What is the fiducial argument and why has it not been accepted? One of the late contributions of R.A. Fisher was fiducial intervals and fiducial principled arguments. This approach however is nowhere near as popular as frequentist or Bayesian principled arguments. 
What is the fiducial argument and why has is not been accepted? 
 A: Fiducial inference sometimes interprets likelihoods as probabilities for the parameter $\theta$. That is, $M(x)L(\theta|x)$, provided that $M(x)$ is finite, is interpreted as a probability density function for $\theta$ in which $L(\theta|x)$ is the likelihood function of $\theta$ and $M(x)=(\int_{-\infty}^{\infty}L(\theta|x)d\theta)^{-1}$. You can see Casella and Berger, pages 291-2, for more details.
A: Just to add to what is said, there was controversy between Fisher and Neyman about significance testing and interval estimation.  Neyman defined confidence intervals while Fisher introduced fiducial intervals.  They argued differently about their construction but the constructed intervals were usually the same.  So the difference in the definitions was largely ignored until it was discovered that they differed when dealing with the Behrens-Fisher problem.  Fisher argued adamantly for the fiducial approach but inspite of his brillance and his strong advocation of the method, there appeared to be flaws and since the statistical community considers it discredited it is not commonly discussed or used.  The Bayesian and frequentist approaches to inference are the two that remain.
A: TL;DR
The fiducial argument has not been accepted because the idea doesn't work.
The fiducial distribution is disguised as something that looks like a probability distribution (and people might have wanted it to behave like a probability distribution) but it is not the same as a probability distribution. It is only a function of probabilities.
You can not do the same thing with a fiducial distribution as with, for instance, a posterior probability density. This is illustrated in the example below where we compute a 80% fiducial interval but get in some cases a 100% coverage.
Example where it doesn't work
In the nice answer by Michael R. Chernick it is mentioned that the logic behind the fiducial distribution started to fail when people tried to apply it in a multidimensional setting like for instance the Behrens Fisher problem. Here we give a one dimensional example that already shows that it does not work.
Let some statistic $X$ be distributed as a Uniform distribution with
$$X \sim \mathcal{U}\left(\theta-0.5\sqrt{1+\theta^2},\theta+0.5\sqrt{1+\theta^2}\right)$$
We can plot the cumulative distribution function (CDF) $F(x;\theta)$ as a function of $x$ and $\theta$ using isolines.

The image shows the CDF as function of $x$ and $\theta$

*

*In vertical direction, for fixed $\theta$, the function describes the CDF of the observation $x$ which is a random variable.
We can compute the probability density function as $$\frac{\partial}{\partial x} F(x;\theta)$$


*In horizontal direction, for fixed $x$, the function describes the fiducial distribution for the estimation of $\theta$
We can compute the fiducial density as $$\frac{\partial}{\partial\theta} F(x;\theta)$$
For example the fiducial density, when we observed $x=0.1$ looks like.

The points in red are inside the 80% interval [-0.396 , 0.475], it is the 80% probability mass with the highest density.
The problem is the following: the probability statements by the fiducial distribution only 'work' when we consider the same quantiles, independent from the observation $x$. However when change the quantiles as function of $x$ then the content of the probability statements entailed by the fiducial distribution are changed and become false. Therefore, the distribution can not be used in a similar way as a probability density. An example when this happens is for instance like the example above when we compute a highest density interval in combination in combination with a fiducial density that doesn't have the same shape for different observations $x$ (which makes us select different quantiles).
We can see this in the plot when $\theta = 0$ for that case we get a 100% coverage by the 80% highest density interval instead of a 80% coverage. This is not what you would expect when the fiducial distribution could be used as a probability density for $\theta$.
R-code for the two plots:
#### parameters for drawing
d = 0.01
t = seq(-2,2.6,d)
t2 = seq(3.4,4,d)
tm = 3
grey = rgb(0.3,0.3,0.3)

### empty canvas
plot(-10,-10, xlim = c(-2,4), ylim = c(-3,5),
     xlab = expression(theta), ylab = "x", main = "example of highest density 80% feducial interval")

### add isolines 
for (q in c(0:10)*0.1) {
   lines(t,t+(q-0.5)*sqrt(1+t^2), col = grey)
   lines(t2,t2+(q-0.5)*sqrt(1+t2^2), col = grey)
   text(tm,tm+(q-0.5)*sqrt(1+tm^2),
        bquote(F(x*";"*theta)==.(q)), col = grey, cex = 0.6, srt = 15+35*q)
}

fiducial = function(x, plotting = TRUE, alpha = 0.8) {
  dt = 0.001
  ### domain of fiducial distribution 
  tmin = (1/3) * (4*x - sqrt(4*x^2 + 3))
  tmax = (1/3) * (4*x + sqrt(4*x^2 + 3))
  ts = seq(tmin, tmax, dt)

  ### compute distribution
  f = (ts*x + 1)/(ts^2+1)^1.5

  ### calculate highest density region by ordering densities
  ord = order(f)       #order
  p = cumsum(f[ord])*dt   #cumulative probability 
  sel = which(p<1-alpha) # select complement of highest alpha% density
  output = range(ts[ord][-sel]) ## range of the highest alpha% density interval

  ### example plot of density
  if(plotting == TRUE) {
     plot(ts,f, col = 1 + (ts >= output[1]) * (ts <= output[2]))
  }
  output
}

### compute intervals as function of observed x

xs = seq(-4,5,0.01)
low = c()    ### empty array that will be filled
high = c()
for (x in xs) {
   interval = fiducial(x, plotting = FALSE)
   low = c(low, interval[1])
   high = c(high, interval[2])
}

### add curves for fiducial interval
lines(low[low<2.6],xs[low<2.6], lwd =2)
lines(high[high<2.6],xs[high<2.6], lwd =2)

### add example interval for observation X = 0.1
int = fiducial(0.1, plotting = 0)
#lines(c(-2,2),c(0.1,0.1), lty = 2)
lines(int,c(0.1,0.1), lwd =2, col = 2)
points(int, c(0.1,0.1), pch = 21, col = 2, bg = 0, cex = 0.7)
text(0.8,0.1,"example interval if x=0.1", pos = 4, col = 2)

### example plot
fiducial(0.1)
title("example fiducial distribution if x = 0.1 \n highest 80% density is highlighted in red")

A: In a large undergraduate class of engineering intro stats
at Georgia Tech,
when discussing confidence intervals for the population mean
with variance known, one student asked me (in the language of MATLAB):
"Can I calculate the interval as > norminv([alpha/2,1-alpha/2], barX, sigma/sqrt(n))?"
In translation: could he take $\frac{\alpha}{2}$ and $1-\frac{\alpha}{2}$ quantiles of
a normal distribution centered at $\bar X$ with scale $\frac\sigma{\sqrt{n}}$?
I said – of course YES, pleasantly surprised that he naturally 
arrived to the concept fiducial distribution.
