What's the probability that the next test will exceed the previous maximum? I take a sample of 10 from a population that I suspect is non-normal (but it is continuous). I need to calculate the probability that the next sample will be larger than the maximum of the previous 10 points (given the previous 10 observations).
I'm an engineer with some extra statistical knowledge. I was taught from a frequentist perspective. Am I correct in thinking I can solve this problem with a Bayesian credible interval? 
Added: I did consider a nonparametric prediction interval, but since that is a frequentist's perspective it doesn't give a probability of the next sample, but that the probability applies to the procedure used to calculate the interval (see this explanation of prediction intervals paragraph 21).
 A: Given a sample $X_1, \dotsc, X_n$ from the same population (distribution) (not even independence is necessary, it is enough with symmetry (or exchangeability), that all permutations have the same distribution).  Then the probability that the next sampled value $X_{n+1}$ from that same distribution will be the maximum is $\frac1{n+1}$.   You can argument directly from symmetry: $X_{n+1} > \max_{i=1,\dotsc, n} X_i$ has a probability that is unchanged by permuting the indices, so $X_3 > \max_{i\not = 3} X_i$ and so on, so the commom probability must be $\frac1{n+1}$.
This leads to a theory of records: Let $X_1, X_2, X_3, \dotsc $ be a sequence of independent (or exchangeable ...) random variables, in this case, we must have exchangeability for the finite subsequences $X_1, \dotsc, X_n$ for all $n$. Let
$$ 
   T_n=\begin{cases} 1 ~\text{if $X_n$ is the record observation (max) up to time $n$,} \\
 0 ~\text{otherwise} \end{cases}
$$
Then $P(T_n=1)=\frac{1}{n}$, and so the expected number of records up to time $n$ is
$\sum_{i=1}^n \frac1{i}$ which is diverging in $n$, so there will never be a last record, for instance. That sum is (for moderate $n$) close to $\ln n$, so the expected number of records in the first thousand trials is approximately 6.9 (exact 7.49).
