[Samuel Pepys](https://en.wikipedia.org/wiki/Samuel_Pepys "samuel p...
Harvard Professor Joe Blitzstein introducing the Newton-Pepys probl...
Newton and Pepys exchanged a series of 6 letters in late 1693. Pepy...
Samuel Pepys was a gambling man, and the Newton-Pepys problem is re...
### The Problem * A. Throwing 6 dice, and getting at least one 6. ...
Am I correct in saying that De Moivre's approximation here is just ...
Probability table of obtaining **n** or more 6 when **n dice** are ...
A binomial distribution with parameters $n$ and $p$ is the discrete...
Newton gave Pepys the correct answer to the problem, although his e...
There is a problem with this part of Newton's argument. He assumes ...
Statistical Science
2006, Vol. 21, No. 3, 400–403
DOI: 10.1214/088342306000000312
© Institute of Mathematical Statistics, 2006
Isaac Newton as a Probabilist
Stephen M. Stigler
Abstract. In 1693, Isaac Newton answered a query from Samuel Pepys
about a problem involving dice. Newton’s analysis is discussed and atten-
tion is drawn to an error he made.
On November 22, 1693, Samuel Pepys wrote a let-
ter to Isaac Newton posing a problem in probability.
Newton responded with three letters, first answering
the question briefly, and then offering more informa-
tion as Pepys pressed for clarification. Pepys (1633–
1703) is best known today for his posthumously pub-
lished diary covering the intimate details of his life over
the years 1660–1669, but Newton would not have been
aware of that diary. He would instead have known of
Pepys as a former Secretary of Admiralty Affairs who
had served as President of the Royal Society of Lon-
don from 1684 through November 30, 1686, the same
period when Newton’s great Principia was presented
to the Royal Society and its preparation for the press
begun. But Pepys’ letter did not concern scientific mat-
ters. He sought advice on the wisdom of a gamble.
The three letters Newton wrote to Pepys on this
problem, on November 26 and December 16 and 23,
1693, are almost all we have bearing on Newton and
probability. Some of the letters were published with
other private correspondence in Pepys (1825,Vol.2,
pages 129–135; 1876–1879, Vol. 6, pages 177–181)
and more completely in Pepys (1926,Vol.1,pages
72–94). The letters were cited in a textbook by Chrys-
tal (1889, page 563), where he gave Pepys’ problem as
an exercise, but they were little known until they were
brought to a wide public attention when selections
were reprinted with commentary independently by Dan
Pedoe (1958, pages 43–48), Florence David (1959;
1962, pages 125–129) and Emil D. Schell (1960).
These authors and several others, notably Chaundy and
Bullard (1960), Mosteller (1965, pages 6, 33–35) and
Stephen M. Stigler is Ernest Dewitt Burton Distinguished
Service Professor of Statistics, Department of Statistics,
University of Chicago, Chicago, Illinois 60637, USA
(e-mail: stigler@galton.uchicago.edu).
Gani (1982) have discussed the problem Pepys posed
and Newton’s solution. Others accorded it briefer no-
tice, including Sheynin (1971), who dismissively rele-
gated it to a footnote; Westfall (1980, pages 498–499),
who gave unwarranted credence to the excuse Pepys
opened his first letter with, that the problem had some
connection to a state lottery; and Gjertsen (1986, pages
427–428). But none of these or any other writer seems
to have noted that a major portion of Newton’s solution
is wrong. The error casts an interesting light on how
Newton thought about the matter, and it seems useful
to revisit the question.
Since Pepys’ original statement was, as Newton no-
ticed, somewhat ambiguous, I will state the problem in
paraphrase as it emerged in the correspondence:
Which of the following three propositions has the
greatest chance of success?
A. Six fair dice are tossed independently and at least
one “6” appears.
B. Twelve fair dice are tossed independently and at
least two “6”s appear.
C. Eighteen fair dice are tossed independently and
at least three “6”s appear.
As it emerged in the correspondence, Pepys initially
thought that the third of these (C) was the most prob-
able, but when Newton convinced him after repeated
questioning by Pepys that in fact A was the most prob-
able, Pepys ended the correspondence and announced
he would, using Mosteller’s (1965, page 35) colorful
later term, welsh on a bet he had made.
Newton stated the solution three times during the
correspondence: first he gave a simple logical reason
for concluding that A is the most probable, then he re-
ported a detailed exact enumeration of the chances in
each of the three cases, and finally he returned to the
logical argument and gave it in more detail.
Newton’s exact enumeration was elegant and flaw-
less; it is equivalent to the solution as might be pre-
sented in an elementary class today. Newton worked
from first principles assuming no knowledge of the bi-
nomial distribution; we can now express what he found
by this calculation in terms of a random variable X
with a Binomial (N, p) distribution as follows:
A. P(X1) = 31031/46656 =0.665 when N =6
and p =1/6.
B. P(X 2) = 1346704211/2176782336 = 0.619
when N =12 and p = 1/6.
C. Here Newton simply stated that, “In the third
case the value will be found still less.
In fact,
P(X3) = 60666401980916/101559956668416
= 0.597
when N = 18 and p = 1/6, as another of Pepys’ cor-
respondents (a Mr. George Tollet) found after much la-
bor, while trying to duplicate Newton’s results (Pepys,
1926, Vol. 1, pages 92–94).
Pepys had originally thought that C was the most
probable; Newton’s logical arguments and his careful
enumeration of chances pointed in the contrary direc-
tion. But while the conclusion Newton reached is cor-
rect, only the enumeration stands up under scrutiny. To
understand why, it will help to develop a heuristic un-
derstanding of why A is the most probable.
Pepys’ problem amounts to a comparison of three
Binomial (N, p) distributions with p = 1/6, namely
those with N = 6, 12 and 18. He desired a ranking
of P(X Np) for the three cases. Now, in all Bino-
mial distributions where the mean Np is an integer, Np
is also the median of the distribution (and indeed the
mode as well). This is always true, surprisingly even
in cases like those under study here, where the dis-
tributions are quite skewed and asymmetric. This is a
byproduct of a proof that for any N and any p, the dif-
ference between the mean and median of a binomial
distribution is strictly less than ln(2)<0.7 (Hamza,
1995). So when the mean Np is an integer the two
must agree, and this implies in particular that in all
these cases,
and P(X Np)
and so in each case P(X Np) exceeds 1/2bya
fraction of the probability P(X = Np).Infact,inthe
cases Pepys considered we have to a fair approxima-
tion P(X Np) 1/2 +(0.4)P (X = Np). The rank-
ing Newton calculated then reflects the fact that the
size of the modal probability for a binomial distribu-
tion, P(X = Np), decreases as N increases and the
distribution spreads out, p being held constant. In-
deed, as De Moivre would find by the 1730s, P(X =
Np) is well approximated by 1/
(2πNp(1 p))
N when p = 1/6. So in particular, the proba-
bilities in A, B, C are about 1/2 + (0.4)(1.07)/
an approximation that would give values 0.67, 0.62,
0.60, which agree with the exact values to two places.
Chaundy and Bullard (1960) provide a cumbersome
rigorous proof that this sequence is decreasing, in some
Note that this approximation depends crucially upon
the probabilities P(X 1), P(X 2) and P(X 3)
of A, B, C being P(X Np) [i.e. P(X E(X))]
for the three respective distributions, and the result de-
pends upon this as well. Franklin B. Evans observed
this sensitivity already in 1961, finding, for example,
that P(X 1|N = 6,p = 1/4) = 0.8220 <P(X
2|N = 12,p = 1/4) = 0.8416 (Evans, 1961). That is,
the ordering of A and B that Newton found for fair dice
can fail for weighted dice, and indeed will tend to fail
when p is sufficiently greater than 1/6, even though
they be tossed fairly and independently.
In his first letter to Pepys on November 26, 1693,
Newton had been content to give a short logical argu-
ment for why the chance of A must be the largest. He
dissected the problem carefully, and made it clear that
the proposition required that in each case at least the
given number of “6”s should be thrown. Newton then
restated the question and gave an apparently clear argu-
ment as to why the chance for A had to be the largest:
“What is the expectation or hope of A to
throw every time one six at least with six
“What is the expectation or hope of B to
throw every time two sixes at least with
twelve dyes?
“What is the expectation or hope of C to
throw every time three sixes at least with 18
And whether has not B and C as great an
expectation or hope to hit every time what
they throw for as A hath to hit his what he
throws for?
“If the question be thus stated, it appears
by an easy computation that the expectation
of A is greater than that of B or C; that is,
the task of A is the easiest. And the reason
is because A has all the chances of sixes
on his dyes for his expectation, but B and
C have not all the chances on theirs. For
when B throws a single six or C but one or
two sixes, they miss of their expectations.
(Pepys, 1926, Vol. 1, 75–76; Schell, 1960)
Newton’s conclusion was of course correct but the
argument is not. It is easy for us to see that it cannot
work because the argument applies equally well for
weighted dice, and as we now know, the conclusion
fails if, for example, p is 1/4. Any correct argument
must explicitly use the fact that 1, 2, 3 are the expec-
tations for A, B, C, and Newton’s does not. His enu-
meration did do so, but A would equally well have “all
the chances of sixes on his dyes” even if the chance of
a“6”is1/4. Newton’s proof refers only to the sample
space and makes no use of the probabilities of different
outcomes other than that the dice are thrown indepen-
dently, and so it must fail. But Newton does casually
use the word “expectations”; might he not have had
something deeper in mind? His subsequent correspon-
dence confirms that he did not.
In his third letter of December 23, 1693, Newton re-
turned to this argument and expanded slightly on it.
He personified the choices by naming the player faced
with bet A “Peter” and the player faced with bet B
“James. He then considered a “throw” to be six dice
tossed at once, so then Peter was to make (at least) one
“6” in a throw, while James was to make (at least) two
“6”s in two throws.
Newton then wrote, As the wager is stated, Peter
must win as often as he throws a six [i.e., makes at
least one “6” among the six dice], but James may of-
ten throw a six and yet win nothing, because he can
never win upon one six alone. If Peter flings a six (for
instance) four times in eight throws, he must certainly
win four times, but James upon equal luck may throw
a six eight times in sixteen throws and yet win nothing.
For as the question in the wager is stated, he wins not
upon every single throw with a six as Peter doth, but
only upon every two throws wherein he throws at least
two sixes. And therefore if he flings but one six in the
two first throws, and one in the two next, and but one
in the two next, and so on to sixteen throws, he wins
nothing at all, though he throws a six twice as often as
Peter doth, and by consequence have equal luck with
Peter upon the dyes. (Pepys, 1926, Vol. 1, page 89;
Schell, 1960)
Here we can see more clearly how Newton was led
astray: Even though in the first letter he had care-
fully pointed out that “throwing a six” must be read as
“throwing at least one six, here he confused the two
statements. His argument might work if “exactly one
six” were understood, but then it would not correspond
to the problem as he and Pepys had agreed it should be
understood. Indeed, Peter will not necessarily register
“throw” of six dice, he wins the same as with just one.
Newton reduced the problem to single “throws” where
each throw is a Binomial (N = 6,p = 1/6), and he lost
sight of the multiplicity of outcomes that could lead to
a win. Many of Peter’s wins (those with at least two
“6”s, which occurs in about 40% of the wins) would be
wins for James as well. And in some of James’s wins
(those with at least two “6”s in one-half of tosses and
none in the other half, about 28% of James’s wins) Pe-
ter would not have done so well on “equal luck” (he
would have won but half the time). Evidently to make
Newton’s argument correct would take as much work
as an enumeration!
Newton’s logical argument failed, but modern prob-
abilists should admire the spirit of the attempt. It was a
simple appeal to dominance, a claim that all sequences
of outcomes will favor Peter at least as often as they
will favor James. It had to fail because the truth of the
proposition depends upon the probability measure as-
signed to the sequences and the argument did not. But
this was 1693, when probability was in its infancy.
Why has apparently no one commented upon this
error before? There are several possible explanations,
and no doubt each held for at least one reader. (1) The
letters were read superficially, with no attempt to parse
the somewhat archaic language of the logical proof,
which after all points in the right direction. (2) The
language was puzzling and unclear to the reader (and
Newton was not available to ask), but it was accepted
since he was, after all, Isaac Newton, and the calcu-
lation clearly showed he was sound on the important
fundamentals. (3) The reader may even have seen that
it was not a satisfactory argument, but drew back from
accusing Newton of error, particularly since he got the
numbers right.
In a sense the argument is more interesting be-
cause it is wrong. Newton was thinking like a great
probabilist—attempting a “eureka” proof that made the
issue clear in a flash. When successful, this is the high-
est form of mathematical art. That it failed is no em-
barrassment; a simple argument can be wonderful, but
it can also create an illusion of understanding when the
matter is, as here, deeper than it appears on the surface.
If Newton fooled himself, he evidently took with him
a succession of readers more than 250 years later. Yet
even they should feel no embarrassment. As Augus-
tus De Morgan once wrote, “Everyone makes errors in
probabilities, at times, and big ones. (Graves, 1889,
page 459)
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torical note. Ann. Sci. 13 137–147. (This is the volume for the
year 1957; this third issue, while nominally dated September
1957, was published April 1959, as stated in the volume Table
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You write “In the case of the Newton-Pepys problem we need to find the probability $P(k)$ of throwing at least $k$ sixes with $6n$ dice,” and you go on to say: \[ P(X=k)=\binom{6n}{k} \left( \frac{1}{6} \right)^k \left(\frac{5}{6}\right)^{6n−k} \] But this is the probability $P(k)$ of throwing _exactly_ $k$ sixes with $6n$ dice, whereas what _we_ want is $P(X \geq k)$ for $6k$ dice. To sum these for all $k ~.. 6k$ would be tedious in the extreme and the solution given prior to this comment would be computationally preferable, for small $k$. @StevePowel Thanks for the heads up! Good point. I had a small typo in my equation. Just revised my previous annotation. @Micael Thanks for the corrections. There is still a small slip in the equation, because it should say $P(X \geq n) = \cdots $ and not $P(X = k) = \cdots$. It would be good to just mention that you employ the complement strategy, too. Samuel Pepys was a gambling man, and the Newton-Pepys problem is related to a wager he planned to make (he planned to stake 10 pounds, the equivalent to about $1500 today on a similar bet). In the correspondence he exchanged with Isaac Newton he wanted to know which of the following three scenarios was the most probable: * A. Throwing 6 dice and getting at least one six. * B. Throwing 12 dice and getting at least two sixes. * C. Throwing 18 dice and getting at least three sixes. Initially **Pepys thought that C was the most probable** but Newton showed him that ** A is the most probable scenario**. We can easily show that Newton was right and A is in fact the most probable scenario. Today this seems like a trivial problem but the field of [probability theory](https://en.wikipedia.org/wiki/Probability_theory) was in it's infancy. Probability table of obtaining **n** or more 6 when **n dice** are thrown. The **probability decreases with increasing n**. ![probability table](http://i.imgur.com/V9Ad2Wb.png "probability table") Newton gave Pepys the correct answer to the problem, although his explanation was wrong. He imagined that A, B and C were tossing their dice in groups of six. For Newton A was most favorable because it required a 6 in only one toss, while B and C required a 6 in each of their tosses. This explanation wrongly assumes that a group of dice does not produce more than one 6, so it does not actually correspond to the original problem posed by Pepys. If Newton's explanation was correct then A would always be more probable independently of p (the probability of a dice rolling to 6). If you start playing with biased dice (changing $p$) Newton's explanation no longer works. The figure below shows the probability of A, B and C winning the game. The ranking is initially unchanged ($P(A)>P(B)>P(C)$). As you can see as the die bias increases, the ranking of the games inverts, and with a highly biased die the games where more dice are rolled are more likely to win ($P(C)>P(B)>P(A)$), thus Newton's explanation no longer holds true. ![Biased dice](http://www.datagenetics.com/blog/february12014/bias.png "biased dice") Am I correct in saying that De Moivre's approximation here is just a special case of the normal approximation to the binomial distribution? This would explain why $P(X=Np) = \frac{1}{2\pi\sigma}$. Harvard Professor Joe Blitzstein introducing the Newton-Pepys problem (Harvard class Stats110): [![Harvard Lecture](http://i.imgur.com/GHL4T5J.png)](https://www.youtube.com/watch?v=P7NE4WF8j-Q&feature=youtu.be&t=17m47s) @RobertAndrewMartin You are correct. In this specific case de Moivre–Laplace theorem shows that the probability function of the number of "successes" observed in a series of n independent throws converges to the probability density function of the normal distribution. You can learn more about it here: [De Moivre–Laplace theorem](https://en.wikipedia.org/wiki/De_Moivre–Laplace_theorem) There is a problem with this part of Newton's argument. He assumes that you can only get one 6 per throw, which does not correspond to this problem. Assume that Peter and James throw groups of six dice. James does not need to throw the dice twice to win. He can win with the first throw by getting two or more 6. Here Newton makes the **wrong argument that you cannot get more than one 6 with one group of dice.** [Samuel Pepys](https://en.wikipedia.org/wiki/Samuel_Pepys "samuel pepys"), was a famous diarist of 17th Century London. He kept a **his private diary from 1660 until 1669 which is one of the most important sources for the English Restoration period**. Pepys was born in modest circumstances and educated as a scholarship student at Cambridge. He was a naturally curious person and undertook at the age of 29 to learn arithmetic. He found the multiplication tables particularly challenging, and used to wake up early and stay up late to study them. During the restoration of the Stuart monarchy, **Pepys was one of the most influential men in England**. As Secretary of the Affairs of the Admiralty he presided over the recovery of the British Navy and helped make it dominant over those of France and the Netherlands, which had dealt Britain an ignominious naval defeat in 1667. **Despite his lack of science training as a student, he had natural curiosity and became quite interested in science.** ![Samuel Pepys](https://upload.wikimedia.org/wikipedia/commons/thumb/2/21/Samuel_Pepys.jpg/1024px-Samuel_Pepys.jpg "Samuel Pepys") A binomial distribution with parameters $n$ and $p$ is the discrete probability distribution of the number of successes in **a sequence of n independent experiments**, with **success probability p**: $$P(X=k)={\binom {n}{k}}p^{k}(1-p)^{n-k}$$ for k = 0, 1, 2, ..., n, where: $${\binom {n}{k}}={\frac {n!}{k!(n-k)!}}$$ is the binomial coefficient. In the case of the Newton-Pepys problem we need to find the probability **P(k) of throwing at least n sixes with 6n dice**. By applying the binomial distribution to this problem we can write: $$P(X=k)=1 - \sum_{k=0}^{n-1} {\binom {6n}{k}}\left(\frac{1}{6}\right)^{k}\left(\frac{5}{6}\right)^{6n-k}$$ !["probability plot"](http://i.imgur.com/KBYA0hN.png) Figure: Probability of rolling at least *k* six when we roll *6k* dice. The probability decreases as k increases. Newton and Pepys exchanged a series of 6 letters in late 1693. Pepys in London and Newton in Cambridge discussed a problem about gambling odds. Here are the links to the letters Pepys and Newton exchanged: You can read the **transcripts of their correspondence** here (chronological order): [Pepys to Newton - 11/22/1693](https://webspace.yale.edu/chem125/125/history99/2Pre1800/SPepysINewton/PepyNewtonPDFs/431P2N_112293.pdf) [Newton to Pepys - 11/26/1693](https://webspace.yale.edu/chem125/125/history99/2Pre1800/SPepysINewton/PepyNewtonPDFs/432N2P_112693.pdf) [Pepys to Newton - 12/9/1693](https://webspace.yale.edu/chem125/125/history99/2Pre1800/SPepysINewton/PepyNewtonPDFs/433P2N_12993.pdf) [Newton to Pepys - 12/16/1693](https://webspace.yale.edu/chem125/125/history99/2Pre1800/SPepysINewton/PepyNewtonPDFs/434N2P_121693.pdf) [Pepys to Newton - 12/21/1693](https://webspace.yale.edu/chem125/125/history99/2Pre1800/SPepysINewton/PepyNewtonPDFs/435P2N_122193.pdf) ### The Problem * A. Throwing 6 dice, and getting at least one 6. * B. Throwing 12 dice, and getting at least two 6. * C. Throwing 18 dice, and getting at least three 6. In the case of the combinatorial approach one computes the number of favorable outcomes and divide by the number of possible outcomes to get the probability of each scenario. Each dice has 6 faces so if you throw $N$ dice we have a total of $6^N$ possible outcomes. Instead of computing the probability of each event X we compute the complement probability. Throwing dice are independent events and one can write: $\bar{A}$ is the event of not getting a 6 when we throw the 6 dice. $D_i$ the event of not throwing a six for a given throw i. For the first scenario we want: \begin{eqnarray} \bar{A} = D_1 \cap D_2 ... \cap D_6 \end{eqnarray} These events are independent and thus one can write that the probability of not getting a six for i throws is: \begin{eqnarray*} P(\bar{A}) = \Pi \, P(D_i) \end{eqnarray*} The probability of *scenario A* then becomes: \begin{eqnarray} P(A) &=& 1 - P(\bar{A}) = 1 - \left(\frac{5}{6}\right)^6 \nonumber \\ &=& 0.665 \label{res1} \end{eqnarray} For scenarion B we do the same considerations as for scenario A. Let's denote $B_1$ the event of not getting any 6 and $B_2$ the event of getting exactly one 6. \begin{eqnarray*} \bar{B} = B_1 \cup B_2 \end{eqnarray*} These are disjoinct events and we can then write that: \begin{eqnarray} P(B) = 1 - P(\bar{B}) = 1 - ( P(B_1) + P(B_2) ) \end{eqnarray} We can compute the probability of $B_1$ from equation 1: \begin{eqnarray*} P(B_1) = \Pi \, P(D_i) = \left(\frac{5}{6}\right)^{12} \end{eqnarray*} For $B_2$ we need to compute the probability of getting one 6 and 11 non-6. We have 12 dice thus the probability of throwing one 6 is $12 1/6$. We have 11 dice left that cannot roll to a 6 and thus: \begin{eqnarray*} P(B_2) = 12 \left(\frac{1}{6}\right) \left(\frac{5}{6}\right)^{11} \end{eqnarray*} The probability of event B can be written: \begin{eqnarray} P(B) &=& 1 - P(\bar{B}) = 1 - \left( \left(\frac{5}{6}\right)^{12} + 2 \left(\frac{5}{6}\right)^{11} \right) \nonumber \\ &=& 1 - \frac{17}{6} \left(\frac{5}{6}\right)^{11} \\ &=& 0.619 \label{res2} \end{eqnarray} For scenarion C we consider the outcomes of B and we need to consider one additional type of outcome. Let's denote $C_3$ the event of getting exactly two 6. \begin{eqnarray*} \bar{C} = C_1 \cup C_2 \cup C_3 \end{eqnarray*} These are disjoinct events and we can then write that: \begin{eqnarray} P(C) = 1 - P(\bar{C}) = 1 - ( P(C_1) + P(C_2) + P(C_3) ) \end{eqnarray} For $C_1$ (no 6): \begin{eqnarray*} P(C_1) = \left(\frac{5}{6}\right)^{18} \end{eqnarray*} For $C_2$ (exactly one 6): \begin{eqnarray*} P(C_2) = 18 \left(\frac{1}{6}\right) \left(\frac{5}{6}\right)^{17} \end{eqnarray*} For $C_3$ we need to consider the events where we get exactly two 6. With 18 dice we have $\frac{18*17}{2}$ ways of getting exactly two sixes and we do not want to roll any 6 on the remaining 16 dice. \begin{eqnarray*} P(C_3) = \frac{18*17}{2} \left(\frac{1}{6}\right)^{2} \left(\frac{5}{6}\right)^{16} \end{eqnarray*} One can now write the total probability for scenarion C as: \begin{eqnarray} P(C) &=& 1 - \left( \left(\frac{5}{6}\right)^{18} + 3 \left(\frac{5}{6}\right)^{17} + 17/4 \left(\frac{5}{6}\right)^{16} \right) \nonumber \\ &=& 1 - \left( \left(\frac{5}{6}\right)^{18} + 3 \left(\frac{5}{6}\right)^{17} + 17/4 \left(\frac{5}{6}\right)^{16} \right) \nonumber \\ &=& 0.597 \label{res3} \end{eqnarray} From \ref{res1}, \ref{res2} and \ref{res3} we could easily determine that: \begin{eqnarray*} P(A) > P(B) > P(C) \end{eqnarray*}