Isaac Newton as a Probabilist

0.11 MB·

by  Stephen M. Stigler

StatisticalScience
2006,Vol.21,No.3,400–403
DOI:10.1214/088342306000000312
(cid:13)c InstituteofMathematicalStatistics,2006
Isaac Newton as a Probabilist
Stephen M. Stigler
7 Abstract. In1693, IsaacNewton answeredaqueryfromSamuelPepys
0 about a problem involving dice. Newton’s analysis is discussed and
0
attention is drawn to an error he made.
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n
a
OnNovember 22, 1693, Samuel Pepys wrote alet- problem as an exercise, but they were little known
J
tertoIsaacNewton posingaprobleminprobability. until they were brought to a wide public attention
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Newtonrespondedwiththreeletters,firstanswering whenselections werereprintedwith commentary in-

] thequestionbriefly,andthenofferingmoreinforma- dependentlybyDanPedoe(1958,pages43–48),Flo-
T
tion as Pepys pressed for clarification. Pepys (1633– rence David (1959; 1962, pages 125–129) and Emil
S
1703)isbestknowntodayforhisposthumouslypub- D. Schell (1960). These authors and several oth-
.
h
lished diary covering the intimate details of his life ers, notably Chaundyand Bullard (1960), Mosteller
t
a over the years 1660–1669, but Newton would not (1965, pages 6, 33–35) and Gani (1982) have dis-
m
have been aware of that diary. He would instead cussed the problem Pepys posed and Newton’s so-
[ have known of Pepys as a former Secretary of Ad- lution. Others accorded it briefer notice, including

1 miralty Affairs who had served as President of the Sheynin (1971), who dismissively relegated it to a
v Royal Society of Londonfrom 1684 throughNovem- footnote; Westfall (1980, pages 498–499), who gave
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ber 30, 1686, the same period when Newton’s great unwarranted credence to the excuse Pepys opened
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0 Principia waspresentedtotheRoyalSocietyandits his firstletter with, that the problem had some con-
1 preparation for the press begun. But Pepys’ letter nection to astate lottery; andGjertsen (1986,pages
0
did not concern scientific matters. He sought advice 427–428). But none of these or any other writer
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0 on the wisdom of a gamble. seems to have noted that a major portion of New-
/ ton’s solution is wrong. The error casts an interest-
h
t 1. PEPYS’ PROBLEM ing light on how Newton thought about the matter,
a
and it seems useful to revisit the question.
m
The three letters Newton wrote to Pepys on this Since Pepys’ original statement was, as Newton
v: problem, on November 26 and December 16 and 23, noticed, somewhat ambiguous, I will state the prob-
Xi 1693, are almost all we have bearing on Newton lem in paraphrase as it emerged in the correspon-
and probability. Some of the letters were published dence:
r
a with other private correspondence in Pepys (1825, Which of the following three propositions has the
Vol. 2,pages 129–135; 1876–1879,Vol. 6,pages177– greatest chance of success?
181) and more completely in Pepys (1926, Vol. 1,
A. Six fair dice are tossed independently and at
pages 72–94). The letters were cited in a textbook
least one “6” appears.
by Chrystal (1889, page 563), where he gave Pepys’
B. Twelve fair dice are tossed independently and
at least two “6”s appear.
Stephen M. Stigler is Ernest Dewitt Burton
C. Eighteenfairdicearetossedindependentlyand
Distinguished Service Professor of Statistics,
at least three “6”s appear.
Department of Statistics, University of Chicago,
Chicago, Illinois 60637, USA e-mail: As it emerged in the correspondence,Pepys initially
[email protected]. thought that the third of these (C) was the most
probable, but when Newton convinced him after re-
This is an electronic reprint of the original article
peated questioning by Pepys that in fact A was the
published by the Institute of Mathematical Statistics in
Statistical Science, 2006, Vol. 21, No. 3, 400–403. This mostprobable,Pepysendedthecorrespondenceand
reprint differs from the original in pagination and announced he would, using Mosteller’s (1965, page
typographic detail. 35) colorful later term, welsh on a bet he had made.
1
2
S.STIGLER
2. NEWTON’S SOLUTION is a byproduct of a proof that for any N and any
p, the difference between the mean and median of a
Newton stated thesolution threetimes duringthe
binomial distribution is strictly less than ln(2)<0.7
correspondence: first he gave a simple logical reason
(Hamza, 1995). So when the mean Np is an integer
for concluding that A is the most probable, then he
the two must agree, and this implies in particular
reportedadetailedexactenumerationofthechances
that in all these cases,
in each of the three cases, and finally he returned to
the logical argument and gave it in more detail. 1 1
P(X Np) and P(X Np) ,
2 2
Newton’sexactenumerationwaselegantandflaw- ≥ ≥ ≤ ≥
less; it is equivalent to the solution as might be pre- and so in each case P(X Np) exceeds 1/2 by a
≥
sented in an elementary class today. Newton worked fraction of the probability P(X =Np). In fact, in
from first principles assuming no knowledge of the the cases Pepys considered we have to a fair ap-
binomial distribution; we can now express what he proximation P(X Np) 1/2+(0.4)P(X =Np).
≥ ≈
found by this calculation in terms of a random vari- TherankingNewtoncalculatedthenreflectsthefact
able X with a Binomial (N, p) distribution as fol- that the size of the modal probability for a binomial
lows: distribution, P(X =Np), decreases as N increases
and the distribution spreads out, p being held con-
A. P(X 1)=31031/46656=0.665 when N =6
≥ stant.Indeed,asDeMoivrewouldfindbythe1730s,
and p=1/6.
P(X = Np) is well approximated by
B. P(X 2) = 1346704211/2176782336 = 0.619
≥ 1/p(2πNp(1 p)) 1.07/√N when p=1/6. So in
when N =12 and p=1/6. − ≈
particular, the probabilities in A, B, C are about
C. Here Newton simply stated that, “In the third
1/2+(0.4)(1.07)/√N,anapproximationthatwould
case the value will be found still less.”
give values 0.67, 0.62, 0.60, which agree with the
In fact, exact values to two places. Chaundy and Bullard
(1960) provide a cumbersome rigorous proof that
P(X 3)=60666401980916/101559956668416
≥ this sequence is decreasing, in some generality.
=0.597 Note that this approximation depends crucially
upon the probabilities P(X 1), P(X 2) and
when N = 18 and p = 1/6, as another of Pepys’ ≥ ≥
P(X 3) of A, B, C being P(X Np) [i.e. P(X
correspondents (a Mr. George Tollet) found after ≥ ≥ ≥
E(X))] for the three respective distributions, and
much labor, while trying to duplicate Newton’s re-
the result depends upon this as well. Franklin B.
sults (Pepys, 1926, Vol. 1, pages 92–94).
Evansobservedthissensitivityalreadyin1961,find-
Pepys had originally thought that C was the most
ing, for example, that P(X 1N =6,p = 1/4) =
probable; Newton’s logical arguments and his care-
≥ |
0.8220<P(X 2N =12,p=1/4)=0.8416(Evans,
ful enumeration of chances pointed in the contrary
≥ |
1961). That is, the ordering of A and B that New-
direction. But while the conclusion Newton reached
ton found for fair dice can fail for weighted dice,
is correct, only the enumeration stands up under
and indeed will tend to fail when p is sufficiently
scrutiny. To understand why, it will help to develop
greater than 1/6, even though they be tossed fairly
aheuristicunderstandingofwhyAisthemostprob-
and independently.
able.
3. A HEURISTIC VIEW 4. NEWTON’S LOGICAL ARGUMENT
Pepys’ problem amounts to a comparison of three In his first letter to Pepys on November 26, 1693,
Binomial (N, p) distributions with p=1/6, namely Newton had been content to give a short logical ar-
thosewithN =6,12and18.Hedesiredarankingof gumentforwhythechanceof Amustbethelargest.
P(X Np)forthethreecases.Now,inallBinomial Hedissectedtheproblemcarefully,andmadeitclear
≥
distributions where the mean Np is an integer, Np that the proposition required that in each case at
is also the median of the distribution (and indeed least the given number of “6”s should be thrown.
the mode as well). This is always true, surprisingly Newton then restated the question and gave an ap-
even in cases like those under study here, where the parently clear argument as to why the chance for A
distributionsarequiteskewedandasymmetric.This had to be the largest:
3
ISAACNEWTON AS A PROBABILIST
“What is the expectation or hope of A to least one “6” among the six dice], but James may
throw every time one six at least with six often throw a six and yet win nothing, because he
dyes? can never win upon one six alone. If Peter flings a
“What is the expectation or hope of B to six(forinstance)fourtimesineightthrows,hemust
throw every time two sixes at least with certainly win four times, butJames uponequal luck
twelve dyes? may throw a six eight times in sixteen throws and
“What is the expectation or hope of C to yet win nothing. For as the question in the wager is
throw every time three sixes at least with stated, he wins not upon every single throw with a
18 dyes? six as Peter doth, but only upon every two throws
“And whether has not B and C as great whereinhethrowsatleasttwosixes.Andthereforeif
an expectation or hope to hit every time he flings butone six in the two firstthrows, and one
what they throw for as A hath to hit his in the two next, and butone in the two next, and so
what he throws for?
on to sixteen throws, he wins nothing at all, though
“If the question bethus stated, it appears
he throws a six twice as often as Peter doth, and by
by an easy computation that the expecta-
consequence have equal luck with Peter upon the
tion of A is greater than that of B or C;
dyes.” (Pepys, 1926, Vol. 1, page 89; Schell, 1960)
that is, the task of A is the easiest. And
Here we can see more clearly how Newton was led
thereasonisbecauseAhasallthechances
astray: Even though in the first letter he had care-
of sixes on his dyes for his expectation,
fullypointedoutthat“throwingasix”mustberead
but B and C have not all the chances on
as “throwing at least one six,” here he confused the
theirs.ForwhenBthrowsasinglesixorC
two statements. His argument might work if “ex-
butoneortwosixes,theymissoftheirex-
actly one six” were understood, but then it would
pectations.” (Pepys, 1926, Vol. 1, 75–76;
not correspond to the problem as he and Pepys had
Schell, 1960)
agreed it should be understood. Indeed, Peter will
Newton’s conclusion was of course correct butthe not necessarily register a gain with every “6”: if he
argument is not. It is easy for us to see that it can- has two or more in the first “throw” of six dice, he
not work because the argument applies equally well wins the same as with justone. Newton reduced the
for weighted dice, and as we now know, the con- problem to single “throws” where each throw is a
clusion fails if, for example, p is 1/4. Any correct Binomial (N =6,p=1/6), and he lost sight of the
argument must explicitly use the fact that 1, 2, 3 multiplicity of outcomes that could lead to a win.
are theexpectations for A, B, C, andNewton’s does Many of Peter’s wins (those with at least two “6”s,
not.Hisenumerationdiddoso,butAwouldequally which occurs in about 40% of the wins) would be
well have “all the chances of sixes on his dyes” even winsfor James as well. And in some of James’s wins
if the chance of a “6” is 1/4. Newton’s proof refers (those with at least two “6”s in one-half of tosses
only to the sample space and makes no use of the and none in the other half, about 28% of James’s
probabilities of different outcomes other than that wins) Peter would not have done so well on “equal
the dice are thrown independently, and so it must luck” (he would have won but half the time). Ev-
fail. ButNewton does casually usetheword “expec- idently to make Newton’s argument correct would
tations”;mighthenothavehadsomethingdeeperin
take as much work as an enumeration!
mind?His subsequentcorrespondenceconfirms that
he did not.
5. CONCLUSION
In his third letter of December 23, 1693, Newton
returned to this argument and expanded slightly on Newton’slogicalargumentfailed,butmodernprob-
it. He personified the choices by naming the player abilists should admire the spirit of the attempt. It
faced with bet A “Peter” and the player faced with was a simple appeal to dominance, a claim that all
bet B “James.” He then considered a “throw” to be sequences of outcomes will favor Peter at least as
six dice tossed at once, so then Peter was to make often as they will favor James. It had to fail because
(at least) one “6” in a throw, while James was to thetruth of the proposition dependsuponthe prob-
make (at least) two “6”s in two throws. ability measure assigned to the sequences and the
Newtonthenwrote,“Asthewagerisstated,Peter argument did not. But this was 1693, when proba-
must win as often as he throws a six [i.e., makes at bility was in its infancy.
4
S.STIGLER
Why has apparently no one commented uponthis David, F. N. (1962). Games, Gods and Gambling. Griffin,
error before? There are several possible explana- London.
tions,andnodoubteachheldforatleastonereader. Evans, F. B. (1961). Pepys, Newton, and Bernoulli trials.
Reader observations on recent discussions, in the series
(1) The letters were read superficially, with no at-
Questions and answers. Amer. Statist. 15 (1) 29.
tempt to parse the somewhat archaic language of
Gani,J.(1982).Newtonon“aquestiontouchingyedifferent
the logical proof, which after all points in the right odds upon certain given chances upon dice.” Math. Sci. 7
direction. (2) The language was puzzling and un- 61–66. MR0642167
clear to the reader (and Newton was not available Gjertsen, D.(1986).The Newton Handbook. Routledgeand
Kegan Paul, London.
to ask), but it was accepted since he was, after all,
Graves, R. P. (1889). Life of Sir William Rowan Hamilton
Isaac Newton, and the calculation clearly showed 3. Hodges, Figgis, Dublin. Reprinted 1975 by Arno Press,
he was sound on the important fundamentals. (3) New York.
The reader may even have seen that it was not a Hamza,K.(1995).Thesmallestuniformupperboundonthe
satisfactory argument, but drew back from accus- distancebetweenthemeanandthemedianofthebinomial
and Poisson distributions. Statist. Probab. Lett. 23 21–25.
ing Newton of error, particularly since he got the
MR1333373
numbers right.
Mosteller, F. (1965). Fifty Challenging Problems in Prob-
In a sense the argument is more interesting be- ability with Solutions. Addison–Wesley, Reading, MA.
cause it is wrong. Newton was thinking like a great MR0397810
probabilist—attemptinga“eureka”proofthatmade Pedoe, D. (1958). The Gentle Art of Mathematics. Macmil-
lan,NewYork.(ReprintsthefirsttwoofNewton’sletters.)
theissueclear inaflash.Whensuccessful,thisisthe
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highest form of mathematical art. That it failed is Pepys, S. (1825). Memoirs of Samuel Pepys, Esq. FRS 1,
no embarrassment; a simple argument can be won- 2. Henry Colburn, London. (Reprints the first of Pepys’
derful, but it can also create an illusion of under- letters and two of Newton’s replies.)
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Pepys, S.(1926).Private Correspondence and Miscellaneous
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no embarrassment. As Augustus De Morgan once Pepys Cockerell 1, 2. G. Bell and Sons, London. [This is
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