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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}
{\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.]

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