Discussion on goat's car

A very classic problem, suitable for testing thinking ~

Car and goats problem is also known as Monty Hall problem or three problems. This problem comes from the American TV and entertainment program Let's Make a Deal. The problem is named by the host Monty Hall ). The problem is: contestants will see three closed doors, one of which is followed by a car, and the other with a car is selected to win the car, there is a goat in each of the other two doors. When the contestant chooses a door but does not start it, the host opens one of the remaining two doors to reveal one goat. Afterwards, the host will ask the contestants if they want to change to another closed door. The constraints are as follows: the contestant will be asked whether to keep his original choice, or if he chooses the other one, which strategy has a high chance of correct change or not.

This is a very interesting question. I saw it in a blog for the first time. Later I saw some people mentioned the same problem in the face of Goldman Sachs (selecting a certain color ball ). I have discussed this issue many times with my classmates, but I have not reached an agreement. Sometimes the deeper I think, the more confused I get. Isn't it 1/2 if it's not changed? This is the most confusing place...

Thought 1: reverse thinking:

If the host asks you if you want to change it, you will never change it. You have participated in the change for 1000 times, and the host asks you for the change for 1000 times. Well, the prompt from the host has no effect on you. That is to say, the host opens one of the remaining two courses, which has no effect on you. What does it mean? The only thing you do is to choose one from the three. if you participate 1000 times, you will probably win 333 prizes. The result is that the probability of winning is 1/3.

So if you change it, you will be entitled to a 2/3 probability prize. This is inferred from the opposite.

Idea 2: direct method, equivalent push conversion. I think this is the best way to understand ..

Let's assume that you have participated in 1000 such events. It will be very interesting to switch the problem to the following. You have chosen one of the three doors. Now the host asks you: give you a chance, and you can give up one door in your hand, choose the two doors on my side (both of them belong to you, and there may be a car, a sheep, or two sheep in the back ), but you have to give me a sheep behind your two doors. Don't you change it?

Change, of course, you can choose two doors at a time. The probability is as high as 2/3.

Thought 3: Under what circumstances is 1/2, why do people think 1/2 is not 1/2 here ??

Many hesitant people are thinking, why not 1/2? Because one goat is excluded, the remaining two choices should be 1/2. Then we will discuss why it is 1/2 instead of 1/2. How can it be?

After you select, the host will exclude a door (for example, on the 3rd, you choose the 1st). Then you can make a selection between the 1st and 2nd. If you do not know whether to change, you decide to vote for a coin (that is, re-random selection), you choose 1 on the front, and 2 on the back.In this case, the probability of your car being in the middle is 1/2.But if you don't pay for coins,But does not change every time (Repeat 1000 times). What is the difference between the three and you? The probability of winning the prize is 1/3.

Train of Thought 4: columns and other probability tables

The probability here is critical.

Assume that the number of the door is 1, 2, 3, and the vehicle appears after the door is 1/3 of the probability of equality. The first choice of the contestant is equal probability. The host is an expert, because he can always choose a goat behind the rest of the door based on the player's choice. The host prompts are not listed here. A total of 3 (contestant selection) x 2 (change or not) x 3 (vehicle distribution) = 18 (CASE)

When the number is changed, the probability of winning is 2/3... Do not change 1/3 ....

There should be 18 cases in the table, a total of 36 experiments (considering the host prompts, that is, each row is twice)

First Selection	Vehicle Distribution	Change or not	Final Choice	Winning?
1	1	Yes	2/3	No
2	1	Yes	1	Yes
3	1	Yes	1	Yes
1	1	No	1	Yes
2	1	No	2	No
3	1	No	3	No
1	2	Yes	2	Yes
2	2	Yes	1/3	No
3	2	Yes	2	Yes
1	2	No	1	No
2	2	No	2	Yes
3	2	No	3	No
1	3	Yes	3	Yes
2	3	Yes	3	Yes
3	3	Yes	1/2	No
1	3	No	1	No
2	3	No	2	No
3	3	No	3	Yes

First think about these ~ These are some of my personal thoughts, which are not strict in mathematics.

Follow-up:

Many people leave a message saying that it is 1/2. My analysis is not strict. They didn't make sense, but I still agree with the 2/3 answer. In fact, my classmates and I have never reached a clear agreement on this issue. At that time, even after the American TV broadcast, there was a doctor who sent a letter to refute this question, which was twice the probability. Here I reference A matrix67 blog: (the original article is here)

A person named S. K. Stein wrote a book named strength in numbers. The book cited the Monty Hall dilemma problem when talking about how Mathematicians solve the problem step by step. He said in his book: If, after thinking some more about the question, you still are not sure about the answer and are not ready to explain it, then do the following. (keep
In mind that just citing experimental data is not an explanation. The data may convince you that something is true, but they do not explain it .)
Get one more canister and perform a similar experiment, using four canisters instead of three. put a wad of paper in one canister. after your friend chooses a canister, look in the remaining three and show the friend two empty canisters. the friend then
Faces a choice between the two other canisters. Carry out the same experiments as before. Think over the results you get. What do they suggest? Do you see a way to explain ain what happens?
Please Ming these experiments not only gives you some clues, it also slows you down from the common frenzy of everyday life, so you can focus on just one thing for a period of time.
If you still do not see how to explain ain what is going on, then use ten can-isters. put the wad in one of them. after your friend chooses a canister, look in the other nine. show your friend eight empty canisters out of those nine and remove all eight. again
That leaves just two canisters. Conducting CT a similar experiment.
I am confident that you will solve this problem, so confident that I do not include the answer anywhere in the book, not even in fine print upside down hidden in the back matter. you mill probably, along the way, calculate the fraction of times that switching
Will pick the car and the fraction of times that not switching will pick the car. using these fractions, you will be able to explain the brainteaser completely. then you will have to admit that you can think mathematically. you just needed the opportunity.

Simply put, Stein wants to express this meaning.He suggested that those who could not think of the answer to the Monty Hall dilemma question should not use the mathematical method to solve the problem. They should first perform several experiments to learn some perceptual knowledge. Call a friend a host in the game and put one thing in one of the three jars. Try it several times more. If the experiment has not yet been inspired, he proposed a very enlightening deformation of the experiment: the rule remains unchanged, but the three can is changed to four. Your friend will open the other two empty jars after you select one and ask if you want to change one. If there is no inspiration, simply turn four cans into ten. If you do not have any idea, you will be completely stupid. Think about it. If three games become ten, the chance of selecting a car is even more slim. After the host opens the other eight door with sheep, you will be stupid if you don't change it. If it was me, I would not hesitate to switch to another door. Yes, as one sheep runs out, I will be more and more excited. I thought, the rest of the door must be a car. Here is a very basic idea: If I select a sheep first, it will become a car after I change it. If I select a car first, it will become a sheep after I change the door. Since most of the first choice is sheep, why don't I change it?
According to this idea, we get that in the Monty Hall dilemma problem, the probability of selecting a car for the first time is 1/3. Obviously, the probability of a car in another door is 2/3. Therefore, I have a 2/3 chance to get the car, but not a 1/3 chance to get the car.
This problem should have ended here, but there is another question: why does the host's chance of changing his choice when opening a door? In fact, the reason is very simple, and the probability itself is not changed, just because the host has a choice when opening the door. This reduces the number of possible cases.