Suppose 100 balls where it is known that 1 is red and 99 are blue. Let y=1 if I select the red ball and 0 otherwise. All of this implies the marginal probability Pr(y=1) = 1/100.

Now I notice some balls are shaped differently. With certainty I can then identify 20 balls that are not red. Given this info (call it x), the conditional probability that I choose the red ball would be Pr(y=1|x) = 1/80.

What is confusing to me is how a Bayesian would use this information to obtain a posterior. In this case, x is presumably a likelihood (or at least info for updating my beliefs), 1/100 is the marginal probability...but what is then the prior? Is the prior simply the marginal? How would I represent x as a likelihood?

More generally: I see how we can use Bayes' rule to update our beliefs if the evidence we observe is treated as a draw from the unknown posterior distribution. However,if the information we receive is a KNOWN quantity, how do we use this in Bayes' rule?

  • 2
    $\begingroup$ A Bayesian starts with a parametrised distribution for a observation (for example the number of red balls selected in a single selection with the parameter being the probability of selecting a red ball) and a prior distribution for that parameter (perhaps a conjugate beta distribution), and then updates the prior distribution to a posterior distribution for the parameter based on the likelihood function for an observed observation. That is not how you have structured your question - instead you have given a point prior (breaking Cromwell's rule) and then changed it to a different point prior. $\endgroup$
    – Henry
    Oct 22, 2022 at 21:03

1 Answer 1


In order to apply Bayesian analysis (or indeed, any other method of inference) you have to have something unknown in the problem that you want to infer from your observed data. Your question is not very clear about this part, but I'm going to consider the case where it is unknown whether or not the shape-information $X$ was given and you observed the outcome of $Y$. (If this is not unknown then your problem does not appear to have any unknown parameter, in which case statistical inference is not needed at all.) This would encapsulate the situation of a third-party who does not know if you had the shape-information or not, but observes whether or not you select a red ball.

In this case $X$ is your unknown parameter and the probability of selecting a red ball is:

$$p(Y=1 | X=x) = \begin{cases} \frac{1}{100} & & \text{if } x = 0, \\[12pt] \frac{1}{80} & & \text{if } x = 1. \\ \end{cases}$$

Under a Bayesian analysis we would set some prior probability that the shape-information was present, which we can denote as $\pi \equiv \mathbb{P}(X=1)$. Applying Bayes' rule then gives the posterior probabilities:

$$\begin{align} \mathbb{P}(X=1|Y=0) &= \frac{\mathbb{P}(Y=0|X=1) \cdot \pi}{\mathbb{P}(Y=0|X=1) \cdot \pi + \mathbb{P}(Y=0|X=0) \cdot (1-\pi)} \\[6pt] &= \frac{\tfrac{79}{80} \cdot \pi}{\tfrac{79}{80} \cdot \pi + \tfrac{99}{100} \cdot (1-\pi)} \\[6pt] &= \frac{395 \cdot \pi}{395 \cdot \pi + 396 \cdot (1-\pi)} \\[10pt] &= \frac{395 \cdot \pi}{396 -\pi}, \\[12pt] \mathbb{P}(X=1|Y=1) &= \frac{\mathbb{P}(Y=1|X=1) \cdot \pi}{\mathbb{P}(Y=1|X=1) \cdot \pi + \mathbb{P}(Y=1|X=0) \cdot (1-\pi)} \\[6pt] &= \frac{\tfrac{1}{80} \cdot \pi}{\tfrac{1}{80} \cdot \pi + \tfrac{1}{100} \cdot (1-\pi)} \\[6pt] &= \frac{5 \cdot \pi}{5 \cdot \pi + 4 \cdot (1-\pi)} \\[10pt] &= \frac{5 \cdot \pi}{4 + \pi}. \\[6pt] \end{align}$$


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