To explain the fiducial argument we first set up a simple example.  Say that we estimate the mean \(\theta\) of a normal distribution with unit variance over a variable \(X\). We collect a sample \(s\) consisting of measurements \(X_{1}, X_{2}, \ldots X_{n}\). The maximum likelihood estimator for \(\theta\) is the average value of the \(X_{i}\), that is, \(\hat{\theta}(s) = \sum_{i} X_{i} / n\).  Under an assumed true value \(\theta\) we then have a normal distribution for the estimator \(\hat{\theta}(s)\), centred on the true value and with a variance \(1 / \sqrt{n}\). Notably, this distribution has the same shape for all values of \(\theta\). Because of this, argued Fisher, we can use the distribution over the estimator \(\hat{\theta}(s)\) as a stand-in for the distribution over the true value \(\theta\). We thus derive a probability distribution \(P(\theta)\) on the basis of a sample \(s\), seemingly without assuming a prior probability.