中国DOS联盟

This is a very classic probability problem:
There are three doors, and one of them has a car. If you choose the correct one, you can get this car.
When the test taker selects one door, the host opens another empty door.
The question is whether the test taker should change the choice.
Assume that the host knows the door where the car is located.

No need to replace, it's very simple. If the selected one has no car, the host won't give you another chance.

Hehe. Obviously, that host just wanted to see if the candidate's determination was strong enough~

DD, it's not to be said like that. KO said it's a probability problem. Here's my analysis.
If the one chosen has no car, then if the host lets you choose again, you have a 50% probability. If he doesn't let you choose again, you only have a 0% probability.
If the one chosen has a car, if the host lets you choose again, you only have a 50% probability. If he doesn't let you choose again, you have a 100% chance.
Under normal circumstances, when there's a car, the host will let you choose again. When there's no car, the host won't let you choose again.

It is obvious that before making a choice, the chance of having a car behind each door is 1/3. But after making a choice, and after the host opens an empty door, will the probability of each door change? If it does, how does it change?

When no door has been opened, the probability of each door is 33%. After one door is opened and it's not the one, then the probability of the remaining two doors is raised to 50%.

The sender: warreni (Fat Cat), posting area: IQDoor
Subject: Re: Could any expert explain the thinking behind that probability problem in detail?
Posted: BBS Tsinghua University Station (Fri Jul 27 01:44:26 2001)

This is my understanding:
You initially choose one door (A), and the probability that there is a car behind it is 1/3
The probability that there is a car behind the other two doors B and C is 2/3
No matter which empty door the host opens among B and C, the above two probabilities will not change
Suppose the host opens B, which is empty, then the probability that there is a car behind C is 2/3
So you should change your choice

Sender: crystal41923 (Brain not working well), Message Area: IQDoor
Subject: Re: Which expert can explain the detailed idea of that probability problem?
Posted: BBS Tsinghua University Station (Fri Jul 27 13:19:03 2001)

You can understand it clearly by dividing it.

Because the probability that the car is in door A, B, and C is all 1/3.

Then we can get: If the car is in A, the probability of getting the car by changing the choice is 0.
If the car is in B, the probability of getting the car by changing the choice is 1.
If the car is in C, it is the same as B.
So the probability of getting the car by changing the choice is 2/3.

The sender: ufo (Tiger sits alone listening to the wind howling), in the forum area: IQDoor
Subject: Re: Which expert can explain in detail the thinking of that probability problem?
Posted: BBS Tsinghua Station (Fri Jul 27 23:28:14 2001)

There is a problem with a reasoning.
【In reply to warreni (Fat Cat)】
: I think like this:
: You initially choose a door (A), the probability that there is a car in it is 1/3
: And the probability that there is a car in the other two doors B, C is 2/3
: No matter which empty door among B, C the host opens, the above two probabilities will not change
: Suppose the host opens B, which is empty, then the probability that there is a car in C is 2/3
This step I think is not valid.

The premise that the probability of having a car in B C is 2/3 earlier is based on three unknowns; when you know that there is no car in B, the unknowns are only A and C two, how do you explain that the probability of having a car in B/C is still 2/3?

Imagine the problem becomes like this: There are three doors A B C, it is known that there is a car in one of the three doors, and there is no car in B, then will you choose A or C?
: So you should change the choice

The sender: warreni (Fat Cat), in the forum area: IQDoor
Subject: Re: Which expert can explain the train of thought of that probability problem in detail?
Posted: BBS Tsinghua University Station (Sat Jul 28 23:41:48 2001)

What is the relationship between probability and the number of unknowns?
As long as what the host does is a definite event and does not change the position of the car,
the sum of the probabilities that there is a car in B and C should not change

【From ufo (Tiger sits alone listening to the wind roar)】
: There is a problem in the reasoning.
: I think this step is not valid.
: The premise that the probability that there is a car in B and C is 2/3 mentioned earlier is based on there being three unknowns; when you already know that there is no car in B, there are only two unknowns left, A and C. How can you say that the probability that there is a car in B/C is still 2/3?

: Imagine the question is changed to this: There are three doors A, B, C. It is known that there is a car in one of the three doors, and there is no car in B. Then will you choose A or C?

There is no car in B, that is, the probability that there is a car in B is zero, and the probabilities that there is a car in A and C are 1/2 respectively

Sender: take (The black dream afraid of being awakened), Board: IQDoor
Subject: Re: Which expert can explain the train of thought of that probability problem in detail?
Posted: BBS Tsinghua Station (Sat Jul 28 23:58:02 2001)

Is there a problem with this consideration?
You mean that under this premise, the probability that there is a car in (B+C) remains unchanged, still 2/3,
but you "forcefully" changed the probability in C alone from 1/3 to 2/3.
Then I can also imagine like this: I ensure that the probability that there is a car in (A+B+C) remains unchanged, always 1,
and after finding that there is a car in B, the probabilities of A and C are both changed to 1/2.

In fact, if we consider it completely from the perspective of probability, for the lottery participant, the host has an equal chance of opening door B and door C. So the probability that there is a car in C is 2/3 × 1/2, or 1/3

Suggest you understand it this way: Suppose you do 300 experiments. If you always choose A each time, you should win the prize about 100 times according to probability. But if you always first choose A and then change your choice, you should win the prize about 200 times (because there are only these two situations). Also, because the host should have an equal chance to open B and C, when you change your choice, you should win the prize about 100 times for B and 100 times for C respectively. Up to this point, you will find that the probabilities of having a car behind doors A, B, and C are equal, all 1/3.

No, this must be taking advantage.

The probability of the door you initially chose remains unchanged, but after the host opens a door, the probability of the other unselected doors changes. This is not a matter of changing from 33% to 50%, but from 33% to 66%.

I can give an extreme example. If there are 10,000 doors. You initially chose one. Of course, you won't be so accurate, actually only a 1/10,000 chance. But the host opens the other 9,998 doors, then do you change?

Nonsense, of course you change. Intuitively, you know that the probability of the other door has become extremely high. Actually it's 9999/10000.

People who choose not to change are because they have been misled by 50%.