Say you have two urns with a large number of red and blue marbles each and you know the proportion of red and blue marbles in each urn. Now we choose one urn at random (but don't know which) and repeatedly draw a marble in hopes of deciding which urn we have selected.

Bayes' Theorem is a natural fit of course, but I wonder, is there a frequentist approach to deciding which urn was selected?


1 Answer 1


Bayes' Theorem is just a simple identity in probability theory for two random variables that is independent of what your fundamental interpretation of statistics would be. So both a frequentist and a Bayesian would define two random variables, $X$ as the binary random variable that describes which urn is selected and another random variable $Y$ that describes the sampled marbles, and then would try to obtain $p(X|Y)$, maybe via Bayes' Theorem (if we are also given the prior $p(X)$).

The difference between frequentists and Bayesians is not that the frequentists don't use Bayes' theorem, but that (roughly) Bayesians consider parameters as random variables which frequentists do not.

But in this example, both frequentists and Bayesians would be happy to consider both $X$ and $Y$ as proper random variables.

  • $\begingroup$ Thanks for that, that makes sense. I guess a better question then is there a way to solve the problem without using Bayes' Theorem? $\endgroup$ Commented Sep 22, 2022 at 21:55
  • $\begingroup$ If the prior is given, and there are not any other constraints or special requirements, then nothing will beat Bayes' Theorem. $\endgroup$
    – frank
    Commented Sep 23, 2022 at 2:39

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