Review questions
The mathematical problem comes from an entertainment program. One program participants and a moderator, there are three closed doors in front of the participants, two doors behind which is empty, a door is left over after a Ferrari sports car.
Moderator knew behind which door sports car, but participants do not know. At this time, participants candidates a door, if the choice is behind that door sports car, sports car given to the participants as a reward.
Until this problem is very simple: a total of three doors, participants were randomized to choose, be sure chance of winning is 1/3.
Here is the crux of the matter, When selecting participants, not being the first to open the door, then host the remaining two doors among an open, is Buddhism .
At this time, the host gave participants the opportunity to re-select: You can insist on the door just selected (is number two in the figure), you can also change another door did not open the door (it is No. 1 in the figure).
If you are a game participant, how do you choose the winning rate higher? Winning rate and how much?
Incredible answer
Just see the problem, but also quite objectionable:
Such questions have to ask it? There are three doors when winning rate is 1/3; now rule out the door, leaving two doors a second election, for door or gate does not change, the winning rate of 50% should all fishes it?
However, the correct answer is "counter-intuitive" of:
In other door winning rate is 2/3
The award-winning not-for-gate rate is 1/3
What's the hell? This is simply incredible ah!
"When the last remaining two doors when winning at this time of the discussion should be a separate event, and how to select participants before, and the moderator opens Buddhism these things should be completely independent of fishes it? Since it is an independent event then a second election, is it 50% winning rate is not it? "
First need to be clear, the rate on the award-winning "for the door," we are not talking about a separate incident, must be the first choice as a base. In the probabilistic them, this is called a conditional probability .
So, in the end what is it an isolated incident?
举个例子，假如游戏的参与者本来是小灰，当小灰选择一扇门，而主持人打开一扇空门之后，不明真相的小红从外面跑了进来。小红并不知道当初小灰选择的是哪一扇门，只知道剩下两扇关闭的门中，有一扇门藏有奖励。
那么此时对于小红来说，无论选择哪一扇门，获奖率都是50%，因为小红是在做独立的选择，而不是基于第一次的选择来”换门”。
这才是所谓的 “独立事件”。
从多个角度来思考
那么，在“换门”的情况下，获奖率2/3又是怎么来的呢？
小灰上周的漫画里，利用了基于“贝叶斯理论”的思想来分析换与不换的获奖率：
直白地讲，就是把第一次选择和第二次选择的所有情况进行细化，分析出每一种情况下的条件概率，再把这些概率进行加总，得到了最终的结果：
不换门的获奖率 = （1/3 X 100%）+（1/3 X 0%）+（1/3 X 0%）=1/3
换门的获奖率 = （1/3 X 0%）+（1/3 X 100%）+（1/3 X 100%）=2/3
用代码来验证
上面所说的都仅仅是理论分析，我们不妨用代码来实际检验一下。
1 package test; 2 3 import java.util.ArrayList; 4 import java.util.Arrays; 5 import java.util.List; 6 import java.util.Random; 7 8 /** 9 * @author zsh 10 * @site qqzsh.top 11 * @create 2019-08-02 9:28 12 * @description 三门问题 13 * 这个数学问题来源于一个娱乐节目。节目中有一位参与者和一位主持人，在参与者的面前有三扇关闭的门，其中两扇门的后面是空的，剩下一扇门后是一辆法拉利跑车。 14 * 主持人知道哪一扇门后面有跑车，但参与者不知道。此时让参与者人选一扇门，如果选择的是后面有跑车的那扇门，跑车就作为奖励送给参与者。 15 * 下面是问题的重点，当参与者进行选择以后，暂时先不打开这扇门，接下来主持人把剩下两扇门当中的一扇打开，是空门。 16 * 此时主持人给了参与者重新选择的机会：可以坚持刚才选择的门（在图中是2号门），也可以换另一扇没有打开的门（在图中是1号门）。 17 * 如果你是游戏参与者，你怎样选择的获奖率更大？获奖率又是多少？ 18 */ 19 public class threeDoorsTest { 20 public static void main(String[] args) { 21 //换门的获奖总次数 22 int changeWinCount = 0; 23 //不换门的获奖总次数 24 int unChangeWinCount = 0; 25 Random random = new Random(); 26 for (int i = 0; i < 100000; i++) { 27 List<Integer> doors = new ArrayList<>(Arrays.asList(0,1,2)); 28 int bonusDoor = random.nextInt(3); 29 int selectedDoor = random.nextInt(3); 30 //主持人打开一扇空门 31 for (int j = 0; j < doors.size(); j++) { 32 if (doors.get(j) != selectedDoor && doors.get(j) != bonusDoor){ 33 doors.remove(j); 34 break; 35 } 36 } 37 //获得换门的序号,此时集合中就剩两个元素 38 int changedDoor = doors.get(0); 39 if (changedDoor == selectedDoor){ 40 changedDoor = doors.get(1); 41 } 42 //如果不换门有奖，unChangeWinCount加1；如果换门有奖，changeWinCount加1 43 if (selectedDoor == bonusDoor){ 44 unChangeWinCount++; 45 }else if (changedDoor == bonusDoor){ 46 changeWinCount++; 47 } 48 } 49 System.out.println("不换门获奖总次数："+unChangeWinCount+"，所占比例："+(float)unChangeWinCount/100000); 50 System.out.println("换门获奖总次数："+changeWinCount+"，所占比例："+(float)changeWinCount/100000); 51 } 52 }
运行结果：
数据结果显而易见，不换门获奖的比例占了约1/3，换门获奖的比例占了2/3。
写在最后
三门问题真的是一个非常有意思的数学问题。在上个世纪的美国，这个问题刚刚被提出的时候，也遭到过许多人的质疑，这些质疑者中有教师，有学者，甚至有数学家。后来人们经过了许多次实验，才逐渐达成共识。
质疑精神是值得鼓励的，有了质疑才能让思想进一步完善。
最后，让我们来致敬一下 “三门问题” 的提出者，集才华和美貌于一身的天才人物 玛丽莲·沃斯·莎凡特。