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Newton–Pepys problem
From Wikipedia, the free encyclopedia
Jump to navigationJump to searchThe 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.
请你通过模拟的方法找到问题的答案。
要求:
1. 说明模拟方法和过程;
2. 提交代码,并说明如何运行程序,得到问题答案。
-}
import System.Random
import Text.Printf
main::IO()
main=
do
let n=1000
a<-getProbablity eventA n
b<-getProbablity eventB n
c<-getProbablity eventC n
putStrLn(printf "P(A)=%.7f,P(B)=%.7f,P(C)=%.7f" a b c)
getProbablity::(IO Bool)->Int->IO Float
getProbablity event n=do
xs<-happen event n
return ((fromIntegral xs)/(fromIntegral n))
happen::(IO Bool)->Int->(IO Int)
happen event 0=return 0
happen event n=
do
e<-event
y<-happen event (n-1)
if e==True
then return (1+y)
else return y
getDices::Int->IO [Int]
getDices 0=return []
getDices x=do
y<-randomRIO(1,6)
ys<-getDices(x-1)
return (y:ys)
eventA::IO Bool
eventA=
do
xs<-getDices 6
return ((countSix xs)>0)
eventB::IO Bool
eventB=
do
xs<-getDices 12
return ((countSix xs)>1)
eventC::IO Bool
eventC=
do
xs<-getDices 18
return ((countSix xs)>2)
countSix::[Int]->Int
countSix []=0
countSix (x:xs)=
if x==6 then (countSix xs)+1
else countSix xs