We give a decision-theoretic justification of a heuristic principle suggested by Robbins. Suppose a sequence $X_1, X_2, \cdots$ of independent random variables can be observed; they are known to be identically distributed with unknown mean and variance $u, \sigma^2(u)$ respectively. We assume further that the common distribution is of exponential type, so that the sample mean $\bar{X}$ is a sufficient statistic. The problem is to estimate $u$ taking into account the cost of sampling, which is assumed to be linear in the sample size, $n$. For the case of a fixed-size sample, with squared-error loss function, the minimum risk (expected loss) is obtained, after rescaling if necessary, as $\min_nE(n + A(\bar{X} - u)^2) = \min_n(n + A\sigma^2(u)/n) = 2A^{\frac{1}{2}}\sigma(u)$, and is attained at $n_0 = A^{\frac{1}{2}}\sigma(u)$. Since $u$ is unknown, this optimum value is not available; and for any fixed $n$, if $\sigma(u)$ is unbounded, the risk $n + A\sigma^2(u)/n$ can be much larger than $2A^{\frac{1}{2}}\sigma(u)$. For calibration of the performance of sequential stopping rules, Robbins advocated consideration of the "regret" $E(n + A(\bar{X} - u)^2) - A^{\frac{1}{2}}\sigma(u),$ and several authors have constructed procedures with uniformly bounded regret. Two questions arise, which we settle here (in a certain asymptotic sense, and after making certain smoothness assumptions). First, can any procedure have strictly negative regret, for all $u$? Second, if a procedure has uniformly bounded regret, is it necessarily close to being optimum, in the sense that for each (suitably smooth) sequence of prior distributions on $u$, is it only boundedly worse than the corresponding sequence of Bayes procedures? Our answers are no, and yes, respectively. Several examples are discussed, and analogies pointed out with the fixed-sample-size concepts of the asymptotic optimality of maximum-likelihood estimates, and the super-efficiency phenomenon.