Construct a random variable under given constraints In preparation for a probability examination, I am working on the following problem.
Problem
A box contains three white balls and ten black balls.
Balls are drawn without replacement until all the white balls are drawn.
Let $Y_0$ be the number of draws made.
Construct a random variable $Y: \Omega \to \mathbb{R}$ and a distribution $P$ over $\Omega$ such that $Y$ and $Y_0$ have the same distribution.
My queries


*

*Most importantly, what is a good thought process for approaching and solving this problem?

*Is it necessary to determine the distribution of $Y_0$ in order to solve this problem?  If so, how can we determine the distribution of $Y_0$?

*What is the purpose of constructing the distribution $P$, as required?

 A: Thought process:  If I want to construct a distribution $P$ such that $Y_0$ and $Y$ have the same distribution, then I will have to know the distribution $Y_0$. So yes, that is necessary.
And one might want that distribution $P$ because you would like to write a program that simulates how many draws were needed, but consumed only one uniform random variate on $(0,1)$.  Or maybe yo want that distribution because somebody will ask for the expected number of draws, or the expected square or $n$-th power of the number of draws.
The distribution will of course have a domain of $[3,13]$ since you can't finish without 3 draws and you muust finish by the time all 13 balls have been drawn. 
I will illustrate how to find $P(6)$:
In order for the 6-th draw to be the last white ball, the first 5 draws must consist of precisely two white and three black balls,  and then the next draw, out of 8 balls, must be the remaining white ball.  So
$$
P(6) = \frac{\binom{3}{2}_{\text{white}}\binom{10}{3}_{\text{bloack}} }{\binom{13}{5}} \times \frac18 = \frac{40}{143}\times \frac{1}{8} = \frac{5}{143}
$$
You can probably do the other numbers by following this example.
A: For $\Omega$ take the set of all subsets of set $\{1,2,\dots,13\}$ that have exactly $3$ elements. 
Note that $\binom{13}3$ is the cardinality of $\Omega$. 
As $\sigma$-algebra take $\wp(\Omega)$.
As probability measure take $P$ uniform, i.e. $P(\{\omega\})=\binom{13}3^{-1}$ for each $\omega\in\Omega$. 
Finally prescribe $Y:\Omega\rightarrow\mathbb R$ by $\omega\mapsto\max\omega$. 
Then $Y$ and $Y_0$ have the same distribution. 
Note that this construction does not ask you to determine the distribution of $Y_0$. The elements of $\Omega$ are sets like e.g. $\omega_0=\{2,4,7\}$. This example stands for the event that a white ball was drawn at the draws $2$, $4$ and $7$. Here $Y(\omega)=\max\omega=7$, equalizing the number of draws needed to pick out all white balls.
