Newton–Pepys problem
The Newton–Pepys problem is a probability problem concerning the probability of throwing sixes from a certain number of dice.[1]
In 1693 Samuel Pepys and Isaac Newton corresponded over a problem posed by Pepys in relation to a wager he planned to make. The problem was:
- 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.[2]
Pepys initially thought that outcome C had the highest probability, but Newton correctly concluded that outcome A actually has the highest probability.
Solution
The probabilities of outcomes A, B and C are:[1]
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:
As n grows, P(N) decreases monotonically towards an asymptotic limit of 1/2.
Example in R
The solution outlined above can be implemented in R as follows:
# Probability of one "six" occurring in a fair die. p <- as.numeric(1/6) # Three possible scenarios. scenarios <- c(1, 2, 3) for (i in scenarios) { # Accumulate the probability that fewer than the target # of sixes occurs. x <- 0 # Total number of dice. n <- 6 * i # Calculate the probability for each possible # of sixes that is less than the goal. for (j in 0:(i-1)) { x <- x + dbinom(j, n, p) } print(paste("Probability of at least", i, "six in", n, "fair dice:", 1 - x)) }
which results in:
[1] "Probability of at least 1 six in 6 fair dice: 0.665102023319616" [1] "Probability of at least 2 six in 12 fair dice: 0.618667373732309" [1] "Probability of at least 3 six in 18 fair dice: 0.597345685947723"
Newton's explanation
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.
Generalizations
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 is the probability of having at least k correct selections when throwing exactly n dice. Then the original Newton–Pepys problem can be generalized as follows:
Let be natural positive numbers s.t.
. Is then
not smaller than
for all n, p, k?
Notice that, with this notation, the original Newton–Pepys problem reads as: is ?
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)):[3]
If are positive natural numbers, and
, then
.
If are positive natural numbers, and
, then
.
(from Varagnolo, Pillonetto and Schenato (2013)):[4]
If are positive natural numbers, and
then
.
References
- ↑ 1.0 1.1 Weisstein, Eric W., "Newton-Pepys Problem", MathWorld.
- ↑ Isaac Newton as a Probabilist, Stephen Stigler, University of Chicago
- ↑ Chaundy, T.W., Bullard, J.E., 1960. John Smith’s Problem. The Mathematical Gazette 44, 253-260.
- ↑ D. Varagnolo, L. Schenato, G. Pillonetto, 2013. A variation of the Newton–Pepys problem and its connections to size-estimation problems. Statistics & Probability Letters 83 (5), 1472-1478.