Before in which movie there is this bridge section ~ This article reproduced from Nju_hupeng "zt a classical probability problem of the ultimate solution" today in the essence of the classical probability of the discussion, a long list of posts, although the "standard" solution is there, but the standard solution method in the continuous questioning before convincing enough. After some thinking, I think I found a more convincing method, that is, using Bayesian formula to avoid a priori posterior entanglement. I do not know the logic of the No, welcome to correct me:
Classic topics:
There are three doors, there is a car inside, if you choose the right to get this car,
When the test taker chooses a door, the host opens another door, empty.
Ask the candidate if they want to change their options.
Suppose the host knows the door where the car is located.
The classical solution (the conclusion is correct):
The probability of choosing the right first time is 1/3.
So the probability of a car in another two doors is 2/3.
The moderator pointed out a door, and if you start to pick the wrong (2/3 probability), then the rest of the door 100% has a car
If you choose Right (1/3) for the first time, there are 100% cars left in that door.
So after the host prompts, you do not change the correct probability is 1/3*100%+2/3*0=1/3
The right probability for you to change is 1/3*0+2/3*100%=2/3.
Let me first talk about the problem of the classical solution. The question of this solution is that, now the host has opened a Stargate (and the host is deliberately open the door), in this "message" appears, can also say that the probability of the first choice is 2/3? Does this post-mortem fact not change our view of the priori probabilities? The answer is yes. More specifically, when the host opens a door, the probability of choosing the wrong is not necessarily equal to 2/3.
Start from the beginning. Suppose I chose B door, assuming that the host opened the C door, then under what circumstances would he open C gate?
If A has a car (a priori probability p=1/3), that host 100% opens the C gate (he will obviously not open B);
If B has a car (prior probability P=1/3), then the host has a and C two options, assuming that he opened the probability of K (General K=1/2, but we temporarily set it to a variable);
If C has a car (prior probability P=1/3), then the host open C probability is 0 (as long as he is not silly ...) ）
Known he opened C, that according to the Bayes formula--here P (m| N) indicates the probability of the M event occurring when the N event occurs:
P (b with Car | C open) =p (b with Car | C Open)/P (c open) = (1/3) *k/[(1/3) *1+ (1/3) *k] = K/(k+1)
When is the value equal to 1/3 (i.e. the assumption in the classical solution)? Only when K=1/2. This is generally the case. But if the host has a preference, say he likes to open the right door (assuming C on the right), set K=3/4, then B has the probability of a car becomes 3/5, no longer is 1/3, after the fact changed the estimate of the prior probability!
But this does not change the right choice, we should still re-elect a door, explained as follows:
P (A with car | C open) = P (A has a car | C Open)/P (c open) = (1/3) */[(1/3) *1+ (1/3) *k] =1/(k+1)
and K < 1 (assuming that the host does not have extreme to non-C not selected degree),
So there's always P (b has a car | C Open) < P (A with car | C Open)
A the probability of a car is always bigger than B, we should change the choice.
The focus of this solution is to consider the effect of the fact that C is opened, thus eliminating the disturbance of a priori posteriori.
(to a thief) a classic probability problem
