The Newton–Pepys problem is a probability problem concerning the probability of throwing sixes from a certain number of dice.
In 1693 Samuel Pepys and Isaac Newton corresponded over a problem posed to Pepys by a school teacher named John Smith.[1] The problem was:
Pepys initially thought that outcome C had the highest probability, but Newton correctly concluded that outcome A actually has the highest probability.
The probabilities of outcomes A, B and C are:
| P(A)=1-\left( | 5 |
| 6 |
\right)6=
| 31031 | |
| 46656 |
≈ 0.6651,
| ||||
| P(B)=1-\sum | ||||
| x=0 |
| ||||
| \right) |
\right)12-x=
| 1346704211 | |
| 2176782336 |
≈ 0.6187,
| ||||
| P(C)=1-\sum | ||||
| x=0 |
| ||||
| \right) |
\right)18-x=
| 15166600495229 | |
| 25389989167104 |
≈ 0.5973.
These results may be obtained by applying the binomial distribution (although Newton obtained them from first principles). In general, if P(n) is the probability of throwing at least n sixes with 6n dice, then:
| n-1 | ||
| P(n)=1-\sum | \binom{6n}{x}\left( | |
| x=0 |
| 1 | |
| 6 |
| ||||
| \right) |
\right)6n-x.
As n grows, P(n) decreases monotonically towards an asymptotic limit of 1/2.
The solution outlined above can be implemented in R as follows:
Although Newton correctly calculated the odds of each bet, he provided a separate intuitive explanation to Pepys. He imagined that B and C toss their dice in groups of six, and said that 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 assumes that a group does not produce more than one 6, so it does not actually correspond to the original problem.
A natural generalization of the problem is to consider n non-necessarily fair dice, with p the probability that each die will select the 6 face when thrown (notice that actually the number of faces of the dice and which face should be selected are irrelevant). If r is the total number of dice selecting the 6 face, then
P(r\gek;n,p)
Let
\nu1,\nu2
\nu1\le\nu2
P(r\ge\nu1k;\nu1n,p)
P(r\ge\nu2k;\nu2n,p)
Notice that, with this notation, the original Newton–Pepys problem reads as: is
P(r\ge1;6,1/6)\geP(r\ge2;12,1/6)\geP(r\ge3;18,1/6)
As noticed in Rubin and Evans (1961), there are no uniform answers to the generalized Newton–Pepys problem since answers depend on k, n and p. There are nonetheless some variations of the previous questions that admit uniform answers:
(from Chaundy and Bullard (1960)):[2]
If
k1,k2,n
k1<k2
P(r\gek1;k1n,
| 1 | |
| n |
)>P(r\gek2;k2n,
| 1 | |
| n |
)
If
k,n1,n2
n1<n2
P(r\gek;kn1,
| 1 | |
| n1 |
)>P(r\gek;kn2,
| 1 | |
| n2 |
)
(from Varagnolo, Pillonetto and Schenato (2013)):[3]
If
\nu1,\nu2,n,k
\nu1\le\nu2,k\len,p\in[0,1]
P(r=\nu1k;\nu1n,p)\geP(r=\nu2k;\nu2n,p)