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?
 A: 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.
A: No, even the marginal distributions are not the same. Consider for simplicity $n=2$. Method 1 generates values $a_1$, $1-a_1$ both Uniformly distributed in $(0,\,1)$.
For Method 2, let us find the distribution of $\frac{X}{X+Y}$, where $X$ and $Y$ are independent r.v.'s Uniformly distributed in $(0,\,1)$. It makes no sense to take the uniform distribution on the interval $(0,\,n)$ as the length of the segment here is the scaling parameter only, and the ratio does not depend on it.
$$
P\left(\dfrac{X}{X+Y}<t\right) = P\left(X<\frac{t}{1-t}Y\right).
$$
One can calculate this probability separately for $0<t\leq 1/2$ and for $1/2<t<1$ in unit square. The answer is
$$
P\left(\dfrac{X}{X+Y}<t\right)=\begin{cases}\dfrac{t}{2(1-t)}, & 0<t\leq 1/2\cr
1-\dfrac{1-t}{2t}, & 1/2<t<1\end{cases}
$$
For the second variable, $\dfrac{Y}{X+Y}$, marginal CDF is the same. 
These functions are very different from Uniform CDF's in Method 1. 
For Method 1 one can write joint distribution. In 2nd volume of W.Feller An Introduction to Probability Theory and Its Applications one can find result of B.De Finetti, 1964. See exercise 23 to 1st Chapter. 
For $x_1\geq 0$, ..., $x_n\geq 0$
$$
P(a_1>x_1, a_2-a_1>x_2, \ldots, 1-a_{n-1}>x_n)=(1-x_1-\ldots-x_n)^{n-1}_+
$$
where $(x)_+=\max(x,0)$. Here $a_1\leq a_2\leq\ldots \leq a_{n-1}$ are order statistics for independent Uniform random numbers.
I do not know how joint distribution looks like for Method 2. Doubt that it looks simple.
If we take for Method 2 independent r.v. $X_1,\ldots,X_n$ from the same Exponential distribution, we obtain the same joint distribution of $X_i/(X_1+\ldots+X_n)$, $i=1,\ldots,n$ as in Method 1. 
This result is discussed in the 2nd Vol. of W.Feller's book, in paragraph 3 Chapter III.
