Understanding that $P(0 \le X_i <t)=t\; |\,0 \le t <1$, then
- We have a continuous random variable $X$, that can assume values in $[0,1)$ with uniform probability;
- we make $n=20$ indipendent tests, and record the outcomes of them as the vector $(X_1, \, X_2, \, \cdots , \, X_n)$;
- the space of events $U$ is the $n$-dimensional hypercube of side $1$, with a uniform probability density;
- we want to know the probability of that portion of hypercube in which one component of the vector
is in $[0,t)$, another is in $[t,t+dt)$ and all the remaining in $[t+dt,1)$;
- since the result of a test run is a vector, order is to be taken into account (we are dealing with equi-probable n-tuples, not with equi-probable n-subsets);
- I can choose one event, the first, in $n$ ways, to occur in $[0..t)$ with $P_1=t$
- another event, the second, in $n-1$ ways, to occur at $[t..t+dt)$, with $P_2=dt$
- all the remaining, in one way, to occur at $[t+dt..1)$, with $P=(1-t)^{n-2}+O(dt)$
thus the probability density is $p_2(t)=n(n-1)t(1-t)^{n-2}$, and the average will be
$$
\eqalign{
& E(t_{\,2} ) = n\left( {n - 1} \right)\int_{t = 0}^1 {t^{\,2} \left( {1 - t} \right)^{\,n - 2} dt} =
n\left( {n - 1} \right){\rm B}(3,n - 1) = n\left( {n - 1} \right)\left( {{{2!\left( {n - 2} \right)!} \over {\left( {n + 1} \right)!}}} \right) = \cr
& = {{2!} \over {\left( {n + 1} \right)}} = {2 \over {21}} \cr}
$$
where ${\rm B}(x,y)$ is the Beta Function
Note that for $n=3$ we would get $ E(t_{\,2} ) =1/2$, which is to be expected for symmetry.
Addendum
Note that in case we were looking for the time of occurrence of the third event, then the probability density would be:
$$
p_{\,3} (t) = {{n\left( {n - 1} \right)\left( {n - 2} \right)} \over {2!}}t^{\,2} \left( {1 - t} \right)^{\,n - 3}
$$
The division by $2!$ is due to that $t^2$ is the probability of the first two events to occur in whichever order
while $n(n-1)$ counts the ways that they occur orderly
(and shall then be associated to a probability $\int_{\tau _{\,2} = 0}^{\,t} {\tau _{\,2} \left( {\int_{\tau _{\,1} = 0}^{\tau _{\,2} } {\tau _{\,1} d\tau _{\,1} } } \right)d\tau _{\,2} } $).
Thus in the general case (probability of $m$-th event occurring at $t$) we have:
$$ \bbox[lightyellow] {
p_{\,m} (t) = {{n^{\,\underline {\,m\,} } } \over {\left( {m - 1} \right)!}}t^{\,m - 1} \left( {1 - t} \right)^{\,n - m}
}$$
where $n^{\,\underline {\,m\,} } $ denotes the Falling Factorial.
In fact
$$
\eqalign{
& \int_{t = 0}^1 {p_{\,m} (t)dt} = {{n^{\,\underline {\,m\,} } } \over {\left( {m - 1} \right)!}}\int_{t = 0}^1 {t^{\,m - 1} \left( {1 - t} \right)^{\,n - m} dt} =
{{n^{\,\underline {\,m\,} } } \over {\left( {m - 1} \right)!}}\;{\rm B}(m,n - m + 1) = \cr
& = {{n^{\,\underline {\,m\,} } } \over {\left( {m - 1} \right)!}}{{\left( {m - 1} \right)!\left( {n - m} \right)!} \over {n!}} = 1 \cr}
$$
and
$$
\eqalign{
& \sum\limits_{1\, \le \,m\, \le \,n} {p_{\,m} (t)} = \sum\limits_{1\, \le \,m\, \le \,n} {{{n^{\,\underline {\,m\,} } } \over {\left( {m - 1} \right)!}}t^{\,m - 1} \left( {1 - t} \right)^{\,n - m} } = \cr
& = n\sum\limits_{1\, \le \,m\, \le \,n} {\left( \matrix{
n - 1 \cr
m - 1 \cr} \right)t^{\,m - 1} \left( {1 - t} \right)^{\,\left( {n - 1} \right) - \left( {m - 1} \right)} } = n \cr}
$$
i.e. $P (\text{any event in}[t,t+dt))=n dt/T$ as should be.
Note about your approach
Your calculation is correct, as far as it returns the probability that
two events (whatever) occur before $t$ and the remaining at or after $t$.
Therefore it includes the cases $[t_{1},t_{1}+dt_{1})<[t_{2}+dt_{2}) < [t,1)$
integrated over $t_{1},t_{2}$, so it is somewhat of a cumulative probability.
But to the purpose of calculating $E(t_2)$ this cannot be used, because of two bugs:
- the value $t$ is assigned also to the case $t_2<t$;
- the remaining events are taken to occur always after $t$, not after $t_2$ (so you cannot use the derivative of your probability).
