# The Newton-Pepys Problem – in a nutshell.

One of the most celebrated pieces of correspondence in the history of probability and gambling, and one of which I am particularly fond, involves an exchange of letters between the greatest diarist of all time, Samuel Pepys, and the greatest scientist of all time, Sir Isaac Newton.

The six letters exchanged between Pepys in London and Newton in Cambridge related to a problem posed to Newton by Pepys about gambling odds. The interchange took place between November 22 and December 23, 1693. The ostensible reason for Mr. Pepys’ interest was to encourage the thirst for truth of his young friend, Mr. Smith. Whether Sir Isaac believed that tale or not we shall never know. The real reason, however, was later revealed in a letter written to a confidante by Pepys indicating that he himself was about to stake 10 pounds, a considerable sum in 1693, on such a bet. Now we’re talking!

The first letter to Newton introduced Mr. Smith as a fellow with a “general reputation…in this towne (inferiour to none, but superiour to most) for his maistery [of]…Arithmetick”.

What emerged has come down to us as the aptly named Newton-Pepys problem.

Essentially, the question came down to this:

Which of the following three propositions has the greatest chance of success.

- Six fair dice are tossed independently and at least one ‘6’ appears
- 12 fair dice are tossed independently and at least two ‘6’s appear.
- 18 fair dice are tossed independently and at least three ‘6’s appear.

Pepys was convinced that C. had the highest probability and asked Newton to confirm this.

Newton chose A as the highest probability, then B, then C, and produced his calculations for Pepys, who wouldn’t accept them.

So who was right? Newton or Pepys?

Well, let’s see.

The first problem is the easiest to solve.

What is the probability of A?

Probability that one toss of a coin produces a ‘6’ = 1/6

So probability that one toss of a coin does not produce a ‘6’ = 5/6

So probability that six independent tosses of a coin produces no ‘6’ = (5/6)^{6}

So probability of AT LEAST one ‘6’ in 6 tosses = 1 – (5/6)^{6} = 0.6651

So far, so good.

The probability of problem B and probability of problem C are more difficult to calculate and involve use of the binomial distribution, though Newton derived the answers from first principles, by his method of ‘Progressions’.

Both methods give the same answer, but using the more modern binomial distribution is easier.

So let’s do it, along the way by introducing the idea of so-called ‘Bernoulli trials’.

The nice thing about a Bernoulli trial is that it has only two possible outcomes.

Each outcome can be framed as a ‘yes’ or ‘no’ question (success or failure).

Let probability of success = p.

Let probability of failure = 1-p.

Each trial is independent of the others and the probability of the two outcomes remains constant for every trial.

An example is tossing a coin. Will it lands heads?

Another example is rolling a die. Will it come up ‘6’?

Yes = success (S); No = failure (F).

Let probability of success, P (S) = p; probability of failure, P (F) = 1-p.

So the question: How many Bernoulli trials are needed to get to the first success?

This is straightforward, as the only way to need exactly five trials, for example, is to begin with four failures, i.e. FFFFS.

Probability of this = (1-p) (1-p) (1-p) (1-p) p = (1-p)^{4 }p

Similarly, the only way to need exactly six trials is to begin with five failures, i.e. FFFFFS.

Probability of this = (1-p) (1-p) (1-p) (1-p) (1-p) p = (1-p)^{5} p

More generally, the probability that success starts on trial number n =

(1-p)^{n-1} p

This is a geometric distribution. This distribution deals with the number of trials required for a single success.

But what is the chance that the first success takes AT LEAST some number of trials, say 12 trials?

One method is to add the probability of 12 trials to prob. of 13 trials to prob. of 14 trials to prob. of 15 trials, etc. …………………………

Easier method: The only time you will need ** at least **12 trials is when the first 11 trials are all failures, i.e. (1-p)

^{11}

In a sequence of Bernoulli trials, the probability that the first success takes ** at least **n trials is (1-p)

^{n-1}

Let’s take a couple of examples.

Probability that the first success (heads on coin toss) takes at least three trials (tosses of the coin)= (1-0.5)^{2} = 0.25

Probability that the first success (heads on coin toss) takes at least four trials (tosses of the coin)= (1-0.5)^{3} = 0.125

But so far we have only learned how to calculate the probability of one success in so many trials.

What if we want to know the probability of two, or three, or however many successes?

To take an example, what is the probability of exactly two ‘6’s in five throws of the die?

To determine this, we need to calculate the number of ways two ‘6’s can occur in five throws of the die, and multiply that by the probability of each of these ways occurring.

So, probability = number of ways something can occur multiplied by probability of each way occurring.

How many ways can we throw two ‘6’s in five throws of the die?

Where S = Success in throwing a ‘6’, F = Fail in throwing a ‘6’, we have:

SSFFF; SFSFF; SFFSF; SFFFS; FSSFF; FSFSF; FSFFS; FFSSF; FFSFS; FFFSS

So there are 10 ways of throwing two ‘6’s in five throws of the dice.

More formally, we are seeking to calculate how many ways 2 things can be chosen from 5. This is known as ‘5 Choose 2’, written as:

^{5 }C _{2}= 10

More generally, the number of ways k things can be chosen from n is:

^{n}C _{k} = n! / (n-k)! k!

n! (known as n factorial) = n (n-1) (n-2) … 1

k! (known as k factorial) = k (k-1) (k-2) … 1

Thus, ^{5}C _{2} = 5! / 3! 2! = 5x4x3x2x1 / (3x2x1x2x1) = 5×4/(2×1) = 20/2=10

So what is the probability of throwing exactly two ‘6’s in five throws of the die, in each of these ten cases? p is the probability of success. 1-p is the probability of failure.

In each case, the probability = p.p.(1-p).(1-p).(1-p)

= p^{2} (1-p)^{3}

Since there are ^{5} C _{2 }such sequences, the probability of exactly 2 ‘6’s =

10 p^{2 }(1-p)^{3}

Generally, in a fixed sequence of n Bernoulli trials, the probability of exactly r successes is:

^{n}C _{r} x p^{r} (1-p) ^{n-r}

This is the binomial distribution. Note that it requires that the probability of success on each trial be constant. It also requires only two possible outcomes.

So, for example, what is the chance of exactly 3 heads when a fair coin is tossed 5 times?

^{5}C _{3} x (1/2)^{3} x (1/2)^{2} = 10/32 = 5/16

And what is the chance of exactly 2 sixes when a fair die is rolled five times?

^{5 }C _{2}x (1/6)^{2} x (5/6)^{3} = 10 x 1/36 x 125/216 = 1250/7776 = 0.1608

So let’s now use the binomial distribution to solve the Newton-Pepys problem.

- What is the probability of obtaining
one six with 6 dice?*at least* - What is the probability of obtaining
two sixes with 12 dice?*at least* - What is the probability of obtaining
three sizes with 18 dice?*at least*

First, what is the probability of no sixes with 6 dice?

P (no sixes with six dice) = ^{n} C _{x }. (1/6)^{x} . (5/6)^{n-x, }x = 0,1,2,…,n

Where x is the number of successes.

So, probability of no successes (no sixes) with 6 dice =

n!/(n-k)!k! = 6!/(6-0)!0! x (1/6)^{0} . (5/6)^{6-0} = 6!/6! X 1 x 1 x (5/6)^{6 = }(5/6)^{6}

*Note that: 0! = 1*

*Here’s the proof: n! = n. (n-1)! *

*At n=1, 1! = 1. (1-1)! *

*So 1 = 0!*

So, where x is the number of sixes, probability of at least one six is equal to ‘1’ minus the probability of no sixes, which can be written as:

P (x≥ 1) = 1 – P(x=0) = 1 – (5/6)^{6 }= 0.665 (to three decimal places).

i.e. probability of at least one six = 1 minus the probability of no sixes.

That is a formal solution to Part 1 of the Newton-Pepys Problem.

*Now on to Part 2.*

Probability of at least two sixes with 12 dice is equal to ‘1’ minus the probability of no sixes minus the probability of exactly one six.

This can be written as:

P (x≥2) = 1 – P(x=0) – P(x=1)

P(x=0) in 12 throws of the dice = (5/6)^{12}

P (x=1) in 12 throws of the dice = ^{12} C _{1} . (1/6)^{1} . (5/6)^{11n}C _{k} = n! / (n-k)! k!

So ^{12} C _{1 }

= 12! / (12-1)! 1! = 12! / 11! 1! = 12

So, P (x≥2) = 1 – (5/6)^{12 }– 12. (1/6) . (5/6)^{11 }

= 1 – 0.112156654 – 2 . (0.134587985) = 0.887843346 – 0.26917597 =

= 0.618667376 = 0.619 (to 3 decimal places)

This is a formal solution to Part 2 of the Newton-Pepys Problem.

*Now on to Part 3.*

Probability of at least three sixes with 18 dice is equal to ‘1’ minus the probability of no sixes minus the probability of exactly one six minus the probability of at exactly two sixes.

This can be written as:

P (x≥3) = 1 – P(x=0) – P(x=1) – P(x=2)

P(x=0) in 18 throws of the dice = (5/6)^{18}

P (x=1) in 18 throws of the dice = ^{18} C _{1} . (1/6)^{1} . (5/6)^{17}

^{n}C _{k} = n! / (n-k)! k!

So ^{18} C _{1}

= 18! / (18-1)! 1! = 18

So P (x=1) = 18. (1/6)^{1} . (5/6)^{17}

P (x=2) = ^{18 }C _{2 .} (1/6)^{2} .(5/6)^{16}

^{18 }C _{2 }

_{ }= 18! / (18-2)! 2! = 18!/16! 2! = 18. (17/2)

So P (x=2) = 18. (17/2) (1/6)^{2 }(5/6)^{16}

So P(x=3) = 1 – P (x=0) – (P(x=1) – P (x=2)

P (x=0) = (5/6)^{18}

= 0.0375610365

P (x=1) = 18. 1/6. (0.0450732438) = 0.135219731

P (x=2) = 18. (17/2) (1/36) (0.0540878926) = 0.229873544

So P(x=3) = 1 – 0.0375610365 – 0.135219731 – 0.229873544 =

P(x≥3) = 0.597345689 = 0.597 (to 3 decimal places, )

This is a formal solution to Part 3 of the Newton-Pepys Problem.

So, to re-state the Newton-Pepys problem.

Which of the following three propositions has the greatest chance of success?

- Six fair dice are tossed independently and at least one ‘6’ appears.
- 12 fair dice are tossed independently and at least two ‘6’s appear.
- 18 fair dice are tossed independently and at least three ‘6’s appear.

Pepys was convinced that C. had the highest probability and asked Newton to confirm this.

Newton chose A, then B, then C, and produced his calculations for Pepys, who wouldn’t accept them.

So who was right? Newton or Pepys?

According to our calculations, what is the probability of A? 0.665

What is the probability of B? 0.619

What is the probability of C? 0.597

So Sir Isaac’s solution was right. Samuel Pepys was wrong, a wrong compounded by refusing to accept Newton’s solution. How much he lost gambling on his misjudgement is mired in the mists of history. The Newton-Pepys Problem is not, and continues to tease our brains to this very day.

*References and Links*

Newton and Pepys. DataGenetics. http://datagenetics.com/blog/february12014/index.html

Newton-Pepys problem. Wikipedia. https://en.wikipedia.org/wiki/Newton%E2%80%93Pepys_problem