If you want to reprint, need to indicate the source: http://blog.csdn.net/xiazdong The source of this question is the C.2-9 and C.2-10 in appendix C of the introduction to algorithms, which is a very famous question:
Three Problems, Also known
Montihall
Problem description:
First Statement: if you are a player, the prize of this game is placed behind one of the three backdrop. If you choose the right one, you will win the prize, if you have selected a canvas, but before you set off the canvas, the host sets off one of the other two backdrop to let you know that the canvas is empty, ask if you want to change the backdrop. If you want to change the backdrop, how will your chances change?
Statement 2: a prison guard randomly chooses one of the three criminals to release and the other two will be put to death. The guard knows whether a person will be released, but does not allow the criminal to have any information about his/her status. Let's name the criminals X, Y, and Z respectively. Criminal x privately asked the guard Y or Z which would be killed because he knew that at least one of them would die, and the guard could not disclose any information about his own status. The guard told X that Y would be killed. X is very happy because he thinks that he or Z will be released, which means that he is released with a probability of 1/2. Is he correct? Or is his chance still 1/3? Please explain.
Answer:
Event A: win prize event B: there is no prize in the backdrop set up by the host.
P (A | B) = P (AB)/P (B) = (1/3)/(2/3) = 1/2
This answer is wrong, because the host does not pick up a backdrop and finds that there is no prize, but because he knows the answer, the host's task is to open one of the other two doors without prize, because one of the other two doors must have no prize, therefore, this message does not make any sense to you, because the prize is only behind one door. If you choose door 1, there must be no prize in door 2 and door 3, the host is sure to open a door without prizes, so this action is meaningless.
So if you choose door 1 and Door 1 is the answer, the probability of the host opening door 2 is 1/2, because both door 2 and door 3 can be opened, however, if door 3 is the answer, the probability of the host opening door 2 is 1, because the host can only open a door without prizes and cannot open the door 1 selected by the participants, therefore, only door 2 can be opened.
Correct answer:
Event A: the first option selected by the participant is the answer. Event B: The host tells the participant that one of the other two doors is not the answer. (To put it bluntly, the host actually spoke nonsense)
P (A | B) = P (AB)/P (B) = (1/3)/1 = 1/3.
Therefore, the probability of the answer is 2/3.
Therefore, if you want to change, you tend to change.
Reference: http://www.nowamagic.net/librarys/veda/detail/781http://bbs.csdn.net/topics/190102887http://blog.renren.com/share/250836298/6100805779
