Probability the Nth realization in a sample of N non-identically distributed reals is the maximum of the sample? Say I have a set of $j$ variates, each distributed uniformly on the intervals $[0,d]$ where $0<d_1<d_2<...<d_j$.
I'm interested in the probability that some $d_k$ of the $j$ variates is the maximum for a sample.
I've got results for small $j$ by just integrating the CDF of the maximum of the variates other than the $k$ I'm interested in, multiplied by the PDF of the $k$ I'm interested in, over $[0,\infty]$.
This works well in my CAS up to $j\sim6$, but it gets very slow for greater $j$.
The results up to $j=5$ look like the following:

Here the rows are by value of $j$, and columns are the probability that $d_k$ is the maximum.
I don't yet see a consistent pattern, but I'm convinced there must be a more direct way to generate these end results.
Is there?
 A: As has been mentioned in the comments, the question is a bit confusing because you don’t distinguish clearly between the $d_i$ and the random variables (e.g. where you say “$d_k$ is the maximum” you apparently mean that the corresponding uniform random variable is the maximum). Let’s call the variables $X_i$.
Let $L_k$ denote the event that $X_{k+1}$ to $X_j$ (and hence $X_k$ to $X_j$) lie in $[0,d_k]$ (‘L’ for “low”), $R_k$ the event that $X_k$ is the greatest among $X_k$ through $X_j$ (‘R’ for “record”), and $M_k$ the event that $X_k$ is the greatest among all the $X_i$ (‘M’ for “maximum”).
The key facts to use here are that conditional on $L_k$, the variables $X_k$ to $X_j$ are identically distributed and thus all their orders are equiprobable, and thus the events $R_i$ with $i\ge k$ are conditionally independent conditional on $L_k$. In your expressions for the probabilities, the ratios of the $d_i$ are the probabilities of certain $L_k$, and the numerical factors are the conditional probabilities of certain combinations of the $R_k$.
Let’s go through the results for $j=5$.  We have
\begin{eqnarray}
\mathsf P(M_1)
&=&
\mathsf P(R_1)
\\
&=&
\mathsf P(R_1\mid L_1)\cdot\mathsf P(L_1)
\\
&=&\frac15\cdot\frac{d_1}{d_2}\cdot\frac{d_1}{d_3}\cdot\frac{d_1}{d_4}\cdot\frac{d_1}{d_5}\;,
\end{eqnarray}
since, given $L_1$, by symmetry $X_1$ is a record with probability $\frac15$.
Next,
\begin{eqnarray}
\mathsf P(M_2)
&=&
\mathsf P(R_2)-\mathsf P(R_2\cap R_1)
\\
&=&
\mathsf P(R_2\mid L_2)\cdot P(L_2)-P(R_2\cap R_1\mid L_1)\cdot P(L_1)
\\
&=&
\frac14\cdot\frac{d_2}{d_3}\cdot\frac{d_2}{d_4}\cdot\frac{d_2}{d_5}-\frac14\cdot\frac15\cdot\frac{d_1}{d_2}\cdot\frac{d_1}{d_3}\cdot\frac{d_1}{d_4}\cdot\frac{d_1}{d_5}\;,
\end{eqnarray}
since, given $L_2$, by symmetry $X_2$ is a record with probability $\frac14$.
Now it gets a bit more complicated because we have to subtract multiple terms, which need to represent mutually exclusive events, but the basic idea is the same:
\begin{eqnarray}
\mathsf P(M_3)
&=&
\mathsf P(R_3)-\mathsf P(R_3\cap R_2)-\mathsf P(R_3\cap\overline{R_2}\cap R_1)
\\
&=&
\mathsf P(R_3\mid L_3)\cdot\mathsf P(L_3)-\mathsf P(R_3\cap R_2\mid L_2)\cdot P(L_2)-\mathsf P(R_3\cap\overline{R_2}\cap R_1\mid L_1)\cdot P(L_1)
\\
&=&
\frac13\cdot\frac{d_3}{d_4}\cdot\frac{d_3}{d_5}-\frac13\cdot\frac14\cdot\frac{d_2}{d_3}\cdot\frac{d_2}{d_4}\cdot\frac{d_2}{d_5}-\frac13\cdot\frac34\cdot\frac15\cdot\frac{d_1}{d_2}\cdot\frac{d_1}{d_3}\cdot\frac{d_1}{d_4}\cdot\frac{d_1}{d_5}\;,
\end{eqnarray}
since, given $L_3$, by symmetry $X_3$ is a record with probability $\frac13$.
Perhaps now you can see the pattern. The general expression is
\begin{eqnarray}
\mathsf P(M_k)
&=&
\mathsf P(R_k)-\sum_{i\lt k}\mathsf P\left(R_k\cap\bigcap_{i\lt m\lt k}\overline{R_m}\cap R_i\right)
\\
&=&
P(R_k\mid L_k)\cdot\mathsf P(L_k)-\sum_{i\lt k}\mathsf P\left(R_k\cap\bigcap_{i\lt m\lt k}\overline{R_m}\cap R_i\mid L_i\right)\cdot\mathsf P(L_i)
\\
&=&
\frac1{j-k+1}\prod_{\ell\gt k}\frac{d_k}{d_\ell}-\sum_{i\lt k}\frac1{j-i}\cdot\frac1{j-i+1}\prod_{\ell\gt i}\frac{d_i}{d_\ell}\;.
\end{eqnarray}
