Proving: $X_i-X_{i-1}\sim \operatorname{Geo}\left(\frac{9-i+1}{9}\right)$ I'm trying to solve the following quetsion:

Visitors reach the zero floor (entrance floor) in a 9-floor building (the building has
10 floors in total). In the building there are two elevators:

*

*Elevator A - reaching from zero floor to the odd floors in the building and back $(1,3,5,7,9)$.

*Elevator B - reaching from zero floor to the even floors in the building and back $(2,4,6,8)$.

Each visitor should reach one of the $1,\ldots,9$ floors with equal odds and independently to other visitors.
It was assumed that an infinity of visitors would come to the building, one after the other.
Also assume that all visitors board an elevator appropriate to the floor they are required to reach.
We denote by $X_i$ the number of visitors who came to the building until they first visited a floor they had not yet visited for $i=1,\ldots, 9$. This means that $X_1=1$, since the first visitor to the building visits one of the floors (and since he is the first, he has not necessarily visited it yet).
For example, if the visitors visited the following floors (from left to right in the order of arrival): 33347346, then, as stated $X_1 = 1$. In addition, the fourth visitor is the first to visit the different floor from the 3rd floor (which they visited first), so $X_2 = 4$. Similarly, the fifth visitor is the first to visit the different floor from the two floors chosen before him (floors 3 and 4) and therefore $X_3 = 5$.
How is $X_i-X_{i-1}$ distributed for $i=2,3,\ldots,9$?

In the solution they stated that the solution is:
$$
X_i-X_{i-1}\sim \operatorname{Geo}\left(\frac{9-i+1}{9}\right)
$$
Why is it geometric? Stuck on this question for a while. How do we approach this question?
 A: $X_{i}-X_{i-1}$ counts how many visitors we need to wait until a new floor is visited for the first time.
At $i-1$, exactly $i-1$ different floors have been visited.
The next visitor will visit an already visited floor with probability $$q = \frac{i-1}{9},$$ while (s)he will visit a new floor with probability $$p = \frac{9-(i-1)}{9}.$$ Notice that $q = 1 - p.$
Then:
$$\begin{array}{rcll}
P(X_i-X_{i-1} = 1) & = & p & \text{(the next player will go to a new floor)},\\
P(X_i-X_{i-1} = 2) & = & qp & \text{(the next player will go to}\\
& & & \text{an old floor, and the last to a new floor)},\\
P(X_i-X_{i-1} = 3) & = & q^2p & \text{(the next 2 players will go to}\\
& & & \text{an old floor, and the last to a new floor)}\end{array}$$
In general:
$$
\begin{array}{rcll}
P(X_i-X_{i-1} = k) & = & q^{k-1}p & \text{(the next}~k-1~\text{players will go to} \\
& & & \text{an old floor, and the last to a new floor)}.\end{array}$$
Hence, $P(X_i-X_{i-1} = k)$ follows a geometric distribution with parameter $$p = \frac{9-(i-1)}{9} = \frac{9-i+1}{9} = \frac{10-i}{9}.$$
