You need $\lceil \log_2(f-s) \rceil$ guesses to guarantee to find $n.$ See: en.wikipedia.org/wiki/Binary_search_algorithm#Performance

@AdamRubinson I know that there would be $\lfloor \log_2 (f - s + 1) \rfloor + 1$ guesses maximum for a given range integer range $s$ to $f$. However, I meant on an individual basis. Like for $s = 1$, $f = 20$, and $n = 10$, it would take only 1 guess using binary search.

@NoChance My question appear to be up to date. No, I'm not looking for the average. I'm looking for a function that works for a given range start, range end, and target number as mentioned in the examples. So, it would essentially replace the pseudocode with a mathematical function that is able to compute $g$ from $s$, $f$, and $n$ (as opposed to a programmatic function).

You aren't saying anything about the input, so there can't be a formula. The sorted input arrays $[1,1,1,1,5]$ and $[1,5,5,5,5]$ will need a different number of probes for the same $s$, $f$ and $n$.

Just to clarify what the int(...) is doing, you're setting $m = \lfloor \frac{s + f}{2}\rfloor$, right? In that case, one reduction is $(s, f, n) \mapsto (0, f-s, n-s)$. Now in the case where $f = 2^r-1$ (and $s = 0$), I think the $k$th $m$ will agree with $f$ for the first $k-1$ bits and have $0$s afterwards, so the answer should be something like $r - (\text{trailing zeros in } n)$.

I expect if $f - s + 1$ is not a power of two you will not get a nice formula, and only a recursion that is equivalent to just simulating the process.

@RandyMarsh There is no array in question. In the case of $s = 1$ and $f = 20$, we are performing binary search for $n$ in $1, 2, \ldots, 20$. So, if you want to think of it as an array, you can think of the array as the one created by starting at $s$ and increasing by 1 until you get to $f$ if that makes sense.

@ronno Yes, that is correct. The only reason why I used int(...) instead of floor(...) was to do truncation that rounds towards 0 as opposed to floor that rounds towards -\infty. So, if \frac{s + f}{2} happened to be negative, then it would act as ceil instead. I am fine with the use of floor to int if it works out better mathematically, though.

@ronno Yep, that reduction makes complete sense. A reduction could also be made if $f$ and $n$ are still nonpositive after that reduction by doing $(0, f, n) \mapsto (0, -f, -n)$ (to account for truncation instead of floor). It is unfortunate that the recursion probably wouldn't lead to a nice solution. Thanks for looking into it!

To be clear, I haven't really looked for patterns when $f \ne 2^r - 1$ but it just seems much more complicated

@GigiBayte2 What do you mean by formula? I suspect that no formula exists, but I cannot prove this unless I know what the complete set of allowed formulae is.

@ronno Right, I agree.

@MikeEarnest I am just looking for some closed-form mathematical expression, I suppose. Similar to how there is a closed form expression for the Fibonacci sequence's nth term.

@GigiBayte2 There has to be an array as input or some other structure that records order. Otherwise we're not talking about a binary search. If your input is the integer interval $[1,n]$, then what is exactly the motivation of doing a binary search, since you can detect whether $i\in\mathbb N$ is in it in constant time?

@RandyMarsh It is more of a matter of interest/curiosity/fascination rather than a matter of use in a practical application.