Statistical twosample testing concerns the following question: given observations $X = \lbrace x_1, x_2, \dots, x_m \rbrace$ and $Y = \lbrace y_1, y_2, \dots, y_n \rbrace$ drawn from distributions $p$ and $q$, respectively, can we decide whether $p=q$?
Twosample testing is an active field of research with an enormous amount of literature. Asymptotic theory of twosample tests is typically concerned with establishing consistency, meaning that the probability of declaring $p=q$, when in fact $p \ne q$, goes to zero in the large sample limit.
Distinguishing distributions with finite number of samples
A very nice (negative) result is that it is generally impossible to distinguish two distributions with high probability, by only observing a finite pair of samples of fixed size. The result is also somewhat disappointing for those who are particularly interested in theoretical guarantees of statistical procedures in applied inference, where the ideal of asymptotia is a far reach from the chaotic array of csv files, and other inhomogeneous data sources and formats, sitting on the analyst’s hard drive. Anyway, consider the following scenario, which is from Gretton et al. “A Kernel TwoSample Test.” JMLR. 2012. (a highly recommended paper to read):
Assume we have a distribution $p$ from which have drawn $m$ iid observations. Construct a distribution $q$ by drawing $m^2$ iid observations from $p$ and define a discrete distribution over these $m^2$ observations with probability $m^{2}$ each. It is easy to check that if we now draw $m$ observations from $q$, there is at least a ${m^2 \choose m}\frac{m!}{m^{2m}} > 1 – e^{1} > 0.63$ probability that we thereby obtain an $m$ sample from $p$. Hence, no test will be able to distinguish samples from $p$ and $q$ in this case. The probability of detecting can be made arbitrarily small by increasing the size $m$ of the sample from which we construct $q$.
To understand why this setup implies the impossibility of twosample testing with high probability in the finite setting, we need to think about the frequentist properties of the procedure. Suppose we repeat the experiment a very large number of times, say 900,000 times. Marginalizing over all experiments, the samples drawn from $q$ will follow a different distribution than samples drawn from $p$. However, in roughly 2/3 or 600,000 of the experiments, the observation set drawn from $q$ is a random sample from $p$. Conditioned on being in one of the 600,000 trials where the $m$ samples are from the same distribution, an ideal twosample test should report $p=q$, which is committing a Type II error with arbitrarily large probability.
Now, the authors left the derivation of the combinatorial bound ${m^2 \choose m}\frac{m!}{m^{2m}}$, for the probability that the $m$ draws from $q$ are an $m$ sample from $p$, as an exercise for the reader. It actually makes a nice homework or exam question for a 101 probability course (as do many other byproducts of academic papers, so I have learned). For those interested in elementary probability and combinatorics, here is a candidate derivation.
First note ambiguities in the setup of the experiment: (i) Is $p$ intended to be a continuous distribution, so that there are no duplicates in the $m^2$ samples? (ii) What exactly is the definition of the event “thereby obtain an $m$ sample from $p$”? To make progress, we assume that (i) $p$ is continuous and the $m^2$ samples in $q$ are unique, and (ii) an “$m$ sample from $p$” means that the resampled sequence contains no duplicates (which occurs with probability zero when sampling directly from $p$). With these assumptions, the bound becomes trivial. The size of the sample space is equal to the number of length $m$ strings from $\lbrace x_1, \dots, x_{m^2} \rbrace$, which is $(m^2)^m$. The number of sequences of length $m$ with unique entries is ${m^2 \choose m}m!$. Dividing the latter quantity by the former gives the bound.
If the assumptions above are too strong and $p$ is discrete, then I am no longer sure what the definition of an “$m$ sample from $p$” is. Perhaps someone can let me know if alternative meanings are clear to them. Discreteness of $p$ would indeed provide a much more compelling example, since one can arrive at arbitrary results based on pathologies of real number being uncomputable.
A simpler counterexample with infinite number of samples
To the computer scientist, real numbers are uninteresting. Let us suppose $p$ is a discrete distribution with finite support, and additionally restrict its probability density at each point in the support to be a rational number. One might simplify the above experiment by defining $q_m$ to:

generate a size $m$ sample from $p$, with probability $11/m$;

generate a size $m$ sample from an arbitrary distribution, with probability $1/m$.
Here is an interesting question — is there any test which can have nonzero power for testing $p$ versus $q_m$? Note that it is always the case that $p \ne q_m$. The probability of detection, also known as the power $\beta$ of the statistical test, is $1/m$, which decays to zero as $m \to \infty$. Therefore, the Type II error $(1\beta)$ necessarily goes to one, irrespective of the testing procedure. Stated otherwise, for any testing procedure $F$ and tolerance $\epsilon > 0$, there exists a pair of distributions $(p,q_m)$ such that $\Pr\left[F\; \textrm{declares}\; p \ne q_m \right] = \beta < \epsilon$, where the probability is taken under the sampling distribution of datasets $D_p \sim p, D_q \sim q_m$.
Also note that $q_m$ converges in distribution to $p$ as $m \to \infty$. This construction is rather different than standard Type II error analysis, since our distribution $q_m$ is itself a function of the sample size $m$, so we cannot formally take the limit as $m \to \infty$ to compute the Type II error. The counterexample above should instead be seen as a “meta counterexample”, showing a single adversarial pair of distributions $(p,q_m)$ and finite sample size $m$ which together will trick an arbitrary testing procedure $F$ into obtaining arbitrarily low power.