Distribution of two random variable generating schemes Goal: Generate five numbers from 0 to 1 with sum 1.
Method 1: Generate four numbers in range $(0,1)$ (by uniform distribution) to be the cuts of the interval, i.e. say the four random numbers generated is $a_1<a_2<a_3<a_4$, then the five random numbers are $a_1, a_2-a_1, a_3-a_2, a_4-a_3,1-a_4$
Method 2: Generate five numbers in range $(0, n)$ (by uniform distribution), where $n$ is arbitrary number bigger than zero. And then normalise the sum to be 1, i.e. divide all the numbers by their sum.
My question is, are the distribution of the two methods equivalent? Is any of the two correspondents to some well-known distributions?

Let's play this game in a simpler case. Let $X_1$ and $X_2$ be two independent and $U(0,1)$ distributed random variables.
FOR METHOD 1
Generate random variables $A_1$ and $A_2$ such that $A_1+A_2=1$ the following way:
$$A_1=\begin{cases}X_1&\text{ if }& X_1<X_2\\
X_2&\text{ if }& X_2\le X_1\end{cases} \text{ and let  }\ A_2=1-A_1$$
Let's see the distribution of $A_1$. That is calculate the following probability
$$P(A_1<x)=P(A_1=X_1\cap X_1<x)+P(A_1=X_2\cap X_2<x)=$$
$$=P(X_1<X_2\cap X_1<x)+P(X_2\le X_1\cap X_2<x).$$
So,
$$P(X_1<X_2\cap X_1<x)=\int_0^xP(u<X_2)\ du=\int_0^x1-u\ du=x-\frac12x^2$$
Because of symmetry reasons
$$P(X_2\ge X_1\cap X_2<x)=x-\frac12x^2.$$
Hence, the cdf and the pdf of $A_1$ are
$$2x-x^2\ \text{ and } 2-2x$$
if $0\le x\le1$.
Which is the pdf of the first element of the order statistic of two independent uniformly distributed random variables.
If we have five such variables then the pdf of $A_1$ can be calculated as well. It will not be of uniform distribution either.
