An Essay towards solving a Problem in the Doctrine of Chances

Bayes's preliminary results in conditional probability (especially Propositions 3, 4 and 5) imply the truth of the theorem that is named for him. He states:"If there be two subsequent events, the probability of the second b/N and the probability of both together P/N, and it being first discovered that the second event has also happened, from hence I guess that the first event has also happened, the probability I am right is P/b.". Symbolically, this implies (see Stigler 1982):

${\displaystyle P(B\mid A)={\frac {P(A\cap B)}{P(A)}},{\text{ if }}P(A)\neq 0,}$

which leads to Bayes's Theorem for conditional probabilities:

${\displaystyle \Rightarrow P(A\mid B)={\frac {P(B\mid A)\,P(A)}{P(B)}},{\text{ if }}P(B)\neq 0.}$

However, it does not appear that Bayes emphasized or focused on this finding. Rather, he focused on the finding the solution to a much broader inferential problem:

"Given the number of times in which an unknown event has happened and failed [... Find] the chance that the probability of its happening in a single trial lies somewhere between any two degrees of probability that can be named."[1]

The essay includes an example of a man trying to guess the ratio of "blanks" and "prizes" at a lottery. So far the man has watched the lottery draw ten blanks and one prize. Given these data, Bayes showed in detail how to compute the probability that the ratio of blanks to prizes is between 9:1 and 11:1 (the probability is low - about 7.7%). He went on to describe that computation after the man has watched the lottery draw twenty blanks and two prizes, forty blanks and four prizes, and so on. Finally, having drawn 10,000 blanks and 1,000 prizes, the probability reaches about 97%.[1]

Bayes's main result (Proposition 9) is the following in modern terms:

Assume a uniform prior distribution of the binomial parameter ${\displaystyle p}$. After observing ${\displaystyle m}$ successes and ${\displaystyle n}$ failures,

${\displaystyle P(a<p<b\mid m;n)={\frac {\int _{a}^{b}{n+m \choose m}p^{m}(1-p)^{n}\,dp}{\int _{0}^{1}{n+m \choose m}p^{m}(1-p)^{n}\,dp}}.}$

It is unclear whether Bayes was a "Bayesian" in the modern sense. That is, whether he was interested in Bayesian inference, or merely in probability. Proposition 9 seems "Bayesian" in its presentation as a probability about the parameter ${\displaystyle p}$. However, Bayes stated his question in a manner that suggests a frequentist viewpoint: he supposed that a ball is thrown at random onto a square table (this table is often misrepresented as a billiard table, and the ball as a billiard ball, but Bayes never describes them as such), and considered further balls that fall to the left or right of the first ball with probabilities ${\displaystyle p}$ and ${\displaystyle 1-p}$. The algebra is of course identical no matter which view is taken.

• An Essay towards solving a Problem in the Doctrine of Chances at viaLibri
• An Essay towards solving a Problem in the Doctrine of Chances at the UCLA Department of Statistics