- Newton–Pepys problem
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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?
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- 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.
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.
References
- ^ a b Weisstein, Eric W., "Newton-Pepys Problem" from MathWorld.
- ^ Isaac Newton as a Probabilist, Stephen Stigler, University of Chicago
Categories:- Factorial and binomial topics
- Named probability problems
- Isaac Newton
- Mathematical problems
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