Bill is working in the 38th floor of an 100-floors building. This building has a strange elevator which only has 2 buttons: the green one which takes you to the next floor every time you press it (and if you are at the last floor, it takes you to the 1st) and a red button which takes you to a random floor every time you press it. Once you enter the elevator, the first stop is selected at random, without pressing any button. Bill wants to eventually arrive at his office at the 38th floor. How is he going to do it, with the least number of steps?
I am thinking of a very simplistic approach: Since the first stop is decided at random, depending on the number of the floor where the elevator stops, Bill will decide which button to use: If he is near the 38th and before it, he will keep pressing the green button until he reaches the 38th. If it stops in a floor $>38$, he will press the red button until the elevator stops at some floor $<38$, but I assume the problem involves the expected number of stops etc probabilities, which I am not very familiar with! Any help is really appreciated!
Well, let me first extract from your question an important idea for solving this - you have identified, essentially, a set of reasonable strategies (which I've parameterized in some unknown integer $0 \leq n \leq 37$):
The question we're faced with is finding the best $n$ - for instance, one extreme case would be $n=0$, where he always presses the green button* when it'll eventually lead to the desired floor. On the other extreme, if $n=37$, Bill just presses the red button until he gets to his location. One might wish to justify why the strategy must be of this form - but it's not too hard to see that, since the green button is entirely non-random, it'd be silly to start pressing it and then to later decide, having learned nothing new, to press the red button.
Now, we can do our analysis. Let's imagine $n$ is fixed and evaluate the strategy. What we need to do is to define a series of variables $T_k$ which tells us how many more buttons we expect to press if we find ourselves on the $k^{th}$ floor.
Clearly, $T_{38}=0$. If $k$ is a floor on which we'd choose the green button, we have simply: $$T_k = 1 + T_{k+1}$$ since we need to press the green button, then do whatever we need to do to finish from the $(k+1)^{th}$ floor. So, we'd get $T_{37}=1$ and $T_{36}=2$ provided that $n$ is no more than $35$ - or, more generally $$T_k = (38 - k)$$ if $n < k \leq 38$.
If $k$ is a floor on which we'd choose the red button - so, any floor where $k\leq n$ or $38<k$ - then $$T_k = 1 + \frac{1}{100}\left(T_1+T_2+T_3+\ldots+T_{100}\right)$$ since we use one button press for sure and then end up, with probability $\frac{1}{100}$ each, at some other floor - and then should expect however many button presses it requires to get from that floor on. Then, the probabilistic part of the question is over - the rest is algebra. Note that our goal can be stated to minimize $T_{39}$ (or any floor above $38$), since doing so just takes one more button press than it takes to get to $38$ from a random floor.
Our equations can be simplified by realizing that, if we're going to press the red button, it doesn't really matter which floor we're on right now - that is, all of these values of $T_k$ are equal - let's call them all $$R=T_1=T_2=\ldots = T_{n-1} = T_n = T_{39} = T_{40} = \ldots = T_{100}.$$ There are $62 + n$ floors on which this random strategy holds, and we can thus simplify that $$R = 1 + \frac{62+n}{100}\cdot R + \frac{1}{100}\left(T_{n+1}+T_{n+2}+\ldots + T_{38}\right)$$ where we just pull out every term equal to $R$ from the original sum.
Then a bit of algebra gives $$R = \frac{100}{38 - n}\cdot \left(1 + \frac{1}{100}\left(T_{n+1}+T_{n+2}+\ldots + T_{38}\right)\right)$$ We can also recognize that the inner sum is known as just the sum of the first $38-n$ natural numbers, which is $\frac{(38-n)(37-n)}2$. Substituting this in and cancelling what needs to be cancelled gives: $$R = \frac{100}{38-n} + \frac{37-n}2$$ Note that we could get this result faster by realizing that we'll spend some amount of time moving randomly then have to go one step at a time. Formally, you could phrase this in random variables by writing that the time taken is $T$ which is $T_R + T_G$ where $T_R$ is the number of presses of the red button and $T_G$ is the number of presses of the green button. $T_R$ follows a geometric distribution with $p=\frac{38-n}{100}$. $T_G$ follows a uniform distribution on $\{0,1,2,\ldots,37-n\}$. The means of these distributions are $\frac{100}{38-n}$ and $\frac{37-n}2$ respectively. However, if you're uncertain of how to approach these questions, it's often better to work from first principles and not involve fancy constructions using random variables and linearity of expectation, as this fast path does.
However we arrive here, we finish by minimizing that expression for $R$ over $n$. We don't need to be clever here - there's not so many values and if you try all of them, you find the best value is $n=24$ and that $R=13+\frac{9}{14}$, meaning that if you press the green button on floors $25$ to $38$, you will, on average, require $13+\frac{9}{14}$ presses to get from any floor on which we'd press the red button to floor $38$. In terms of the original question, since the first random movement is free (and forced), this means Bill will need $12+\frac{9}{14}$ button presses on average.
(*I'm assuming the floors are $1,2,3,\ldots,100$ not $0,1,2,\ldots,99$ - you'll need to check your nearest skyscrapper to see if your country uses zero based indexing of floors)