# The Three-door probl...

1.  The problem setting

“三个门标记为1,2,3，随机选取一个门放置奖品。游戏开始： 首先，选手任选一个门，如1号。然后，知道奖品在哪个门后的主持人会打开另外两个门中的一个，使得打开的那个门不含有奖品，如3号。此时，选手有更换选择的机会，如放弃原先选择的1号门，改选2号门。问：选手改选对自己是否有利？”

2.  The problem solving

• 把上述问题形式化如下。

H = 1 表示1号门后有奖品；H = 2 表示2号门后有奖品；H = 3 表示3号门后有奖品。

D = 2 表示主持人打开2号门； D = 3 表示主持人打开3号门。

• 由题意可得如下已知概率。

P(H=1) = 1/3;  P(H=2) = 1/3;  P(H=3) = 1/3;

P(D=2|H=1) = 1/2;  P(D=3|H=1) = 1/2;

P(D=2|H=2) = 0;  P(D=3|H=2) = 1;

P(D=2|H=3) = 1;  P(D=3|H=3) = 0.

• 选手所关心的概率是：P(H=1|D=3)， P(H=2|D=3)。由贝叶斯定理：

P(H=1|D=3) = P(H=1) * P(D=3|H=1) / P(D=3) = (1/3 * 1/2) / P(D=3);

P(H=2|D=3) = P(H=2) * P(D=3|H=2) / P(D=3) = ( 1/3 * 1 ) / P(D=3);

P(H=1|D=3)  = 1/3

P(H=2|D=3)  = 2/3

P(H=2|D=3) / P(H=1|D=3) = 2 : 1

3. 模拟实验 1） 三种选择：

There are three possible ways the game could play out: 

• You pick goat 1. Monty shows you goat 2. You switch. You win the car. 
• You pick goat 2. Monty shows you goat 1. You switch. You win the car. 
• You pick the car. Monty shows you a goat. You switch. You get the other goat.

So by switching, you win two out of three times.

2）极限情况：

To consider the limiting case of many doors, say a thousand of them.

• You pick door #1. Like every other door, it's a one-in-a-thousand long shot. Then Monty opens 998 doors with goats behind all of them, leaving only yours and one other, say door #723.
• ''Now,'' he asks, ''would you like to keep what's behind door #1, or switch to door #723?''

Can you see he's practically telling you where the car is? Switch to #723! Or do you really have so much faith in your guessing ability that you believe #1 is worth keeping?

3）要点：主持人事先知道奖品在哪里，他总是为你排除掉一些山羊，剩下奖品。

Intuition

A well-known example is the three-box problem. A contestant in a game is
told that one of X, Y, or Z contains a prize. The contestant chooses X but does
not open it. The host then opens Y and shows that it is empty. Should the contestant
change to Z or stay with X?

Intuition says that it doesn't matter, as the
probability of X containing the prize is 1/2. Careful analysis of the alternatives
shows that Z contains the prize with probability 2/3, but when this problem was
presented many mathematicians publicly argued that 1/2 was correct.

In a variation of this error, suppose that person P has tossed two coins.
Person Q asks if one of them is heads, and P says yes. Then the intuitive estimate
of the probability that the other coin is heads is 1/2, on the basis that
the status of one coin is independent of the other, but again this is wrong. The
correct response is 1/3. The reason is that there are four possible configurations: