What are the values of the ​average length? A segment of length $1$ is broken into three pieces. We consider that the breakpoints are two random variables $X, Y$ independent and uniformly distributed $U([0, 1])$. 
We want to approximate average lengths of the three segments obtained. For this, simulate $N = 5000$ independent achievements of the couple $(X, Y)$. What are the values of ​​average length? I'm wondering if I can use SLLN. But as I see there many cases, for example if $X<Y$. Is there another solution?
 A: Here is a simulation of a million iterations of your experiment in R
statistical software. First, I simulated $U_1$ and $U_2,$ independently
$Unif(0,1),$ for the two 'cuts' in $(0,1).$ Then I found $V = \min(U_1,U_2)$
for each iteration, and then $W = \max(U_1, U_2).$ Finally, for the
lengths of left, middle and right segments are $L = V,\,M = W - V,$ and
$R = 1 - W,$ respectively. (I hope this progression of computations, including
computation of the minimum and maximum,
clears up some of the ambiguity in your question.)
The most basic findings, all of which can be confirmed analytically, are
as follows:
The means, SDs and histograms of each segment $L, M$ and $R$ are consistent with
the distribution $Beta(1, 2),$ which has density $f(x) = 2(1 - x),$
for $0 < x < 1$ (see Wilipedia on 'beta distribution'). Also, the three pairwise correlations are all $-0.5;$
that is, $Cor(L,M) = Cor(L,R) = Cor(M,R) = -0.5.$
With a million iterations, these results should be accurate to two or
three decimal places. [Note: Neither the random variables $V$ and $W$ nor
the random variables $L, M,$ and $R$ are to be confused with order statistics when three independent observations are
taken from $Unif(0,1).$]
m = 10^6;  u1 = runif(m);  u2 = runif(m)
w = pmax(u1,u2);  v = pmin(u1,u2)
L = v;  M = w - v;  R = 1 - w
mean(L);  mean(M);  mean(R)
## 0.3334269      # aprx E(L) = 1/3
## 0.3337332 
## 0.3328399
sd(L); sd(M); sd(R)
## 0.2358886      # aprx SD(L) = 0.2357023
## 0.2360321
## 0.2358404
cor(L,M); cor(L,R); cor(M,R)
## -0.5005081
## -0.4992892
## -0.5002024
mean(u1 < u2)
## 0.499321       # aprx P(U1 < U2) = 1/2

The same simulation, but with $m = 100,000$ iterations gives the following selected
graphs. (A scatterplot with a million points can be too crowded with points for
clarity.)
 par(mfrow=c(1,2))  # two panels per plot
   hist(M, prob=T, col="skyblue2", main="Sim. Dist'n of M with Exact PDF")
     curve(dbeta(x, 1, 2), col="blue", lwd=2, add=T)
   plot(L, R, pch=".", main="Sim. Joint Dist'n of L and R")
 par(mfrow=c(1,1))  # return to single-panel plots


Finally, you mentioned the LLN. As the number of iterations increases,
the accuracy of findings such as those above tends to improve. The following
plot shows a trace of the convergence of $\bar L_n$ (the sample mean of $n$
left segments) to $1/3$ as
the number $n$ of iterations increases. [The vertical window is restricted
to the interval $(.30, .35)$ in order to show some detail as the trace
approaches $1/3$.]
n = 1:m;  L.bar = cumsum(L)/n
plot(n, L.bar, type="l", ylim=c(.30,.35), lwd=2)
  abline(h=1/3, col="green2")


Addendum: Per Comment of @JanDvorak, here is a plot with traces
of three additional (independently simulated) traces of $\bar L_n.$
Notice that traces can be quite different for small $n,$ but are
much the same for large $n$ as they approach $1/3.$ (R code omitted:
traces beyond the initial one are plotted on the same axes with lines
function.)

