\documentclass[12pt]{article} \usepackage{amsmath} \usepackage{epsfig} \usepackage[T4,T1]{fontenc} \newcommand{\s}{{\fontencoding{T4}\selectfont\m{s}}} % Ligatures with long s \newcommand{\sa}{\s a} \newcommand{\soc}{\s c} \newcommand{\sd}{\s d} \newcommand{\se}{\s e} \newcommand{\sh}{\s h} \newcommand{\si}{\s i} \newcommand{\sol}{\s l} \newcommand{\sm}{\s m} \newcommand{\so}{\s o} \newcommand{\sop}{\s p} \newcommand{\sq}{\s q} \newcommand{\sls}{\s\s} % \renewcommand{\sls}{\s\!\s} \newcommand{\sos}{\s s} \newcommand{\st}{\s t} \newcommand{\su}{\s u} \newcommand{\sw}{\s w} \newcommand{\sy}{\s y} % Ligature ct \newcommand{\ct}{ct} \begin{document} \thispagestyle{empty} \renewcommand{\thefootnote}{\fnsymbol{footnote}} Congre\s s\\ \textit{Congre\s s}\\ \noindent\hangindent=1em {\Large LII. \textit{An E{\sls}ay towards {\so}lving a Problem in the Do{\ct}rine of Chances. By the late Rev.\ Mr.}\ Bayes, \textit{communicated by Mr.}\ Price, \textit{in a letter to} John Canton, \textit{M.~A.\ and F.~R.~S.}} \bigskip Dear Sir, \bigskip \noindent {\footnotesize Read Dec.\ 23, 1763.} I now {\se}nd you an e{\sls}ay which I have found among the papers of our decea{\se}d friend Mr.\ Bayes, and which, in my opinion, has great merit, and well de{\se}rves to be pre{\se}rved. Experimental philo{\so}phy, you will find, is nearly intere{\st}ed in the {\su}bje{\ct} of it; and on this account there {\se}ems to be particular rea{\so}n for thinking that a communication of it to the Royal Society cannot be improper. He had, you know, the honour of being a member of that illu{\st}rious Society, and was much e{\st}eemed by many as a very able mathematician. In an introdu{\ct}ion which he has writ to this E{\sls}ay, he {\sa}ys, that his de{\si}gn at fir{\st} in thinking on the {\su}bje{\ct} of it was, to find out a method by which we might judge concerning the probability that an event has to happen, in given circum{\st}ances, upon {\su}ppo{\si}tion that we know nothing concerning it but that, under the {\sa}me circum{\st}ances, it has happened a certain number of times, and failed a certain other number of times. He adds, that he {\so}on perceived that it would not be very difficult to do this, provided {\so}me rule could be found, according to which we ought to e{\st}imate the chance that the probability for the happening of an event perfe{\ct}ly unknown, {\sh}ould lie between any two named degrees of probability, antecedently to any experiments made about it; and that it appeared to him that the rule mu{\st} be to {\su}ppo{\se} the chance the {\sa}me that it {\sh}ould lie between any two equidifferent degrees; which, if it were allowed, all the re{\st} might be ea{\si}ly calculated in the common method of proceeding in the do{\ct}rine of chances. Accordingly, I find among his papers a very ingenious {\so}lution of this problem in this way. But he afterwards con{\si}dered, that the \textit{po{\st}ulate} on which he had argued might not perhaps be looked upon by all as rea{\so}nable; and therefore he cho{\se} to lay down in another form the propo{\si}tion in which he thought the {\so}lution of the problem is contained, and in a \textit{Scholium} to {\su}bjoin the rea{\so}ns why he thought it {\so}, rather than to take into his mathematical rea{\so}ning any thing that might admit di{\sop}ute. This, you will ob{\se}rve, is the method which he has pur{\su}ed in this e{\sls}ay. Every judicious per{\so}n will be {\se}n{\si}ble that the problem now mentioned is by no means merely a curious {\sop}eculation in the do{\ct}rine of chances, but nece{\sls}ary to be {\so}lved in order to a {\su}re foundation for all our rea{\so}nings concerning pa{\st} fa{\ct}s, and what is likely to be hereafter. Common {\se}n{\se} is indeed {\su}fficient to {\sh}ew us that, form the ob{\se}rvation of what has in former in{\st}ances been the con{\se}quence of a certain cau{\se} or a{\ct}ion, one may make a judgement what is likely to be the con{\se}quence of it another time. and that the larger number of experiments we have to {\su}pport a conclu{\si}on, {\so} much more the rea{\so}n we have to take it for granted. But it is certain that we cannot determine, at lea{\st} not to any nicety, in what degree repeated experiments confirm a conclu{\si}on, without the particular di{\soc}u{\sls}ion of the beforementioned problem; which, therefore, is nece{\sls}ary to be con{\si}dered by any that would give a clear account of the {\st}rength of \textit{analogical} or \textit{indu{\ct}ive rea{\so}ning}; concerning, which at pre{\se}nt, we {\se}em to know little more than that it does {\so}metimes in fa{\ct} convince us, and at other times not; and that, as it is the means of [a]cquainting us with many truths, of which otherwi{\se} we mu{\st} have been ignorant; {\so} it is, in all probability, the {\so}urce of many errors, which perhaps might in some mea{\su}re be avoided, if the force that this {\so}rt of rea{\so}ning ought to have with us were more di{\st}in{\ct}ly and clearly under{\st}ood. The{\se} ob{\se}rvations prove that the problem enquired after in this e{\sls}ay is no le{\sos} important than it is curious. It may be {\sa}fely added, I fancy, that it is al{\so} a problem that has never before been {\so}lved. Mr.\ De Moivre, indeed, the great improver of this part of mathematics, has in his \textit{Laws of chance}\footnote{See Mr.\ De Moivre's \textit{Do{\ct}rine of Chances}, p.\ 243, \&c. He has omitted the demon{\st}ration of his rules, but the{\se} have been {\su}pplied by Mr.\ Simp{\so}n at the conclu{\si}on of his treati{\se} on \textit{The Nature and Laws of Chance.}}, after Bernoulli, and to a greater degree of exa{\ct}ne{\sos}, given rules to find the probability there is, that if a very great number of trials be made concerning any event, the proportion of the number of times it will happen, to the number of times it will fail in tho{\se} trials, {\sh}ould differ le{\sos} than by {\sm}all a{\sls}igned limits from the proportion of its failing in one {\si}ngle trial. But I know of no per{\so}n who has {\sh}own how to deduce the {\so}lution of the conver{\se} problem to this; namely, ``the number of times an unknown event has happened and failed being given, to find the chance that the probability of its happening {\sh}ould lie {\so}mewhere between any two named degrees of probability.''\ \, What Mr.~De Moivre has done therefore cannot be thought {\su}fficient to make the con{\si}deration of this point unnece{\sls}ary: e{\sop}ecially, as the rules he has given are not pretended to be rigorou{\sol}y exa{\ct}, except on {\su}ppo{\si}tion that the number of trials are made infinite; from whence it is not obvious how large the number of trials mu{\st} be in order to make them exa{\ct} enough to be depended on in pra{\ct}ice. Mr.~De Moivre calls the problem he has thus {\so}lved, the harde{\st} that can be propo{\se}d on the {\su}bje{\ct} of chance. His {\so}lution he has applied to a very important purpo{\se}, and thereby {\sh}ewn that tho{\se} are much mi{\st}aken who have in{\si}nuated that the Do{\ct}rine of Chances in mathematics is of trivial con{\se}quence, and cannot have a place in any {\se}rious enquiry\footnote{See his Do{\ct}rine of Chances, p.\ 252, \&c.}. The purpo{\se} I mean is, to {\sh}ew what rea{\so}n we have for believing that there are in the con{\st}itution of things fixt laws according to which things happen, and that, therefore, the frame of the world mu{\st} be the effe{\ct} of the wi{\sd}om and power of an intelligent cau{\se}; and thus to confirm the argument taken from final cau{\se}s for the exi{\st}ence of the Deity. It will be ea{\sy} to {\se}e that the conver{\se} problem {\so}lved in this e{\sls}ay is more dire{\ct}ly applicable to this purpo{\se}; for it {\sh}ews us, with di{\st}in{\ct}ne{\sos} and preci{\si}on, in every ca{\se} of any particular order or recurrency of events, what rea{\so}n there is to think that such recurrency or order is derived from {\st}able cau{\se}s or regulations innature, and not from any irregularities of chance. The two la{\st} rules in this e{\sls}ay are given without the dedu{\ct}ions of them. I have cho{\se}n to do this becau{\se} the{\se} dedu{\ct}ions, taking up a good deal of room, would {\sw}ell the e{\sls}ay too much; and al{\so} becau{\se} the{\se} rules, though not of con{\si}derable u{\se}, do not an{\sw}er the purpo{\se} for which they are given as perfe{\ct}ly as could be wi{\sh}ed. They are however ready to be produced, if a communication of them {\sh}ould be thought proper. I have in {\so}me places writ {\sh}ort notes, and to the whole I have added an application of the rules in this e{\sls}ay to {\so}me particular ca{\se}s, in order to convey a clearer idea of the nature of the problem, and to {\sh}ew who far the {\so}lution of it has been carried. I am {\se}n{\si}ble that your time is {\so} much taken up that I cannot rea{\so}nably expe{\ct} that you {\sh}ould minutely examine every part of what I now {\se}nd you. Some of the calculations, particularly in the Appendix, no one can make without a good deal of labour. I have taken {\so} much care about them, that I believe there can be no material error in any of them; but {\sh}ould there be any {\su}ch errors, I am the only per{\so}n who ought to be con{\si}dered as an{\sw}erable for them. Mr.~Bayes has thought fit to begin his work with a brief demon{\st}ration of the general laws of chance. His rea{\so}n for doing this, as he {\sa}ys in his introdu{\ct}ion, was not merely that his reader might not have the trouble of {\se}arching el{\se}where for the principles on which he has argued, but becau{\se} he did not know whither to refer him for a clear demon{\st}ration of them. He has al{\so} make an apology for the peculiar definition he has given of the word \textit{chance} or \textit{probability}. His de{\si}gn herein was to cut off all di{\sop}ute about the meaning of the word, which in common language is u{\se}d in different {\se}n{\se}s by per{\so}ns of different opinions, and according as it is applied to \textit{pa{\st}} or \textit{future} fa{\ct}s. But whatever different {\se}n{\se}s it may have, all (he ob{\se}rves) will allow that an expe{\ct}ation depending on the truth of any \textit{pa{\st}} fa{\ct}, or the happening of any \textit{future} event, ought to be e{\st}imated {\so} much the more valuable as the fa{\ct} is more likely to be true, or the event more likely to happen. In{\st}ead therefore, of the proper {\se}n{\se} of the word \textit{probability}, he has given that which all will allow to be its proper mea{\su}re in every ca{\se} where the word is u{\se}d. But it is time to conclude this letter. Experimental philo{\so}phy is indebted to you for {\se}veral di{\soc}overies and improvements; and, therefore, I cannot help thinking that there is a peculiar propriety in dire{\ct}ing to you the following e{\sls}ay and appendix. That your enquiries may be rewarded with many further {\su}cce{\sls}es, and that you may enjoy every valuable ble{\sls}ing, is the {\si}ncere wi{\sh} of, Sir, \begin{flushright} your very humble {\se}rvant,\quad \\ \medskip {\large Richard Price.} \end{flushright} \begin{flushleft} {\small Newington Green,} \\ {\small \ Nov.\ 10, 1763.} \end{flushleft} \pagebreak \begin{center} P R O B L E M. \end{center} \noindent\hangindent=1em \textit{Given} the number of times ion which an unknown event has happened and failed: \textit{Required} the chance that the probability of its happening in a {\si}ngle trial lies {\so}mewhere between any two degrees of probability that can be named. \begin{center} S E C T I O N\quad I. \end{center} DEFINITION 1. Several events are \textit{incon{\si}{\st}ent}, when if one of them happens, none of the rest can. 2. Two events are \textit{contrary} when one, or other of them must; and both together cannot happen. 3. An event is {\sa}id to \textit{fail}, when it cannot happen; or, which comes to the same thing, when its contrary has happened. 4. An event is {\sa}id to be determined when it has either happened or failed. 5. The \textit{probability of any event} is the ratio between the value at which an expe{\ct}ation depending on the happening of the event ought to be computed, and the value of the thing expe{\ct}ed upon it's happening. 6. By \textit{chance} I mean the same as probability. 7. Events are independent when the happening of any one of them does neither increa{\se} nor abate the probability of the re{\st}. \begin{center} P R O P.\quad1. \end{center} When {\se}veral events are incon{\si}{\st}ent the probability of the happening of one or other of them is the {\su}m of the probabilities of each of them. Suppose there be three {\su}ch events, and which ever of them happens I am to receive N, and that the probability of the 1{\st}, 2d, and 3d are re{\sop}e{\ct}ively $\frac{a}{\text{N}}$, $\frac{b}{\text{N}}$, $\frac{c}{\text{N}}$. Then (by definition of probability) the value of my expe{\ct}ation from the 1{\st} will be $a$, from the 2d $b$, and from the 3d $c$. Wherefore the value of my expe{\ct}ations from all three is in this ca{\se} an expe{\ct}ations from all three will be $a+b+c$. But the {\su}m of my expe{\ct}ations from all three is in this ca{\se} an expe{\ct}ation of receiving N upon the happening of one or other of them. Wherefore (by definition 5) the probability of one or other of them is $\frac{a+b+c}{\text{N}}$ or $\frac{a}{\text{N}}+\frac{b}{\text{N}}+\frac{c}{\text{N}}$. The {\su}m of the probabilities of each of them. Corollary. If it be certain that one or other of the events mu{\st} happen, then $a+b+c=\text{N}$. For in this ca{\se} all the expe{\ct}ations together amounting to a certain expe{\ct}ation of receiving N, their values together mu{\st} be equal to N. And from hence it is plain that the probability of an event added to the probability of its failure (or its contrary) is the ratio of equality. For the{\se} are two incon{\si}{\st}ent events, one of which nece{\sls}arily happens. Wherefore if the probability of an event is $\frac{\text{P}}{\text{N}}$ that of it's failure will be $\frac{\text{N}-\text{P}}{\text{N}}$. \begin{center} P R O P.\quad2. \end{center} If a per{\so}n has an expe{\ct}ation depending on the happening of an event, the probability of the event is to the probability of its failure as his lo{\sos} if it fails to his gain if it happens. Suppo{\se} a per{\so}n has an expe{\ct}ation of receiving N, depending on an event the probability of which is $\frac{\text{P}}{\text{N}}$. Then (by definition 5) the value of his expe{\ct}ation is P, and therefore if the event fail, he lo{\se}s that which in value is P; and if it happens he receives N, but his expe{\ct}ation cea{\se}s. His gain therefore is $\text{N}-\text{P}$. Likewi{\se} {\si}nce the probability of the event is $\frac{\text{P}}{\text{N}}$, that of its failure (by corollary prop.\ 1) is $\frac{\text{N}-\text{P}}{\text{N}}$. But $\frac{\text{P}}{\text{N}}$ is to $\frac{\text{N}-\text{P}}{\text{N}}$ as P is to $\text{N}-\text{P}$, i.e.\ the probability of the event is to the probability of it's failure, as his lo{\sos} if it fails to his gain if it happens. \begin{center} P R O P.\quad3. \end{center} The probability that two {\su}bsequent events will both happen is a ratio compounded of the probability of the 1{\st}, and the probability of the 2d on {\su}ppo{\si}tion the 1{\st} happens. Suppo{\se} that, if both events happen, I am to receive N, that the probability both will happen is $\frac{\text{P}}{\text{N}}$, that the 1{\st} will is $\frac{a}{\text{N}}$ (and consequently that the 1{\st} will not is $\frac{\text{N}-a}{\text{N}}$) and that the 2d will happen upon {\su}ppo{\si}tion the 1{\st} does is $\frac{b}{\text{N}}$. Then (by definition 5) P will be the value of my expe{\ct}ation, which will become $b$ is the 1{\st} happens. Con{\se}quently if the 1{\st} happens, my gain is $b-\text{P}$, and if it fails my lo{\sos} is P. Wherefore, by the foregoing proposition, $\frac{a}{\text{N}}$ is to $\frac{\text{N}-a}{\text{N}}$, i.e.\ $a$ is to $\text{N}-a$ as P is to $b-\text{P}$. Wherefore (componendo inver{\se}) $a$ is to N as P is to $b$. But the ratio of P to N is compounded of the ratio of P to $b$, and that of $b$ to N. Wherefore the {\sa}me ratio of P to N is compounded of the ratio of $a$ to N and that of $b$ to N, i.e.\ the probability that the two {\su}b{\se}quent events will both happen is compounded of the probability of the 1{\st} and the probability of the 2d on {\su}ppo{\si}tion the 1{\st} happens. Corollary. Hence if of two {\su}b{\se}quent events the probability of the 1{\st} be $\frac{a}{\text{N}}$, and the probability of both together be $\frac{\text{P}}{\text{N}}$, then the probability of the 2d on {\su}ppo{\si}tion the 1{\st} happens is $\frac{\text{P}}{a}$. \begin{center} P R O P.\quad4. \end{center} If there be two {\su}b{\se}quent events be determined every day, and each day the probability of the 2d is $\frac{b}{\text{N}}$ and the probability of both $\frac{\text{P}}{\text{N}}$, and I am to receive N if both of the events happen the 1{\st} day on which the 2d does; I {\sa}y, according to the{\se} conditions, the probability of my obtaining N is $\frac{\text{P}}{b}$. For if not, let the probability of my obtaining N be $\frac{x}{\text{N}}$ and let $y$ be to $x$ as $\text{N}-b$ to N. The {\si}nce $\frac{x}{\text{N}}$ is the probability of my obtaining N (by definition 1) $x$ is the value of my expe{\ct}ation. And again, becau{\se} according to the foregoing conditions the 1{\st} day I have an expe{\ct}ation of obtaining N depedning on the happening of both events together, the probability of which is $\frac{\text{P}}{\text{N}}$, the value of this expe{\ct}ation is P. Likewi{\se}, if this coincident {\sh}ould not happen I have an expe{\ct}ation of being rein{\st}ated in my former circum{\st}ances, i.e.\ of receiving that which in value is $x$ depending on the failure of the 2d event the probability of which (by cor.\ prop.\ 1) is $\frac{\text{N}-b}{\text{N}}$ or $\frac{y}{x}$, becau{\se} $y$ is to $x$ as $\text{N}-b$ to N. Wherefore {\si}nce $x$ is the thing expe{\ct}ed and $\frac{y}{x}$ the probability of obtaining it, the value of this expe{\ct}ation is $y$. But the{\se} two la{\st} expe{\ct}ations together are evidently the {\sa}me with my original expe{\ct}ation, the value of which is $x$, and therefore $\text{P}+y=x$. But $y$ is to $x$ as $\text{N}-b$ is to N. Wherefore $x$ is to P as N is to $b$, and $\frac{x}{\text{N}}$ (the probability of my obtaining N) is $\frac{\text{P}}{b}$. Cor. Suppo{\se} after the expe{\ct}ation given me in the foregoing propo{\si}tion, and before it is at all known whether the 1{\st} event has happened or not, I {\s}hould find that the 2d event has happened; from hence I can only infer that the event is determined on which my expe{\ct}ation depended, and have no rea{\so}n to e{st}eem the value of my expe{\ct}ation either greater or le{\sos} than it was before. For if I have rea{\so}n to think it le{\sos}, it would be rea{\so}nable for me to give {\so}mething to be rein{\st}ated in my former circum{\st}ances, and this over and over again as I {\sh}ould be informed that the 2d event had happened, which is evidently ab{\su}rd. And the like ab{\su}rdity plainly follows if you {\sa}y I ought to {\se}t a greater value on my expe{\ct}ation than before, for then it would be rea{\so}nable for me to refu{\se} {\so}mething if offered me upon condition I would relinqui{\sh} it, and be rein{\st}ated in my former circum{\st}ances; and this likewi{\se} over and over again as often as (nothing being known concerning the 1{\st} event) it {\sh}ould appear that the 2d had happened. Notwithstanding therefore that the 2d event has happened, my expe{\ct}ation ought to be e{\st}eemed the {\sa}me as before i.\ e.\ $x$, and con{\se}quently the probability of my obtaining N is (by definition 5) {\st}ill $\frac{x}{\text{N}}$ or $\frac{\text{P}}{b}$\setcounter{footnote}{0}\footnote{What is here {\sa}id may perhaps be a little illu{\st}rated by con{\si}dering that all that can be lo{\st} by the happening of the 2d event is the chance I {\sh}ould have of being rein{\st}ated in my formed circumstances, if the event on which my expe{\ct}ation depended had been determined in the manner expre{\sls}ed in the propostion. But this chance is always as much \text{again{\st}} me as it is \textit{for} me. If the 1{\st} event happens, it is \textit{again{\st}} me, and equal to the chance for the 2d event's failing. If the 1{\st} event does not happen, it is \textit{for} me, and equal al{\so} to the chance for the 2d event's failing. The lo{\sos} of it, therefore, can be no di{\sa}dvantage.}. But after this di{\soc}overy the probability of my obtaining N is the probability that the 1{\st} of two {\su}b{\se}quent events has happen=ed upon the {\su}ppo{\si}tion that the 2d has, who{\se} probabilities were as before {\sop}ecified. But the probability that an event has happened is the {\sa}me as as the probability I have to gue{\sos} right if I gue{\sos} it has happened. Wherefore the following propo{\si}tion is evident. \begin{center} P R O P.\quad5. \end{center} If there be two {\su}b{\se}quent events, the probability of the 2d $\frac{b}{\text{N}}$ and the probability of both together $\frac{\text{P}}{\text{N}}$, and it being 1{\st} di{\soc}overed that the 2d event has sl{\so} happened, from hence I gue{\sos} that the 1{\st} event has al{\so} happened, the probability I am right is $\frac{\text{P}}{b}$\footnote{What is proved by Mr.~Bayes in this and the preceding propo{\si}tion is the {\sa}me with the an{\sw}er to the following que{\st}ion. What is the probability that a certain event, when it happens, will be accompanied with another to be determined at the {\sa}me time? In this ca{\se}, as one of the events is given, nothing can be due for the expe{\ct}ation of it; and, con{\se}quently, the value of an expe{\ct}ation depending on the happening of both events mu{\st} be the {\sa}me with the value of an expe{\ct}ation depending on the happening of one of them. In other words; the probability that, when one of two events happens, the other will, is the {\sa}me with the probability of this other. Call $x$ then the probability of this other, and if $\frac{b}{\text{N}}$ be the probability of the given event, and $\frac{p}{\text{N}}$ the probability of both, becau{\se} $\frac{p}{\text{N}}=\frac{b}{\text{N}}\times x$, $x=\frac{p}{b}=\text{the probability mentioned in the{\se} propo{\si}tions.}$ }. \begin{center} P R O P.\quad6. \end{center} The probability that {\se}veral independent events {\sh}all happen is a ratio compounded of the probabilities of each. For from the nature of independent events, the probability that any one happens is not altered by the happening or failing of any one of the re{\st}, and con{\se}quently the probability that the 2d event happens on {\su}ppo{\si}tion the 1{\st} does is the {\sa}me with its original probability; but the probability that any two events happen is a ratio compounded of the 1{\st} event, and the probability of the 2d on the {\su}ppo{\si}tion on the 1{\st} happens by prop.\ 3. Wherefore the probability that any two independent events both happen is a ratio compounded of the 1{\st} and the probability of the 2d. And in the like manner con{\si}dering the 1{\st} and 2d events together as one event; the probability that three independent events all happen is a ratio compounded of the probability that the two 1{\st} both happen and the probability of the 3d. And thus you may proceed if there be ever {\so} many {\su}ch events; from which the propo{\si}tion is manife{\st}. Cor.\ 1. If there be {\se}veral independent events, the probability that the 1{\st} happens the 2d fails, the 3d fails and the 4th happens, \&c.\ is a ratio compounded of the probability of the 1{\st}, and the probability of the failure of 2d, and the probability of the failure of the 3d, and the probability of the 4th, \&c. For the failure of an event may always be considered as the happening of its contrary. Cor.\ 2. If there be {\se}veral independent events, and the probability of each one be $a$, and that of its failing be $b$, the probability that the 1{\st} happens and the 2d fails, and the 3d fails and the 4th happens, \&c.\ will be $abba$, \&c. For, according to the algebraic way of notation, if $a$ denote any ratio and $b$ another $abba$ denotes the ratio compounded of the ratios $a$, $b$, $b$, $a$. This corollary is therefore only a particular ca{\se} of the foregoing. Definition. If in con{\se}quence of certain data there ari{\se}s a probability that a certain event {\sh}ould happen, its happening or failing, in con{\se}quence of the{\se} data, I call it's happening or failing in the 1{\st} trial. And if the {\sa}me data be again repeated, the happening or failing of the event in con{\se}quence of them I call its happening or failing in the 2d trial; and {\so} again as often as the {\sa}me data are repeated. And hence it is manife{\st} that the happening or failing of the {\sa}me event in {\so} many diffe-[rent] trials, is in reality the happening or failing of {\so} many di{\st}in{\ct} independent events exa{\ct}ly {\si}milar to each other. \begin{center} P R O P.\quad7. \end{center} If the probability of an event be $a$, and that of its failure be $b$ in each {\si}ngle trial, the probability of its happening $p$ times, and failing $q$ times in $p+q$ trials is E~$a^pb^q$ if E be the coefficient of the term in which occurs $a^pb^q$ when the binomial $\overline{a+b\,}\!|^{b+q}$ is expanded. For the happening or failing of an event if different trials are {\so} many independent events. Wherefore (by cor.\ 2.\ prop.\ 6.) the probability that the event happens the 1{\st} trial, fails the 2d and 3d, and happens the 4th, fails the 5th. \&c.\ (thus happening and failing till the number of times it happens be $p$ and the number it fails be $q$) is $abbab$ \&c.\ till the number of $a$'s be $p$ and the number of $b$'s be $q$, that is; 'tis $a^pb^q$. In like manner if you con{\si}der the event as happening $p$ times and failing $q$ times in any other particular order, the probability for it is $a^pb^q$; but the number of different orders according to which an event may happen or fails {\so} as in all to happen $p$ times and fail $q$, in $p+q$ trials is equal to the number of permutations that $aaaa\,bbb$ admit of when the number of $a$'s is $p$ and the number of $b$'s is $q$. And this number is equal to E, the coefficient of the term in which occurs $a^pb^q$ when $\overline{a+b\,}\!|^{p+q}$ is expanded. The event therefore may happen $p$ times and fail $q$ in $p+q$ trials E different ways and no more, and its happening and failing the{\se} {\se}veral different ways are {\so} many incon{\si}{\st}ent events, the probability for each of which is $a^pb^q$, and therefore by prop.\ 1.\ the probability that {\so}me way or other it happens $p$ times and fails $q$ times in $p+q$ trials is E~$a^pb^q$. \pagebreak \begin{center} S E C T I O N\quad II. \end{center} \begin{figure} \begin{center} \includegraphics[height=9cm]{table.jpg} \end{center} \end{figure} Po{\st}ulate.\ 1. Suppo{\se} the {\sq}uare table or plane ABCD to be {\so} made and levelled, that if either of the balls $o$ or W be thrown upon it, there {\sh}all be the {\sa}me probability that it re{\st}s upon any one equal part of the plane as another, and that it mu{\st} nece{\sls}arily re{\st} {\so}mewhere upon it. 2. I {\su}ppose that the ball W {\sh}all be 1{st} thrown, and through the point where it re{\st}s a line $os$ shall be drawn parallel to AD, and meeting CD and AB in $s$ and $o$; and that afterwards the ball O {\sh}all be thrown $p+q$ or $n$ times, and that its re{\st}ing between AD and $os$ after a {\si}ngle throw be called the happening of the event M in a {\si}ngle trial. The{\se} things {\su}ppo{\se}d, Lem.\ 1. The probability that the point $o$ will fall between any two points in the line AB is the ratio of the di{\st}ance between the two points to the whole line AB. Let any two points be named, as $f$ and $b$ in the line AB, and through them parallel to AD draw $fF$, $b$L meeting CD in F and L. Then if the re{\ct}angles C$f$, F$b$, LA are commen{\su}rable to each other, they may each be divided into the {\sa}me equal parts, which being done, and the ball W thrown, the probability it will re{\st} {\so}mewhere upon any number of the{\se} equal parts will be the {\su}m of the probabilities it has to re{\st} upon each one of them, becau{\se} its re{\st}ing upon any different parts of the plance AC are {so} many incon{\si}{\st}ent events; and this {\su}m, becau{\se} the probability it {\sh}ould re{\st} upon any one equal part as another is the {\sa}me, is the probability it {\sh}ould re{\st} upon any one equal part multiplied by the number of parts. Con{\se}quently, the probability there is that the ball W {\sh}ould re{\st} {\so}mewhere upon F$b$ is the probability it has to re{\st} upon one equal part multiplied by the number of equal parts in F$b$; and the probability it re{\st}s {\so}mewhere upon C$f$ or LA, i.e.\ that it dont re{\st} upon F$b$ (becau{\se} it mu{\st} re{\st} {\so}mewhere upon AC) is the probability it re{\st}s upon one equal part multiplied by the number of equal parts in C$f$, LA taken together. Wherefore, the probability it re{\st}s upon F$b$ is to the probability it dont as the number of equal parts in F$b$ is to the number of equal parts in C$f$, LA together, or as F$b$ to C$f$, LA together, or as $fb$ to B$f$, A$b$ together. And \textit{(compendo inver{\se})} the probability it re{\st}s upon F$b$ added to the probability it dont, as $fb$ to A B, or as the ratio of $fb$ to AB to the ratio of AB to AB. But the probability of any event added to the probability of its failure is the ratio of equality; wherefore, the probability if re{\st} upon F$b$ is to the ratio of equality as the ratio of $fb$ to AB to the ratio of AB to AB, or the ratio of equality; and therefore the probability it re{st} upon F$b$ is the ratio of $fb$ to AB. But \textit{ex hypothe{\si}} according as the ball W falls upon F$b$ or nor the point $o$ will lie between $f$ and $b$ or not, and therefore the probability the point $o$ will lie between $f$ and $b$ is the ratio of $fb$ to AB. Again; if the re{\ct}angles C$f$, F$b$, LA are not commen{\su}rable, yet the la{\st} mentioned probability can be neither greater nor le{\sos} than the ratio of $fb$ to AB; for, if it be le{\sos}, let it be the ratio of $fc$ to AB, and upon the line $fb$ take the points $p$ and $t$, {\so} that $pt$ {\sh}all be greater than half $cb$, and taking $p$ and $t$ the neare{\st} points of divi{\si}on to $f$ and $c$ that lie upon $fb$). Then becau{\se} B$p$, $pt$, $t$A are commen{\su}rable, {\so} are the re{\ct}angles C$p$, D$t$, and that upon $pt$ compleating the {\sq}uare AB. Wherefore, by what has been {\sa}id, the probability that the point $o$ will lie between $p$ and $t$ is the ratio of $pt$ to AB. But if it lies between $p$ and $t$ it mu{\st} lie between $f$ and $b$. Wherefore, the probability it {\sh}ould lie between $f$ and $b$ cannot be le{\sos} than the ratio of $fc$ to AB ({\si}nce $pt$ is greater than $fc$). And after the {\sa}me manner you may prove that the forementioned probability cannot be greater than the ratio of $fb$ to AB, it mu{\st} therefore be the {\sa}me. Lem.\ 2. The ball W having been thrown, and the line $os$ drawn, the probability of the event M in a {\si}ngle trial is the ratio of A$o$ to AB. For, in the {\sa}me manner as in the foregoing lemma, the probability that the ball $o$ being thrown {\sh}all re{\st} {\so}mewhere upon D$o$ or between AD and $so$ is the ratio of A$o$ to AB. But the re{\st}ing of the ball $o$ between AD and $so$ after a {\si}ngle throw is the happening of the event M in a {\si}ngle trial. Wherefore the lemma is manife{\st}. \begin{center} P R O P.\quad8. \end{center} If upon BA you ere{\ct} the figure B$ghikm$A who{\se} property is this, that (the ba{\se} BA being divided into any two parts, as A$b$, and B$b$ and at the point of division $b$ a perpendicular being ere{\ct}ed and terminated by the figure in $m$; and $y$, $x$, $r$ repre{\se}nting re{\sop}e{\ct}ively the ratio of $bm$, A$b$, and B$b$ to AB, and E being the coefficient of the term which occurs in $a^pb^q$ when the binomial $\overline{a+b\,}\!|^{p+q}$ is expanded) $y=\text{E}x^pr^q$. I {\sa}y that before the ball W is thrown, the probability the point $o$ {\sh}ould fall between $f$ and $b$, any two points named in the line AB, and withall that the event M {\sh}ould happen $p$ times and fail $q$ in $p+q$ trials, is the ratio of $fghikmb$, the part of the figure B$ghikm$A intercepted between the perpendiculars $fg$, $bm$ rai{\se}d upon the line AB, to CA the {\sq}uare upon AB. \begin{center} D E M O N S T R A T I O N. \end{center} For if not; 1{\st} let it be the ratio of D a figure greater than $fghikmb$ to CA, and through the points $e$, $d$, $c$ draw perpendiculars to $fb$ meeting the curve A$mig$B in $h$, $i$, $k$; the point $d$ being {\so} placed that $di$ {\sh}all be the longe{\st} of the perpendiculars terminated by the line $fb$, and the curve A$mig$B; and the points $e$, $d$, $c$ being {\so} many and {\so} placed that the re{\ct}angles $bk$, $ci$, $ei$, $fb$ taken together {\sh}all differ le{\sos} from $fghikmb$ than D does; all which may be ea{\si}ly done by the help of the equation of the curve, and the difference between D and the figure $fghikmb$ given. Then {\si}nce $di$ is the longe{\st} of the perpendicular ordinates that in{\si}{st} upon $fb$, the re{\st} will gradually decrea{\se} as they are farther and farther from it on each {\si}de, as appears from the constru{\ct}ion of the figure, and con{\se}quently $eb$ is greater than $gf$ or any other ordinate that in{\si}{\st}s upon $ef$. Now if A$o$ were equal to A$e$, then by lem.\ 2.\ the probability of the event M in a {\si}ngle trial would be the ratio of A$e$ to AB, and con{\se}quently by cor.\ Prop.\ 1.\ the probability of it's failure would be the ratio of B$e$ to AB. Wherefore, if $x$ and $r$ be the two forementioned ratios re{\sop}e{\ct}ively, by Prop.\ 7.\ the probability of the event M happening $p$ times and failing $q$ in $p+q$ trials would be E~$x^pr^q$. But $x$ and $r$ being re{\sop}e{\ct}ively the ratios of A$e$ to AB and B$e$ to AB, if $y$ is the ratio of $eb$ to AB, then, by constru{\ct}ion of the figure A$i$B, $y=\text{E}x^pr^q$. Wherefore, if A$o$ were equal to A$e$ the probability of the event M happening $p$ times and failing $q$ times in $p+q$ trials would be $y$, or the ratio of $eb$ to AB. And if A$o$ were equal to A$f$, or were any mean between A$e$ and A$f$, the la{\st} mentioned probability for the {\sa}me rea{\so}ns would be the ratio of $fg$ or {\so}me other of the ordinates in{\si}{\st}ing upon $ef$, to AB. But $eh$ is the greate{\st} of all the ordinates that in{\si}{\st} upon $ef$. Wherefore, upon {\su}ppo{\si}tion the point {\sh}ould lie any where between $f$ and $e$, the probability that the event M happens $p$ times and fails $q$ in $p+q$ trials can't be greater than the ratio of $eh$ to AB. There then being the{\se} two {\su}b{\sq}uent events. the 1{\st} that the point $o$ will lie between $e$ and $f$, the 2d that the event M will happen $p$ times and fail $q$ in $p+q$ trials, and the probability of the 1{\st} (by lemma 1{\st}) is the ratio of $ef$ to AB, and upon {\su}ppo{\si}tion the 1{\st} happens, by what has now been proved, the probability of the 2d cannot be greater than the ratio of $eh$ to AB it evidently follows (from Prop.\ 3.) that the probability both together will happen cannot be greater than the ratio compounded of that of $ef$ to AB and that of $eh$ to AB, which compound ratio is the ratio of $fh$ to CA. Wherefore, the probability that the point $o$ will lie between $f$ and $e$, and the event M will happen $p$ times and fail $q$, is not greater than the raio of $fh$ to CA. And in like, manner the probability the point $o$ will lie between $e$ and $d$, and the event M happen and fail as before, cannot be greater than the raio of $ei$ to CA. And again, the probability the point $o$ will lie between $c$ and $b$, and the event M happen and fail as before, cannot be greater than the ratio of $bk$ to CA. Add now all the{\se} {\se}veral probabilities together, and their {\su}m (by Prop.\ 1.) will be the probability that the point will lie {\so}mewhere between $f$ and $b$, and the event M happen $p$ times and fail $q$ in $p+q$ trials. Add likewi{\se} the corre{\sop}ondent ratios together, and their {\su}m will be the ratio of the {\su}m of the antecedents to their con{\se}quent, i.\ e.\ the ratio of $fb$, $ei$, $ci$, $bk$ together to CA; which ratio is le{\sos} than that of D to CA, becau{\se} D is greater than $fh$, $ei$, $ci$, $bk$ together. And therefore, the probability that the point $o$ will lie between $f$ and $b$, and withal that the event M will happen $p$ times and fail $q$ in $p+q$ times, is \textit{le{\sos}} than the ratio of D to CA; but it was {\su}ppo{\se}d the {\sa}me which is ab{\su}rd. And in like manner, by in{\soc}ribing re{\ct}angles within the figure, as $eg$, $dh$, $dk$, $cm$ you may prove that the la{\st} mentioned probability is \textit{greater} than the ratio of any figure le{\sos} than $fghikmb$ to CA. Wherefore, that probability mu{\st} be the ratio of $fghikmb$ to CA. Cor. Before the ball W is thrown the probability that the point $o$ will lie {\so}mwehere between A and B, or {\so}mewhere upon the line AB, and withal that the event M will happen $p$ times, and fail $q$ in $p+q$ trials is the ratio of the whole figure A$i$B to ZCA. But it is certain that the point $o$ will lie {\so}mewhere upon AB. Wherefore, before the ball W is thrown the probability the event M will happen $p$ times and fail $q$ in $p+q$ trials is the ratio of A$i$B to CA. \begin{center} P R O P.\quad9. \end{center} If before any thing is di{\soc}overed concerning the place of the point $o$, it {\sh}ould appear that the event M had happened $p$ times and failed $q$ in $p+q$ trials, and from hence I gue{\sos} that the point $o$ lies between any two points in the line AB, as $f$ and $b$, and con{\se}quently that the probability of the event M in a {\si}ngle trial was {\so}mewhere between the ratio of A$b$ to AB and that of A$f$ to AB: the probability I am in the right is the ratio of that part of the figure A$i$B de{\soc}ribed as before which is intercepted between perpendiculars ere{\ct}ed upon AB at the points $f$ and $b$, to the whole figure A$i$B. For, there being the{\se} two {\su}b{\se}quent events. the fir{\st} that the point $o$ will lie between $f$ and $b$; the {\se}cond that the event M {\sh}ould happen $p$ times and fail $q$ in $p+q$ trials; and (by cor.\ prop.\ 8.) the original probability of the {\se}cond is the ratio of A$i$B to CA, and (by prop.\ 8.) the probability of both is the ratio of $fghikmb$ to CA; wherefore (by prop.\ 5) it being fir{\st} di{\soc}overed that the {\se}cond has happened, and from hence I gue{\sos} that the fir{\st} has happened al{\so}, the probability I am in the right is the ratio of $fghimb$ to A$i$B, the point which was to be proved. Cor. The {\sa}me things {\su}ppo{\se}d, I gue{\sos} that the probability of the event M lies {\so}mewhere between $o$ and the ratio of A$b$ to AB, my chance to be in the right is the ratio of A$bm$ to A$i$B. \begin{center} \textsc{S c h o l i u m.} \end{center} From the preceding propo{\si}tion it is plain, that in the ca{\se} of {\su}ch an event as I there call M, from the number of trials it happens and fails in a certain number of trials, without knowing any thing more concerning it, one may give a gue{\sos} whereabouts it's probability is, and, by the u{\su}al methods computing the magnitudes of the areas there mentioned {\se}e the chance that the gue{\sos} is right. And that the {\sa}me rule is the proper one to be u{\se}d in the ca{\se} of an event concerning the probability of which we ab{\so}lutely know nothing antecedently to any trials made concerning it, {\se}ems to appear from the following con{\si}deration: viz.\ that concerning {\su}ch an event I have no rea{\so}n to think that, in a certain number of trials, it {\sh}ould rather happen any one po{\sls}ible number of times than another. For, on this account, I may ju{\st}ly rea{\so}n concerning it as if its probability had been at fir{\st} unfixed, and then determined in {\su}ch a manner as to give me no rea{\so}n to think that, in a certain number of trials, it {\sh}ould rather happen any one po{\sls}ible number of times rather than another. But this is exa{\ct}ly the ca{\se} of the event M. For before the ball W is thrown, which determines it's probability in a {\si}ngle trial, (by cor.\ prop.\ 8.) the probability it has to happen $p$ times and fail $q$ in $p+q$ or $n$ trials is the ratio of A$i$B to CA, which ratio is the {\sa}me when $p+q$ or $n$ is given, whatever number $p$ is; as will appear by computing the magnitude of A$i$B by the method\setcounter{footnote}{0} \footnote{It is here proved pre{\se}ntly in art.\ 4.\ by computing in the method here mentioned that A$i$B contra{\ct}ed in the ratio of E to 1 is to CA as 1 to $\overline{n+1}\times\text{E}$; from whence it plainly follows that, antecedently to this contra{\ct}ion, A$i$B mu{\st} be to CA in the ratio of 1 to $n+1$, which is a con{\st}ant ratio when $n$ is given, whatever $p$ is.} of fluxions. And con{\se}quently before the place of the point $o$ is di{\soc}overed or the number of times the event M has happened in $n$ trials, I have not rea{\so}n to think it {\sh}ould rather happen one po{\sls}ible number of times than another. In what follows therefore I {\sh}all take for granted that the rule given concerning the event M in prop.\ 9.\ is al{\so} the rule to be u{\se}d in relation to any event concerning the probability of which nothing at all is known antecedently to any trials made of ob{\se}rved concerning it. And {\su}ch and event I {\sh}all call an unknown event. Cor. Hence, by {\su}ppo{\si}ng the ordinates in the figure A$i$B to be contra{\ct}ed in the ratio of E to one. which makes no alteration in the proportion of the parts of the figure intercepted between them, and applying what is {\sa}id of the event M to an unknown event, we have the following propo{\si}tion, which gives the rules of finding the probability of an event from the number of times it a{\ct}ually happens and fails. \begin{center} P R O P.\quad10. \end{center} If a figure be described upon any ba{\se} AH (Vid.\ Fig.) having for it's equation $y=x^pr^q$; where $y$, $x$, $r$ are respe{\ct}ively the ratios of an ordinate of the figure in{\si}{\st}ing on the ba{\se} at right angles, of the {\se}gment of the ba{\se} intercepted between the ordinate and A the beginning of the ba{\se}, and of the other {\se}gment of the ba{\se} lying between the ordinate and the point H, to the ba{\se} as their ommon con{\se}quent. I {\sa}y then that if an unknown event has happened $p$ times and failed $q$ in $p+q$ trials, and in the ba{\se} AH taking any two points as $f$ and $t$ you ere{\ct} the ordinates $fc$, $t$F at right angles with it, the chance that the probability of an event lies {\so}mewhere between the ratio of A$f$ to AH and that of A$t$ to AH, is the ratio of $t$FC$f$, that part of the before-de{sc}ribed figure which is intercepted between the two ordinates, to ACFH the whole figure in{\si}{\st}ing on the ba{\se} AH. This is evident from prop.\ 9.\ and the remarks made in the foregoing {\soc}holium and corollary. \begin{figure} \begin{center} \includegraphics[height=9cm]{bayebeta.jpg} \end{center} \end{figure} Now, in order to reduce the foregoing rule to pra{\ct}ice, we mu{\st} find the value of the area of the figure de{\soc}ribed and the {\se}veral parts of it {\se}parated, by ordinates perpendicular to its ba{\se}. For which purpo{\se}, {\su}ppo{\se} $AH=1$ and HO the {\sq}uare upon AH $\text{likewi{\se}}=1$, and C$f$ will $\text{be}=y$, and $\text{A}f=x$ and $\text{H}f=r$, becau{\se} $y$, $x$ and $r$ denote the ratios of C$f$, A$f$, and H$f$ re{\sop}e{\ct}ively to AH. And by the equation of the curve $y=x^pr^q$ and (becau{\se}$\text{A}f+f\text{H}=\text{AH}$) $r+x=1$. Wherefore $y=x^p\times\overline{1-x\,}\!|^q=x^p-qx\begin{array}{c} \mbox{\small$p+1$}\\+\end{array} q\times\frac{q-1}{2}\times x^{p+2}-q\times\frac{q-1}{2}\times\frac{q-2}{3} \times x^{p+3}+\text{\&c.}$ Now the ab{\soc}i{\sls}e being $x$ and the ordinate $x^p$ the corre{\sop}ondent area is $\frac{x^{p+1}}{p+1}$ (by prop.\ 10.\ ca\s.\ 1.\ Quadrat.\ Newt.)\setcounter{footnote}{0}\footnote{Tis very evident here, without having recour{\se} to Sir I{\sa}ac Newton, that the fluxion of the area AC$f$ being $y\dot{x}-qx^{p+1}\dot{x}+q\times\frac{q-1}{2}x^{p+2}\dot{x}\text{\ \&c.}$, the fluent or area it{\se}lf is $\frac{x^{p+1}}{p+1}-q\times\frac{x^{p+2}}{p+2}\times q\times\frac{q-1}{2}\times\frac{x^{p+3}}{p+3}\text{\ \&c.}$} and the ordinate being $qx^{p+1}$ the area is $\frac{qx^{p+2}}{p+2}$; and in like manner of the re{\st}. Wherefore, the ab{\soc}i{\sls}e being $x$ and the ordinate $y$ or $x^p-qx^{p+1}+\text{\&c.}$ the corre{\sop}endent area is $\frac{x^{p+1}}{p+1}-\frac{q\times x^{p+2}}{p+2}+ q\times\frac{q-1}{2}\times\frac{x^{p+3}}{p+3}-q\times\frac{q-1}{2} \times\frac{q-2}{3}\times\frac{x^{p+2}}{p+4}+\text{\&c.}$ Wherefore, if $x=\text{A}f=\frac{\text{A}f}{\text{AH}}$, and $y=\text{C}f=\frac{\text{C}f}{\text{AH}}$, then $\text{AC}f=\frac{\text{AC}f}{\text{HO}}=\frac{x^{p+1}}{p+1}- \frac{q}{p+2}\times x^{p+2}+q\times\frac{q-1}{2}\times\frac{x^{p+3}}{p+3}- \text{\&c.}$. From which equation, if $q$ be a {\sm}all number, it is ea{\sy} to find the value of the ratio of AC$f$ to HO.\ and in like manner the value of the ratio of HC$f$ to HO is $\frac{r^{q+1}}{q+1}-p\times\frac{r^{q+2}}{q+2}+ p\times\frac{p-1}{2}\times\frac{r^{q+3}}{q+3}- p\times\frac{p-1}{2}\times\frac{p-2}{3}\times\frac{r^{q+4}}{q+4}$ \&c. which {\se}ries will con{\si}{\st} of a few terms and therefore is to be u{\se}d when $p$ is {\sm}all. 2. The {\sa}me things {\su}ppo{\se}d as before, the ratio of AC$f$ to HO is $\frac{x^{p+1}r^q}{p+1}+\frac{q\times}{p+1}$ $\frac{x^{p+2}r^{q-1}}{p+2}+ \frac{q\times}{p+1}\times\frac{q-1}{p+2}\times\frac{x^{p+3}r^{q-2}}{p+3}+ \frac{q}{p+1}\times\frac{q-1}{p+2}\times\frac{q-2}{p+3}\times \frac{x^{p+4}r^{q-3}}{p+4}+\text{\&c.} +\frac{x^{n+1}}{n+1}\times\frac{q}{p+1}\times\frac{q-1}{p+2}\times\text{\&c.} \times\frac{1}{n}$ where $n=p+q$. For this {\se}ries is the {\sa}me with $\frac{x^{p+1}}{p+1}-q\times\frac{x^{p+2}}{p+2}$ \&c. {\se}t down in Art.\ 1{\st} as the value of the ratio of AC$f$ to HO; as will ea{\si}ly be {\se}en by putting in the former in{\st}ead of $r$ its value $1-x$, and expanding the terms and ordering them according to the powers of $x$. Or, more readily, by comparing the fluxions of the two {\se}ries, and in the former in{\st}ead of $r$ {\su}b{\st}ituting $-\dot{x}$\setcounter{footnote}{0}\footnote{The fluxion of the fir{\st} {\se}ries is $x^pr^q\dot{x}+ \frac{qx^{p+1}r^{p-1}r}{p+1}+\frac{qx^{p+1}r^{q-1}\dot{x}}{p+1}+ q\times\frac{q-1}{p+1}\times\frac{x^{p+2}r^{q-2}\dot{r}}{p+2}+ \frac{q}{p+1}\times\frac{q-1}{p+2}\times x^{p+2}r^{q-2}\dot{x} +\frac{q}{p+1}\times\frac{q-1}{p+2}\times\frac{q-3}{p+3}\times x^{p+3}r^{q-3}\dot{r}$ \&c.\ or, {\su}b{\st}uting $-\dot{x}$ for $r$, $x^pr^q\dot{x}-\frac{qx^{p+1}r^{q-1}\dot{x}}{p+1}+ \frac{qx^{p+1}r^{q-1}\dot{x}}{p+1}-q\times\frac{q-1}{p+1}\times \frac{x^{p+2}r^{q-2}\dot{x}}{p+2}+q\times\frac{q-1}{p+1}\times \frac{x^{p+2}r^{q-2}\dot{x}}{p+2}$ \&c.\ which, as all the terms after the fir{\st} de{\st}roy one another, is equal to $x^pr^q\dot{x}= x^p\times\overline{1-x\,}\!|^q\dot{x}=x^p\dot{x}\times \overline{1-qx+q\times\frac{q-1}{2}x^2\ \text{\&c.}}= x^p\dot{x}-qx^{p+1}\dot{x}+q\times\frac{q-1}{2}x^{p+2}\dot{x}\text{\&c.}= \text{the}$ fluxion of the latter {\se}ries or of $\frac{x^{p+1}}{p+1}- q\times\frac{x^{p+2}}{p+2}$ \&c. The two {\se}ries therefore are the {\sa}me.}. 3. In like manner, the ratio of HC$f$ to HO is $\frac{r^{q+1}x^p}{q+1}+ \frac{p}{q+1}\times\frac{r^{q+2}x^{p-1}}{q+2}+ \frac{p}{q+1}\times\frac{p-1}{q+2}\times\frac{r^{q+3}x^{p-2}}{q+3}+ \text{\&c.}$ 4. If E be the coefficient of that term of the binomial $\overline{a+b\,}\!|^{p+q}$ expanded in which occurs at $a^pb^q$, the ratio of the whole figure ACFH to HO is $\frac{1}{n+1}\times\frac{1}{\text{E}}$, $n$ $\text{being}=p+q$. For, when A$f=\text{AH}$ $x=1$, $r=0$. Wherefore, all the terms of the {\se}ries {\se}t down in Art.\ 2.\ as expre{\sls}ing the ratio of AC$f$ to HO will vani{\sh} except the la{\st}, and that becomes $\frac{1}{n+1}\times\frac{q}{p+1}\times\frac{q-1}{p+2}\times \text{\&c.} \times\frac{1}{n}$. But E being the coefficient of that term in the binomial $\overline{p+q}\!|^n$ expabded in which occurs $a^pb^q$ is equal to $\frac{p+1}{q}\times\frac{p+2}{q-1}\times\text{\&c.} \times\frac{n}{1}$. And, becau{\se} A$f$ is {\su}ppo{\se}d to $\text{become}=\text{AH}$, AC$f=\text{ACH}$. From whence this article is plain. 5. The ratio of AC$f$ to the whole figure ACFH is (by Art.\ 1.\ and 4.) $\overline{n+1}\times\text{E}\times \overline{\frac{x^{p+1}}{p+1}-q\times\frac{x^{p+2}}{p+2} +q\times\frac{q-1}{2}\times\frac{x^{p+3}}{p+3}}$ \&c. and if, as $x$ expre{\sls}es the ratio of A$f$ to AH, X {\sh}ould expre{\sos} the ratio of A$t$ to AH; the ratio of AF$t$ to ACFH would be $\overline{n+1}\times \text{E}\times\frac{\text{X}^{p+1}}{p+1}-q\frac{\text{X}^{p+2}}{p+2}+ q\times\frac{q-1}{2}\times\frac{\text{X}^{p+3}}{p+3}-\text{\&c.}$ and con{\se}quently the ratio of $t$FC$f$ to ACFH is $\overline{n+1}\times \text{E\,X}^d$ into the difference between the two {\se}ries. Compare this with prop.\ 10.\ and we {\sh}all have the following pra{\ct}ical rule. \begin{center} R U L E\quad1. \end{center} If noting is known concerning an event but that it has happened $p$ times and failed $q$ in $p+q$ or $n$ trials, and from hence I gue{\sos} the probability that of its happening in a {\si}ngle time lies {\so}mewhere between any two degrees of probability as X and $x$, the chance I am right in my gue{\sos} is $\overline{n+1}\times\text{E\,X}^d$ into the difference between the {\se}ries $\frac{\text{X}^{p+1}}{p+1}- q\frac{\text{X}^{p+2}}{p+2}+ q\times\frac{q-1}{2}\times\frac{\text{X}^{p+3}}{p+3}-\text{\&c.}$ and the {\se}ries $\frac{x^{p+1}}{p+1}-q\frac{x^{p+2}}{p+2}+ q\times\frac{q-1}{2}\times\frac{x^{p+3}}{p+3}-\text{\&c.}$ E being the coefficient of $a^pb^q$ when $\overline{a+b\,}\!|^n$ is expanded. This is the proper rule to be u{\se}d when $q$ is a {\sm}all number; but if $q$ is large and $p$ {\sm}all, change every where in the {\se}ries here {\se}t down $p$ into $q$ and $q$ into $p$ and $x$ into $r$ or $1-x$, and X into $\text{R}=1-\text{X}$; which will not make any alteration in the difference between the two {\se}rie{\se}s. Thus far Mr.\ Bayes's e{\sls}ay. With re{\sop}e{\ct} to the rule here given, it is further to be observed, that when both $p$ and $q$ are very large numbers, it will not be po{\sls}ible to apply it in pra{\ct}ice on account of the multitude of terms which the {\se}ri{\se}s in it will contain. Mr.\ Bayes, therefore, by an inve{\st}igation which it would be too tedious to give here, has deduced from this rule another, which is as follows. \newpage \begin{center} R U L E\quad2. \end{center} If nothing is known concerning an event but that it has happened $p$ times and failed $q$ om $p+q$ or $n$ trials, and from hence I gue{\sos} that the probability of its happening in a {\si}ngle trial lies between $\frac{p}{n}+z$ and $\frac{p}{n}-z$; if $m^2=\frac{n^3}{pq}$, $a=\frac{p}{n}$, $b=\frac{q}{n}$, E the coefficient of the term which occurs at $a^pb^q$ when $\overline{a+b\,}\!|^n$ is expanded, and $\Sigma=\frac{n+1}{n}\times\frac{\sqrt{2pq}}{\sqrt{n}}\times\text{E}\, a^pb^qX^d$ by the {\se}ries $mz-\frac{m^3z^3}{3}+ \frac{n-2}{2n}\times\frac{m^5z^5}{5}- \frac{\overline{n-2}\times\overline{n-4}}{2n\times3n}\times\frac{m^7z^7}{7}+ \frac{n-2}{2n}\times\frac{n-2}{2n}\times\frac{n-4}{3n}\times\frac{n-6}{4n} \times\frac{n^pz^9}{9}$ \&c. my chance to be in the right is greater than $\frac{2\,\Sigma}{1+2\text{E}\,a^pb^q+2\,\text{E}\,a^pb^q}$\setcounter{footnote}{0} \footnote{In Mr.\ Bayes's manuscript this chance is made to be greater than $\frac{2\,\Sigma}{1+2\,\text{E}\,a^pb^q}$ and le{\sos} than $\frac{2\,\Sigma}{1-2\,\text{E}\,a^pb^q}$. The third term in the two divi{\so}rs, as I have given them, being omitted. But this being evidently owing to a {\sm}all over{\si}ght in the dedu{\ct}ion of this rule, which I have rea{\so}n to think Mr.\ Bayes had him{\se}lf di{\soc}overed, I have ventured to corre{\ct} his copy, and to give the rule as I am {\sa}tisfied it ought to be given.} and le{\sos} than $\frac{2\,\Sigma}{1-2\,\text{E}\,a_pb^q-\frac{2\,\text{E}\,a^pb^q.}{n}}$ And if $p=q$ my chance is $2\,\Sigma$ exa{\ct}ly. In order to render this rule fit for u{\se} in all ca{\se}s it is only nece{\sls}ary to know how to find within {\su}fficient nearne{\sos} the value of E$\,a^pb^q$ and al{\so} of the {\se}ries $mz-\frac{m^3z^3}{3}$\setcounter{footnote}{0}\footnote{A very few terms of this {\se}ries will generally give the hyperbolic logarithm to a {\su}fficient degree of exa{\ct}ne{\sos}. A {\si}milar {\se}ries has been given by Mr.\ De Moivre, Mr.\ Simp{\so}n and other eminent mathematicians in an expre{\sls}ion for the {\su}m of the logarithms of the numbers 1, 2, 3, 4, 5, to $x$, which {\su}m they have a{\sls}erted to be equal to $\frac{1}{2}\log. c +x+\overline{x+\frac{1}{2}}\times\log. x-x+ \frac{1}{12x}-\frac{1}{360x^3}+\frac{1}{1260x^5}$ \&c. $c$ denoting the circumference of a circle whose radius is unity. But Mr.\ Bayes, in a preceding paper in this volume, has demon{\st}rated that, though this expre{\sls}ion will very nearly approach to the value of this {\su}m when only a proper number of the fir{\st} terms is taken, the whole {\se}ries cannot expre{\sos} any quantity at all, beau{\se}, let $x$ be what it will, there will always be a part of the {\se}ries where it will begin to diverge. This ob{\se}rvation, though it does not much affe{\ct} the u{\se} of this {\se}ries, {\se}ems well worth the noticeof mathematicians.}. With respe{\ct} to the former Mr.Bayes has proved that, {\su}pposing K to {\si}gnify the ratio of the quadrantal arc to it's radius, E $a^pb^q$ will be equal to $\frac{\sqrt{n}}{2\sqrt{\text{K}pq}}\times$ by the \textit{ratio} who{\se} \textit{hyperbolic} logarithm is $\frac{1}{12}\times \overline{\frac{1}{n}-\frac{1}{p}-\frac{1}{q}}- \frac{1}{360}\times\overline{\frac{1}{n^3}-\frac{1}{p^3}\frac{1}{q^3}}+ \frac{1}{1260}\times\overline{\frac{1}{n^5}-\frac{1}{p^5}-\frac{1}{q^5}}- \frac{1}{1680}\times\overline{\frac{1}{n^7}-\frac{1}{p^7}-\frac{1}{q^7}}+ \frac{1}{1188}\times\overline{\frac{1}{n^9}-\frac{1}{p^9}-\frac{1}{q^9}}$ \&c.\ where the numeral coefficients may be found in the following manner. Call them A, B, C, D, E, \&c. Then $\text{A}=\frac{1}{2.\,2.\,3}= \frac{1}{3.\,4}$. $\text{B}=\frac{1}{2.\,4.\,5}-\frac{\text{A}}{3}$. $\text{C}=\frac{1}{2.\,6.\,7}-\frac{10\text{B}+\text{A}}{5}$. $\text{D}=\frac{1}{2.\,8.\,9}-\frac{35\text{C}+21\text{B}+\text{A}}{7}$. $\text{E}=\frac{1}{2.\,10.\,11}- \frac{126\text{C}+84\text{D}+36\text{B}+\text{A}}{9}$. $\text{F}=\frac{1}{2.\,12.\,13}- \frac{462\text{D}+330\text{C}+165\text{E}+55\text{B}+\text{A}}{11}$ \&c.\ where the coefficients of B, C, D, E, F, \&c.\ in the values of D, E, F, \&c.\ are the 2, 3, 4, \&c.\ highe{\st} coefficients in $\overline{a+b\,}\!|^7$, $\overline{a+b\,}\!|^9$, $\overline{a+b\,}\!|^{11}$, \&c.\ expanded; affixing in every particular value the lea{\st} of the{\se} coefficients to B, the next in magnitude to the furthe{\st} letter from B, the next to C, the next to the furthe{\st} but one, the next to D, the next to the furthe{\st} but two, and {\so} on\footnote{This method of finding the{\se} coefficients I have deduced from the demon{\st}ration of the third lemma at the end of Mr.\ Simpson's Treati{\se} on the Nature and Laws of Chance.}. With respe{\ct} to the value of the {\se}ries $mz-\frac{m^3z^3}{3}+\frac{n-2}{2n}\times\frac{m^5z^5}{5}$ \&c. he has ob{\se}rved that it may be calculated dire{\ct}ly when $mz$ is le{\sos} than 1, or even not greater than $\sqrt{3}$: but when $mz$ is much larger it becomes impra{\ct}icable to do this; in which ca{\se} he {\sh}ews a way of ea{\si}ly finding two values of it very nearly equal between which it's true value mu{\st} lie. The theorem he gives for this purpo{\se} is as follows. Let K, as before, {\st}and for the ratio of the quadrant arc to its radius, and H for the ratio who{\se} hyperbolic logarithm is $\frac{2^2-1}{2n}-\frac{2^4-1}{360n^3}+\frac{2^6-1}{1260n^5}- \frac{2^8-1}{1680n^7}$ \&c. Then the {\se}ries $mz-\frac{m^3z^3}{3}$ \&c.\ will be greater or le{\sos} than the {\se}ries $\frac{\text{H}n}{n+1} \times\frac{\sqrt{\text{K}}}{\sqrt{2}}-\frac{n}{n+2}\times \frac{\overline{1-\frac{2m^2z^2}{n}\,}|^{\frac{n}{2}+1}}{2mz}+ \frac{n^2}{n+2}\times \frac{\overline{1-\frac{2m^2z^2}{n}\,}|^{\frac{n}{2}+2}} {\overline{n+4}\times4m^3z^3}+ \frac{3n^3}{n+2}\times \frac{\overline{1-\frac{2m^2z^2}{n}\,}|^{\frac{n}{2}+3}} {\overline{n+4}\times\overline{n+6}\times8m^5z^7}+ \frac{2\times5\times n^4}{n+2}\times \frac{\overline{1-\frac{2m^2z^2}{n}\,}|^{\frac{n}{2}+4}} {\overline{n+4}\times\overline{n+6}\times\overline{n+8}\times16m^7z^7}- \text{\&c.}$ continued to any number of terms, according as the la{\st} term has a po{\si}tive or a negative {\si}gn before it. From {\su}b{\st}uting the{\se} values of E $a^pb^q$ and $mz-\frac{m^3z^3}{3}+ +\frac{n-2}{2n}\times\frac{m^5z^5}{5}$ \&c.\ in the 2d rule ari{\se}s a 3d rule, which is the rule to be u{\se}d when $mz$ is of {\so}me con{\si}derable magnitude. \begin{center} R U L E\quad3. \end{center} If nothing is known of an event but that it has happened $p$ times and failed $q$ in $p+q$ or $n$ trials, and from hence I judge that the probability of it's happening in a {\si}ngle trial lies between $\frac{p}{n}+z$ and $\frac{p}{n}-z$ my chance to be right is \textit{greater} than $\frac{\sqrt{\text{K}pq}\times h} {2\sqrt{\text{K}pq}+hn\frac{1}{2}+hn^{-\frac{1}{2}}}\times 2\overline{\text{H}-\frac{\sqrt{2}}{\sqrt{\text{K}}}\times\frac{n+1}{n+2} \times\frac{1}{mz}\times1-\ \overline{{\frac{2m^2z^2}{n}}}\!|\!^{\frac{n}{2}+1}}$ and \textit{le{\sos}} than $\frac{\sqrt{\text{K}pq}\times h} {2\sqrt{\text{K}pq}-hn\frac{1}{2}-hn^{-\frac{1}{2}}}$ multiplied by the 3 terms $2\text{H}-\frac{\sqrt{2}}{\sqrt{\text{K}}}\times\frac{n+1}{n+2} \times\frac{1}{mz}\times\overline{1-\frac{2m^2z^2}{n}\,}\!|^{\frac{n}{2}+1}+ \frac{\sqrt{2}}{\sqrt{\text{K}}}\times\frac{n}{n+2}\times\frac{n+1}{n+4}\times \frac{1}{2m^3z^3}\times\overline{1-\frac{2m^2z^2}{n}\,}\!|^{\frac{n}{2}+2}$ where $m^2$, K, $h$ and H {\st}and for the quantities already explained. \bigskip \begin{center} {\large An A P P E N D I X.} \\ \ \\ CONTAINING \\ \ \\ An Application of the foregoing Rules to {\so}me particular Ca{\se}s \end{center} THE fir{\st} rule gives a dire{\ct} and perfe{\ct} {\so}lution in all ca{\se}s; and the two following rules are only particular methods of approximating to the {\so}lution given in the fir{\st} rule, when the labour of applying it becomes too great. The fir{\st} rule may be u{\se}d inj all ca{\se}s where either $p$ or $q$ are nothing or not too large. The {\se}cond rule may be u{\se}d in all ca{\se}s where $mz$ is le{\sos} than $\sqrt{3}$; and the 3d in all ca{\se}s where $m^2z^2$ is greater than 1 and le{\sos} than $\frac{n}{2}$, if $n$ is an even number and very large. If $n$ is not too large this la{\st} rule cannot be much wanted, becau{\se}, $m$ decrea{\si}ng continually as $n$ is dimini{\sh}ed, the value of $z$ may in this ca{\se} be taken large, (and therefore a con{\si}derable interval had between $\frac{p}{n}-z$ and $\frac{p}{n}+z$) and yet the operation be carried on by the 2d rule; or $mz$ not exceed $\sqrt{3}$. But in order to {\sh}ew di{\st}in{\ct}ly and fully the nature of the pre{\se}nt problem, and how far Mr.\ Bayes has carried the {\so}lution of it; I {\sh}all give the re{\su}lt of this {\so}lution in a few ca{\se}s, beginning with the lowe{\st} and mo{\st} {\si}mple. Let us then fir{\s}t {\su}ppo{\se}, of {\su}ch and event as that called M in the e{\sls}ay, or an event about the probability of which, antecedently to trials, we know nothing, that it has happened \textit{once}, and that it is enquired what conclu{\si}on we may draw from hence with re{\sop}{\ct} to the probability of it's happening on a \textit{{\se}cond} trial. The an{\sw}er is that there would be an odds of three to one for {\so}mewhat more than an even chance that it would happen on a {\se}cond trial. For in this ca{\se}, and in all others where $q$ is nothing, the expre{\sls}ion $\overline{n+1}\times \overline{\frac{\text{X}^{p+1}}{p+2}-\frac{x^{p+1}}{p+1}}$ or $\text{X}^{p+1}-x^{p+1}$ gives the {\so}lution, as will appear from considering the fir{\st} rule. Put therefore in this expre{\sls}ion $\overline{p+1}=2$, $\text{X}=1$. and $x=\frac{1}{2}$ and it will be $1-\overline{1\,}\!|^2$ or $\frac{1}{4}$; which {\sh}ews the chance there is that the probability of an event that has happened once lies {\so}mewhere between 1 and $\frac{1}{2}$; or (which is the {\sa}me) the odds that it is {\so}mewhat more than an even chance that it will happen on a {\se}cond trial\setcounter{footnote}{0}\footnote{There can, I {\su}ppo{\se}, be no rea{\so}n for ob{\se}rving that on this {\su}bje{\ct} unity is always made to {\st}and for certainty, and $\frac{1}{2}$ for an even chance.}. In the {\sa}me manner it will appear that if the event has happened twice, the odds now mentioned will be {\se}ven to one; if thrice, fifteen to one; and in general, if the event has happened $p$ times, there will be an odds of $2^{p+1}-1$ to one, for \textit{more} than an equal chance that it will happen on further trials. Again, {\su}ppo{\se} all I know of an event to be that it has happened ten times without failing, and the enquiry to be what rea{\so}n we {\sh}all have to think we are right if we gue{\sos} that the probability of it's happening in a {\si}ngle trial lies {\so}mewhere between $\frac{16}{17}$ and $\frac{2}{3}$, or that the ratio of the cau{\se}s of it's happening to tho{\se} of it's failure is {\so}me ratio between that of {\si}xteen to one and two to one. Here $p+1=11$, $\text{X}=\frac{16}{17}$ and $x=\frac{2}{3}$ and $\text{X}^{p+1}-x^{p+1}=\overline{\frac{16}{17}\,}\!|^{11}- \overline{\frac{2}{3}\,}\!|^{11}=.5013$ \&c. The an{\sw}er therefore is, that we {\sh}all have very nearly an equal chance for being right. In this manner we may determine in any ca{\se} what conclu{\si}on we ought to draw from a given number of experiments which are unoppo{\se}d by contrary experiments. Every one {\se}es in general that there is rea{\so}n to expe{\ct} an event with more or le{\sos} confidence according to the greater of le{\sos} number of times in which, under given circumstances, it has happened without failing; but we here {\se}e exa{\ct}ly what this rea{\so}n is, on what principles it is is founded, and how we ought to regulate our expe{\ct}ations. But it will be proper to dwell longer on this head. Suppo{\se} a {\so}lid or die or who{\se} number of {\si}des and con{\st}itution we know nothing; and that we are to judge of the{\se} from experiments made in throwing it. In this ca{\se}, it {\sh}ould be ob{\se}rved, that it would be in the highe{\st} degree improbable that the {\so}lid {\sh}ould, in the fir{\st} trial, turn any one {\si}de which could be a{\sls}igned before hand; becau{\se} it would be known that {\so}me {\si}de mu{\st} turn, and that there was an infinity of {\si}des, or {\si}des otherwi{\se} marked, which it was equally likely that it {\sh}ould turn. The fir{\st} throw only {\sh}ews that \textit{it has} the {\si}de then thrown, without giving any rea{\so}n to think that it has any number of times rather than any other. It will appear, therefore, that \textit{after} the fir{\st} throw and not before, we {\sh}ould be in the circum{\st}ances required by the conditions of the pre{\se}nt problem, and that the whole effe{\ct} of this throw would be to bring us into the{\se} circum{\st}ances. That is: the turning the {\si}de fir{\st} thrown in any {\su}b{\se}quent {\si}ngle trial would be an event about the probability or improbability of which e could form no judgment, and of which we {\sh}ould know no more than that it lay {\so}mewhere between nothing and certainty. With the {\se}cond trial then our calculations mu{\st} begin; and if in that trial the {\su}ppo{\se}d {\so}lid turns again the {\sa}me {\si}de, there will ari{\se} the probability of three to one that it has more of that {\so}rt of {\si}des than of \textit{all} others; or (which comes to the {\sa}me) that there is {\so}mewhat in its constitution di{\sop}osing it to turn that {\si}de oftene{\st}: And this probability will increa{\se}, in the manner already explained, with the number of times in which that {\si}de has been thrown without failing It {\sh}ould not, however, be imagined that any number of {\su}ch experiments can give {\su}fficient rea{\so}n for thinking that it would \textit{never} turn any other {\si}de. For, {\su}ppose it has turned the {\sa}me {\si}de in every trial a million of times. In the{\se} circumstances there would be an improbability that it had \textit{le{\sos}} than 1.400,000 more of the{\se} {\si}des than all others; but there would al{\so} be an improbability that it had \textit{above} 1.600,000 times more. The chance for the latter is expre{\sls}ed by $\frac{1600000}{1600001}$ rai{\se}d to the millioneth power {\su}btra{\ct}ed from unity, which is equal to .4647 \&c.\ and the chance for the former is equal to $\frac{1400000}{1400001}$ rai{\se}d to the {\sa}me power, or to .4895; which, being both le{\sos} than an equal chance, proves what I have {\sa}id. But thouth it would be thus improbable that it had \textit{above} 1.600,000 times more or \textit{le{\sos}} than 1.400,000 times \textit{more} of the{\se} {\si}des than of all others, it by no means follows that we have any rea{\so}n for judging that the true proportion in this ca{\se} lies {\so}mewhere between that of 1.600,000 to one and 1.400,000 to one. For he that will take the pains to make the calculation will find that there is nearly the probability expere{\sls}ed by .527, or but little more than an equal chance, that it lies {\so}mewhere between that ot 600,000 to one and three millions to one. It may de{\se}rve to be added, that it is more probable that this proportion lies {\so}mewhere between that of 900,000 to 1 and 1.900,000 to 1 than between any other two proportions who{\se} antecedents are to one another as 900,000 to 1.900,000, and con{\se}quents unity. I have made the{\se} ob{\se}rvattions chiefly becau{\se} they are all {\st}ri{\ct}ly applicable to the events and appearances of nature. Antecedently to all experience, it would be improbable as infinite to one, that any particular event, before-hand imagined, {\sh}ould follow the application of any one natural obje{\ct} to another; becau{\se} there would be an equal chance for any one of an infinity of other events. But it we had once {\se}en any particular effe{\ct}s, as the burning of wood on putting ity into fire, or the falling of a {\st}one on detaching it from all contiguous obje{\ct}s, then the conclu{\si}ons to be drawn from any number of {\su}b{\se}quent events of the {\sa}me kind would be determined in the {\sa}me manner with the conclu{\si}ons ju{\st} mentioned relating to the con{\st}itution of the {\so}lid I have {\su}ppo{\se}d.{---}{---}In other words. The fir{\st} experiment {\su}ppo{\se}d to be ever made on any natural obje{\ct} would only inform us of one event that may follow a particular chance in the circum{\st}ances of tho{\se} obje{\ct}s; but it would not {\su}gge{\st} to us any ideas of uniformity in nature, or give use the lea{\st} rea{\so}n to apprehend that it was, in that in{\st}ance or in any other, regular rather than irregular in its operations. But it the {\sa}me event has followed without interruption in any one or more {\su}b{se}quent experiments, then {\so}me degree of uniformity will be ob{\se}rved; rea{\so}n will be given to expe{\ct} the {\sa}me {\su}cce{\sos} in further experiments, and the calculations dire{\ct}ed by the {\so}lution of this problem may be made. One example here it will not be ami{\sos} to give. Let us imagine to our{\se}lves the ca{\se} of a per{\so}n ju{\st} brought forth into this, world and left to colle{\ct} from his ob{\se}rvations the order and cour{\se} of events what powers and cau{\se}s take place in it. The Sun would, probably, be the fir{\st} obje{\ct} that would engage his attention; but after lo{\si}ng it the fir{\st} night he would be entirely ignorant whether he {\sh}ould ever {\se}e it again. He would therefore be in the condition of a per{\so}n making a fir{\st} experiment about an event entirely unknown to him. But let him {\se}e a {\se}cond appearance or one \textit{return} of the Sun, and an expe{\ct}ation would be rai{\se}d in him of a {\se}cond return, and he might know that there was an odds of 3 to 1 for \text{{\so}me} probability of this. This odds would increa{\se}, as before repre{\se}nted, with the number of returns to which he was witne{\sos}. But no finite number of returns would be {\su}fficient to produce ab{\so}lute or phy{\si}cal certaintly. For let it be {\su}ppo{\se}d that he has {\se}en it return at regular and {\st}ated intervals a million of times. The conclu{\si}ons this would warrant would be {\su}ch as follow{---}{---} There would be the odds of the millioneth power of 2, to one, that it was likely that it would return again at the end of the u{\su}al interval. There would be the probability expre{\sls}ed by .5352, that the odds for this was not \textit{greater} than 1.600,000 to 1; And the probability expre{\sls}ed by .5105, that it was not \textit{le{\sos}} than 1.400,000 to 1. It {\sh}ould be carefully remembered that the{\se} dedu{\ct}ions {\su}ppo{\se} a previous total ignorance of nature. After having ob{\se}rved for {\so}me time the cour{\se} of events it would be found for {\so}me time the cour{\se} of events it would be found that the operations of nature are in general regular, and that the powers and laws which prevali in it are {\st}able and parmanent. The con{\si}deration of this will cau{\se} one or a few experiments often to produce a much {\st}ronger expe{\ct}ation of {\su}cce{\sos} in further experiments than would otherwi{\se} have been rea{\so}nable; ju{\st} as the frequent ob{\se}rvation that things of a {\so}rt are di{sop}po{\se}d together in any place would lead us to conclude, upon di{\soc}overing there any obje{\ct} of a particular {\so}rt, that there are laid up with it many others of the {\sa}me {\so}rt. It is obvious that this, {\so} far from contradi{\ct}ing the foregoing dedu{\ct}ions, is only one particular ca{\se} to which they are to be applied. What has been {\sa}id {\se}ems {\su}fficient to {\sh}ew us what conclu{\si}ons to draw from \textit{uniform} experience. It demon{\st}rates, particularly, that in{\st}ead of proving that events will \textit{always} happen agreeably to it, there will be always rea{\so}n again{\st} this conclusion. In other words, where the cour{\se} of nature has been the mo{\st} con{\st}ant, we can have only rea{\so}n to reckon upon a recusrrency of events proportioned to the degree of this con{\st}ancy, but we can have no reason for thin king that there are no cau{\se}s in nature which will \textit{ever} interfere with the operations the cau{\se}s from which this con{\st}ancy is derived, or no circumstancce of thw world in which it will fail. And if this is true, {\su}ppo{\si}ng our only \textit{data} derived from experience, we {\sh}all find additional rea{\so}n for thinking thus if we apply other principles, or have recour{\se} to {\su}ch con{\si}derations as rea{\so}n, independently of experience, can {\su}gge{\st}. But I have gone further than I intended here; and it is time to turn our thoughts to another branch of this {\su}bje{\ct}: I mean, to ca{\se}s where an experiment has {\so}metimes {\su}cceeed and {\so}metimes failed. Here, again, in order to be as plain and explicit as po{\sls}ible, it will be proper to put the following ca{\se}, which is the easie{\st} and {\si}mple{\st} I can think of. Let us then imagine a per{\so}n pre{\se}nt at the drawing of a lottery, who knows nothing of its {\soc}heme or of the proportion of \textit{Blanks} to \textit{Prizes} in it. Let it further be {\su}ppo{\se}d, that he is obliged to infer this from the number of \textit{blanks} he hears drawn compared with the number of \textit{prizes}; and that it is enquired what conclu{\si}ons in the{\se} circum{\st}ances he may rea{\so}nably make. Let him fir{\st} hear \textit{ten} blanks drawn and \textit{one} prize, and let it be enquired what chance he will have for being right if he gus{\sls}es that the proportion of \textit{blanks} to \textit{prizes} in the lottery lies {\so}mewhere between the proportions of 9 to 1 and 11 to 1. Here taking $\text{X}=\frac{11}{12}$, $x=\frac{9}{10}$, $p=10$, $q=1$, $n=11$, $\text{E}=11$, the required chance, according to the fir{\st} rule, is $\overline{n+1}\times\text{E}$ into the differences between $\overline{\frac{\text{X}^{p+1}}{p+1}-\frac{q\text{X}^{p+2}}{p+2}}$ and $\overline{\frac{x^{p+1}}{p+1}-\frac{qx^{p+2}}{p+2}}=12\times11\times \overline{\frac{\overline{\frac{11}{12}\,}\!|^{11}}{11}- \frac{\overline{\frac{11}{12}}\!|^{12}}{12}}- \overline{\frac{\overline{\frac{9}{10}\,}\!|^{11}}{11}- \frac{\overline{\frac{9}{10}}\!|^{12}}{12}}=.07699$ \&c. There would therefore be an odds of about 923 to 76 \textit{again{\st}} his being right. Had he gue{\sls}ed only in general there were le{\sos} than 9 blanks to a prize, there would have been a probability of his being right equal to .6589, or the odds of 65 to 34. Again. {\su}ppo{\se} that he has heard 20 \textit{blanks} drawn and 2 \textit{prizes}; what chance will he have for being right if he makes the {\sa}me gue{\sos}? Here X and $x$ being the {\sa}me, we have $n=22$, $p=20$, $q=2$, $\text{E}=231$, and the required chance equal to $\overline{n+1}\times\text{E}\times \overline{\frac{\text{X}^{p+1}}{p+1}-q\frac{\text{X}^{p+2}+}{p+2} q\times\frac{q-1}{2}\times\frac{\text{X}^{p+3}}{p+3}}$ \\ $-\overline{\frac{x^{p+1}}{p+1}-\frac{qx^{p+2}}{p+2}+q\times\frac{q-1}{2}\times \frac{x^{p+3}}{p+3}}=.10843$ \&c. He will, therefore, have a better chance for being right in the former in{\st}ance, the oddes again{\st} him now being 892 to 108 or about 9 to 1. But {\sh}ould b=he only gue{\sos} in general, as before, that there were le{\sos} than 9 blanks to a prize, his chance for being right will be wor{\se}; for in{\st}ead of .6589 or an odds of near two to one, it will be .584, or an odds of 584 to 415. Suppo{\se}, further, that he has heard 40 \textit{blanks} drawn and 4 \textit{prizes}; what will the before-mentioned chances be? The an{\sw}er here is .1525, for the former of the{\se} chances; and .527, for the latter. There will, therefore, now be an odds of only $5\frac{1}{2}$ to 1 again{\st} the proportion of blanks to prizes lying between 9 to 1 and 11 to 1; and but little more than an equal chance that it is le{\sos} than 9 to 1. Once more. Suppo{\se} he has heard 100 \textit{blanks} drawn and 10 \textit{prizes}. The an{\sw}er here may {\st}ill be found by the fir{\st} rule; and the chance for a proportion of blanks to prizes \textit{le{\sos}} than 9 to 1 will be .44109, and for a proportion \textit{graeter} than 11 to 1 .3082. It would therefore be likely that there were not \textit{fewer} than 9 or \textit{more} than 11 blanks to a prize. But at the {\sa}me time it will remain unlikely \setcounter{footnote}{0}\footnote{I {\su}ppo{\se} no attentive person will find any difficulty in this. It is only {\sa}ying that, {\su}ppo{\si}ng the interval between nothing and certainty divided into a hundred equal chances, there will be 44 of them for a le{\sos} proportion of blanks to prizes than 9 to 1, 31 for a greater than 11 to 1; in which it is obvious that, though though one of the{se} {\su}ppo{\si}tions mu{\st} be true, yet, having each of them more chances again{\st} them than more them, they are all {\se}parately unlikely.} that the true proportion {\sh}ould lie between 9 to 1 and 11 to 1, the chance for this being .2506 \&c. There will therefore be {\st}ill an odds of near 3 to 1 again{\st} this. From the{\se} calculations it appears that, in the circum{\st}ances I have {\su}p\-po{\se}d, the chance for being right in gue{\sls}ing the proportion of \textit{blanks} to \textit{prizes} to be nearly the {\sa}me with that of the number of \textit{blanks} drawn in a given time to the number of prizes drawn, is continually increa{\si}ng as the{\se} numbers increa{\se}; and therefore, when they are con{\si}derably large, this conclu{\si}on may be lloked upon as moreally certain. By parity of rea{\so}n, it follows univer{\sa}lly, with re{\sop}e{\ct} to every element about which a great number of experiments has been made, that the cau{se}s of its happening bear the {\sa}me proportion to the cau{\se}s of its failing, with the number of happenings to the number of failures; and that. if an event who{\se} cas{\se}s are {\su}ppo{\se}d to be known, happens oftener or {\se}ldomer to be known, happens oftener or {\se}ldomer than is agreeable to this conclu{\si}on, there will be rea{\so}n to believe that there are {\so}me unknown cau{\se}s which di{\st}urb the operations of the known ones. With re{\sop}e{\ct}, therefore, particularly to the cour{\se} of events in nature, it appears, that there is demon{\st}rative evidence to prove that order of events which we ob{\se}rve, and not from any of the powers of chance\setcounter{footnote}{0}\footnote{See Mr.\ De Moivre's Do{\ct}rine of Chances, pag.\ 250.}. This is ju{\st} as evident as it would be, in the ca{\se} I have in{\si}{\st}ed on. that the rea{\so}n of drawing 10 times more \textit{blanks} than \textit{prizes} in millions of trials, was, that there were in the wheel about {\so} many more \textit{blanks} than \textit{prizes}. But to proceed a little further in the demon{\st}ration of this point. We have {\se}en that {\su}ppo{\si}ng a per{\so}n, ignorant of the whole {\soc}heme of a lottery, {\sh}ould be led to conje{\ct}ure, from hearing 100 \textit{blanks} and 10 prizes drawn, that the proportion of \textit{blanks} to \textit{prizes} in the lottery was {\so}mewhere between 9 to 1 and 11 to 1, the chance for his being right would be .2506 \&c. Let now enquire what this chance would be in {\so}me higher ca{\se}s. Let it be {\su}ppo{\se}d that \textit{blanks} have been drawn 1000 times, and prizes 100 times in 1100 trials. In this ca{\se} the powers of X and $x$ ri{\se} to high, and the number of terms in the two {\se}rie{\se}s $\frac{\text{X}^{p+1}}{p+1}- \frac{q\text{X}^{p+1}}{p+2}$ \&c.\ and $\frac{x^{p+1}}{p+1}- \frac{qx^{p+2}}{p+2}$ \&c.\ become {\so} numerous that it would require immen{se} labout to obtain the an{\sw}er by the fir{st} rule. 'Tis nece{\sls}ary, therefore, to have recour{\se} to the {\se}cond rule. But in order to make u{\se} of it, the interval between X and $x$ mu{\st} be a little aleterd. $\frac{10}{11}-\frac{9}{10}$ is $\frac{1}{110}$, and therefore the interval between $\frac{10}{1\phantom{1}}-\frac{1}{110}$ and $\frac{10}{11}+\frac{1}{110}$ will nearly be the {\sa}me with the interval between $\frac{9}{10}$ and $\frac{11}{12}$, only {\so}mewhat larger. If then we make the que{\st}ion to be; what chance there would be ({\su}ppo{\si}ng no more known than that blanks have been drawn 1000 times and prizes 100 times in 1100 trials) that the probability of drawing a blank in a {\si}ngle trial would lie {\so}mewhere between $\frac{10}{11}-\frac{1}{110}$ and $\frac{10}{11}+\frac{1}{110}$ we {\sh}all have a que{\st}ion of the {\sa}me kind with the previous que{\st}ions, and deviate but little from the limits a{\sls}igned in them. The an{\sw}er, according to the {\se}cond rule, is that this chance is greater than \\ $\frac{2\,\Sigma}{1-2\,\text{E}\,a^pb^q+\frac2\,\text{E}\,a^pb^q}{n}$ and le{\sos} than $\frac{2\,\Sigma}{1-2\,\text{E}\,a^pb^q-\frac2\,\text{E}\,a^pb^q}{n}$, E being $\frac{n+1}{n}\times\frac{\sqrt{2pq}}{\sqrt{n}}\times\text{E}\, a^pp^q\times mz-\frac{m^3z^3}{3}+\frac{n-2}{2n}\times\frac{m^5z^5}{5}$ \&c. By making here $1000=p$ $100=q$ $1100=n$ $\frac{1}{110}=z$, $m=\frac{\sqrt{n^3}}{\sqrt{pq}}=1.048808$, $\text{E}\,a^pb^q=\frac{b}{2}\times\frac{\sqrt{n}}{\text{K}pq}$, $b$ being the ratio who{\se} hyperbolic logarithm is $\frac{1}{12}\times \overline{\frac{1}{n}-\frac{1}{p}-\frac{1}{q}}-\frac{1}{360}\times \overline{\frac{1}{n^3}-\frac{1}{p^3}-\frac{1}{q^3}}+\frac{1}{1260}\times \overline{\frac{1}{n^5}-\frac{1}{p^5}-\frac{1}{q^5}}$ \&c.\ and K the ratio of the quadrantal to radius; the former of the{\se} expre{\sls}ions will be found to be .7953, and the latter .9405 \&c. The chance enquired after, therefore, is greater than .7953, and le{\sos} than .9405. That is; there will be an odds for being right in gue{\sls}ing that the proportion of blanks to prizes lies \textit{nearly} between 9 to 1 and 11 to 1, (or \textit{exa{\ct}ly} between 9 to 1 and 1111 to 99) which is greater than 4 to 1, and le{\sos} than 16 to 1. Suppo{\se}, again, that no more is known than that \textit{blanks} have been drawn 10,000 times and \textit{prizes} 1000 times in 11000 trials; what will the chance now mentioned be? Here the {\se}cond as well as the fir{\st} rule becomes u{\se}le{\sos}, the value of $mz$ being {\so} great as to render it {\soc}arcely po{\sls}ible to calculate dire{\ct}ly the {se}ries $\overline{mz-}\frac{m^3z^3}{3}+\frac{n-2}{2n}\times \frac{m^5z^{\phantom{5}}}{5}-\text{\&c.}$ The third rule, therefore, mu{\st} be u{\se}d; and the information it gives us is, that the required chance is greater than .97421, or more than an odds of 40 to 1. By calculations {\si}milar to the{\se} may be determined univer{\sa}lly, what expe{\ct}ations are warranted by any experiments, according to the different number of times in which they have {\su}cceeded and failed; or what {\sh}ould be thought of the probability that any particular cau{\se} in nature, with which we have any acquaintance, will or will not, in any {\si}ngle trial, produce an effe{\ct} that has been conjoined with it. Mo{\st} persons, probably, might expe{\ct} that the chances in the {\sop}ecimen I have given would have been greater than I have found them. But this only {\sh}ews how liable we are to be in error when we judge on this subje{\ct} independently of calculation. One thing, however, {\sh}ould be remembered here; and that is, the narrowne{\sos} of the interval between $\frac{10}{11}+\frac{1}{110}$ and $\frac{10}{11}-\frac{1}{110}$. Had this interval been taken a little larger, there would have been a con{\si}derable difference in the re{\su}lts of the calculations. Thus had it been taken double, or $z=\frac{1}{55}$, it would have been found in the fourth in{\st}ance that in{\st}ead of odds again{\st} that in{\st}ead of odds again{\st} there were odds for being right in judging that the probability of drawing a blank in a {\si}ngle trial lies between $\frac{10}{11}+\frac{1}{55}$ and $\frac{10}{11}-\frac{1}{55}$. The foregoing calculations further {\sh}ew us the u{\se}s and defe{\ct}s of the rules laid down in the e{\sls}ay. 'Tis evident that the two la{\st} rules do not give us the required chances within {\su}ch narrow limits as could be wi{\sh}ed. But here again it {\sh}ould be con{\si}dered, that the{\se} limits become narrower and narrower as $q$ is taken larger in re{\sop}{ct}of $p$; and when $p$ and $q$ are equal, the exa{\ct} {\so}lution is given in all ca{\se}s by the {\se}cond rule. The{\se} two rules therefore afford a dire{\ct}ion to our judgment that may be of con{\si}derable u{\se} till {\so}me per{\so}n {\sh}all di{\soc}over a better approximation to the value of the two {\se}ries's in the fir{\st} rule\setcounter{footnote}{1}\footnote{Since this was written I have found a method of con{\si}derably improving the approximation in the 2d and 3d rules by demon{\st}rating the expre{\sls}ion $\frac{2\,\Sigma}{1+2\,\text{E}\,a^pb^q+\frac{2\,\text{E}\,a^pb^q}{n}}$ comes almo{\st} as near to the true value wanted as there is rea{\so}n to de{\si}re, only always {\so}mewhat le{\sos}. It {\se}ems nece{\sls}ary to hint this here; though the proof of it cannot be given.}. But what mo{\st} of all recommends the {\so}lution in this \textit{E{\sls}ay} is, that it is compleat in tho{\se} ca{\se}s where information is mo{st} wanted, and where Mr.\ De Moivre's {\so}lution of the inver{se} problem can give little of no dire{\ct}ion; I mean, in all ca{\se}s where either $p$ or $q$ are of no con{\si}derable magnitude. In other ca{\se}s, or when $p$ and $q$ are very con{\si}derable, it is not difficult to perceive the truthe of what has been here demon{st}rated, or that there is rea{\so}n to believe in general that the chances for the happening of an event are to the chances for its failure in the {\sa}me \textit{ratio} with that of $p$ to $q$. But we {\sh}all be greatly deceived if we judge in this manner when either $p$ or $q$ are {\sm}all. And tho' in {\su}ch ca{\se}s the \textit{Data} are not {\su}fficient to di{\soc}over the exa{\ct} probability of an event. yet it is very agreeable to be able to find the limits between which it is rea{\so}nable to think it mu{\st} lie, and al{\so} to be able to determine the preci{\se} degree of a{\sls}ent which is due to any conclu{\si}ons or a{\sls}ertions relating to them. \bigskip\bigskip \noindent [\textit{Philosophical Transactions of the Royal Society of London} \textbf{53} (1763), 370--418.] \end{document} %