搜搜考试网用python编写“生日悖论”的解决方法
来源：学生作业帮 编辑：搜搜考试网作业帮 分类：综合作业 时间：2024/05/10 21:30:17
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.
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.
import randomdef 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：
亲测能用：