Bayes v. Frequentist — Bayes’s Billiards (continued)

Introduction | Coin Problem | Bayes’s Rule: History | Bayes’s Rule: Derivation | Billiards Problem | Billiards (continued) | Endnotes | References

Billiards Problem (continued)

I like to think of the balls that Bayes imagines throwing as pool balls with ball $W$ as the (white) cue ball, and ball $O$ as the (orange) 5-ball.

So, here is the billiards problem:

A cue ball $W$ is tossed from the right end onto a billiards table and ends up at an unknown distance $\theta$ from the right. Then an (orange) 5-ball $O$ is tossed and ends up nearer to the right end than the cue ball $W$ . This is arbitrarily called a “success”. Given this one success, what is the probability that the cue ball $W$ made it more than halfway across the table? In other words what is $P(0.5 < \theta <1)$ ?

Again, the billiards problem is more difficult than the coin problem. We have moved from a discrete uniform probability distribution $P(\theta)$ to a continuous uniform probability density function $p(\theta)$ . I will use upper case $P(\theta)$ for the discrete probability distribution, also called the probability mass function (PMF), and I will use lower case $p(\theta)$ for the continuous probability density function (PDF). If you are new to PDFs, they take some getting used to. Like a probability, a probability density is always greater than or equal to 0, $p(\theta) \geq 0$ , but it doesn’t have to be less than 1. In a figure, probability is no longer represented by the height of a discrete point but by the area under the continuous PDF ( $P(\theta_a < \theta < \theta_b) = \int_{\theta_a}^{\theta_b} p(\theta) d\theta$ ). For the PDF to be valid, the area under it over the entire range of $\theta$ must equal 1 ( $\int_{-\infty}^{+\infty} p(\theta) d\theta = 1$ ).

This is the uniform probability density function (PDF) for the variable $\theta$ prior to any trials. The probability that $\theta$ is between any two values is the area under the PDF between those two values.

This is the posterior probability density function (PDF) for the variable $\theta$ after seeing one success. The probability that $\theta$ is between any two values is the area under the PDF between those two values. The area under the curve from $\theta = 0.5$ to $\theta = 1$ is $\frac{3}{4}$ .

The remainder of this article is available in the pdf. Link to the pdf.

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Introduction | Introduction | Coin Problem | Bayes’s Rule: History | Bayes’s Rule: Derivation | Billiards Problem | Billiards (continued) | Endnotes | References