用python编写“生日悖论”的解决方法that if 23 people are selected at random,there is better than 50% chance that at least two of them will have the same birthday (not considering the birth year).You are to write a Python function to simu

用python编写“生日悖论”的解决方法that if 23 people are selected at random,there is better than 50% chance that at least two of them will have the same birthday (not considering the birth year).You are to write a Python function to simu
用python编写“生日悖论”的解决方法
that if 23 people are selected at random,there is better than 50% chance that at
least two of them will have the same birthday (not considering the birth year).
You are to write a Python function to simulate selecting n people at random
and checking the probability of having at least two people with the same
birthday.You should ignore the leap years and assume 365-day years.To be
more specific,devise a Python function,call it bdp(n,k),that once invoked,will
select n numbers with replacement from the set of numbers 1 through 365
inclusive,determine if two or more of the numbers selected are the same (call
it a hit),and repeat this task k times,and finally return the percentage of the
hits.Plot your results similar to the graph in the above URL.

用python编写“生日悖论”的解决方法that if 23 people are selected at random,there is better than 50% chance that at least two of them will have the same birthday (not considering the birth year).You are to write a Python function to simu

import random

def bdp(n,k):

\x09    cv = []

\x09    for i in range(k):

\x09\x09        m = []

\x09\x09        for j in range(n):

\x09\x09\x09            m.append(random.randint(1,365))

\x09\x09        counter = 0

\x09\x09            for k1 in m:

\x09\x09\x09                for k2 in m:

\x09\x09\x09\x09                    if k1 == k2:

\x09\x09\x09\x09\x09                        counter += 1

\x09\x09        cv.append(float(counter/2)/float(n))

\x09        ss = 0

\x09        for i in cv:

\x09\x09            ss += i

        return ss/float(len(cv))

亲测能用：