 1, n - m - 1) &\text{if } m < n \\ 1 + \text{binary_search}(m - 1, n) &\text{otherwise} \end{cases} \\ \text{where } m = \left \lfloor \frac{f}{2} \right \rfloor $$