I know this is a tired question, but I have seen no clear answers.

I saw a related post but didn't see how it applied to the general case. In a 3b1b video, it is alluded to that given some known number of trials and successes, the probability of the probability in a binomial distribution can be calculated (or perhaps inferred, guessed at?). Using Baye's Theorem I find difficulty; the theorem relates two probabilities of two variables, but there are three variables here: the probability of a success (which we want to find), the number of trials and the number of successes.

E.g. if a car factory makes $100$ cars and finds $2$ defects, what is the most likely probability of a defect? How do we find the expected value of this thing?

By hand one could go through many test probabilities and see which has $2$ defects as its most likely outcome but that feels wrong - and would only work in the expected value case, anyway. The general probability of a probability is a mystery to me.

Many thanks to anyone who can even tell me what this sort of process is even called.

  • 1
    $\begingroup$ maybe you're thinking to something like maximum likelihood estimation? $\endgroup$
    – Tortar
    Commented Jul 19, 2021 at 20:03
  • 1
    $\begingroup$ Not sure what you are after. For the example you gave, you should easily be able to confirm (numerically) that $p=.02$ maximizes the probability that you observe $2$ instances out of $100$. What else? Beyond that, unless you have some a priori distribution on the possible probabilities, I don't think you can get more detail. $\endgroup$
    – lulu
    Commented Jul 19, 2021 at 20:19

2 Answers 2


Let $p$ be the probability of success in one trial; this quantity is unknown to you. Given the outcome of $n$ trials ($n$ is non-random and fixed), you are interested in understanding

what does this information tells me about $p$?

In statistics there are broadly two frameworks for answering this question.

Frequentist statistics

In frequentist statistics, $p$ is a fixed, non-random, unknown quantity. Note that because this is a non-random quantity, it does not make sense to ask things like "what is the probability that $p$ is larger than $0.6$?"

One ubiquitous approach for answering the above question is maximum likelihood. This is almost what you described here, but not quite.

By hand one could go through many test probabilities and see which has 2 defects as its most likely outcome.

In your binomial scenario, if $X$ is the number of successes you observed in $n$ trials, and you observed $X=x$, then (as you described) you can compute $L(p) := \binom{n}{x} p^x(1-p)^{n-x}$ for many values of $p$ and find the $p$ that makes this quantity large. The function $L$ is called the likelihood function, and this procedure of finding the $p$ that makes $L$ the largest is called maximum likelihood estimation. It is important to note that $L$ is a function of $p$ (with $x$ fixed at whatever number of successes you observed), rather than a function of $x$ with $p$ fixed (this would be the PMF of the binomial distribution).

What you proposed is to find $p$ such that $2$ is the mode of the $\text{Binomial}(n, p)$ distribution (i.e. $P(X=2) \ge P(X=k)$ for all $k \ne 2$). This is not quite the same as the above.

Bayesian statistics

In Bayesian statistics, $p$ is a random quantity with its own distribution, called a prior distribution. In your binomial example, $p$ could follow a $\text{Uniform}[0, 1]$ distribution. It could also follow some other distribution, depending on your prior belief in what $p$ could be. The prior distribution must be chosen by you, but as you get more and more data, the influence of this initial choice becomes less and less as you rely more and more on the data.

Since $p$ is random in this framework, it now makes sense to ask questions like "what is the probability that $p$ is larger than $0.6$?"

Now that $p$ is random, the model is now that $X$ conditioned on $p$ is binomial, e.g. $P(X=x \mid p=0.2) = \binom{n}{x} (0.2)^x (0.8)^{n-x}$.

To understand what your data $X=x$ says about the unknown $p$, you can consider the posterior distribution of $p$, that is, the conditional distribution of $p$ given $X=x$. This can be computed using Bayes's Rule. Here is an example of the posterior probability that $p=0.4$ in the case where $p$ follows a discrete distribution.

$$f(0.4) = P(p=0.4 \mid X=x) = \frac{P(X=x \mid p=0.4) P(p=0.4)}{\sum_{p'} P(X=x \mid p=p') P(p=p')}.$$

One Bayesian approach to answer the question is Maximum A Posteriori estimation, which chooses $p$ to maximize the above function $f$.

  • $\begingroup$ What is $X$ in the Bayesian statistics section? Thank you for your effort! $\endgroup$
    – FShrike
    Commented Jul 19, 2021 at 20:35
  • $\begingroup$ @IdioticShrike $X$ is still the number of successes in the $n$ trials. $\endgroup$
    – angryavian
    Commented Jul 19, 2021 at 20:45

It depends on how you want to model the problem. If you assume the data are binomial, then you have a likelihood function $f(2|p)={100\choose 2}p^2(1-p)^{100-2}$ and you can maximize it (maximum likelihood estimation). If you assume another distribution such as Poisson, you get $f(2|\lambda)=\frac{e^{-\lambda}\lambda^2}{2!}$ and you can again find the global maximum in terms of $\lambda$.

There is another way to do these involving prior and posterior distributions. This involves assigning a prior density $g$ to $p$ (e.g. density of a Beta distribution) and a distribution to the data conditional on $p$ (e.g. Binomial). Then you can find the posterior distribution of $p$ by the well-known formula $g(p|2)\propto g(p)f(2|p)$. After getting $g(p|2)$ then you can find the Bayes Estimator (which is the mean of the distribution) or the median of the distribution, as well as calculate probabilities that $p$ lies in certain intervals.

If the sample size is very large it won't matter much what prior you assign to $p$ as the posterior will be much like the likelihood function of the data multiplied by a constant needed to turn it into a pdf. Furthermore the posterior distribution as $n$ is very large will be approximately the normal distribution with mean $\hat p$ and variance $\frac{\text{Var}(\hat p)}n$ where $\hat p$ is the MLE.

The MLE also has its own distribution (note the difference between maximum likelihood estimator and maximum likelihood estimate, where the latter is with the observed data plugged in). As $n$ is very large, it will be about normal with mean $p$ and variance $c/n$ for some $c$. The distribution of the MLE, posterior distribution, and the normal estimate of the posterior will be about the same as $n\rightarrow \infty$.


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