Probability Problem -- Monti Hall Problem

Monti Hall

Mathematical game problems originating from Game Theory:

Once upon a time, a person was given a chance to guess the prize. There were three doors in front of him: A, B, and C. One of them was hidden behind the door, there are no prizes behind the other two doors. The winner does not know which door has a prize, but the host knows which door has a prize. The winner first chooses door A. Instead of opening door A immediately, the host Opens Door B, so that the winner can see that there is no prize behind door B. At this time, the host will give the winner a chance to re-choose and say, "Do you choose a, do you want to change to C ?", So the question came out: If you are the winner, do you choose A or C? Why?

Various proofs:

1. When contestants move to another door rather than continuing to maintain their original choice, the chances of winning the car will double. There are three possible cases, all of which have equal possibilities (1/3 ):

The contestant picked goat 1 and the host picked goat 2. Conversion will win the car.

The contestant picked goat 2 and the host picked goat 1. Conversion will win the car.

The contestant picked a car and the host picked either of the two goats. Conversion will fail.

In the first two cases, contestants can win the car by switching their options. The third case is the only one where contestants win by keeping their original selections. There are two of the three cases that win through conversion selection, so the probability of winning through conversion selection is 2/3.

2. After the contestant selects a door (A), the probability of the car behind the door is 1/3, and the probability of the remaining two doors (B and C) as a whole is 2/3. When the host opens the remaining two doors, one with goat's door (B), the probability of the remaining two doors (B and C) is only determined by the unopened door (c) therefore, the probability of that door (c) is 2/3.

3. the calculation of this problem requires the knowledge of conditional probability. If a and B are two events and P (B)> 0, P (A then B) is called) = P (A ∩ B)/P (B) is the conditional probability of event a occurring under event B.

Calculate the probability of winning the first prize:

If event a is set to {a winning prize} and Event B is set to {B won't win}, P (A) = 1 \ 3, P (B) = 2 \ 3. If the gamer chooses gate A and the goat is behind Gate B, the probability P (A branch B) = P (A branch B)/P (B) = P (a) * P (B)/P (B) = 1/3.

Therefore, under this condition, the probability of winning a is 1/3.

Calculate the probability of winning the C course:

If event a is set to {BC post prize} and B is set to {B do not win}, P (A) = 2 \ 3, P (B) = 2 \ 3. If the gamer chooses gate A and door B is followed by a goat, the conditional probability P (A branch B) = P (A branch B)/P (B) = P (a) * P (B)/P (B) = 2/3, because door B is followed by goat

Therefore, the probability of winning the C course is equal to the probability of winning the post-BC Course. It is equal to 2/3.

It can be seen that the probability of winning the C course is higher, and C should be selected.

From Baidu encyclopedia-Monti Hall

Another similar problem comes from Yunfeng's blog probability issue.

I don't know what it is, so I have to wait for someone to get it. There are four boxes left, three boxes of beef and one box of mushrooms. According to the current analysis, the probability of my meal being beef is 75%. So I took a box of Beef rice. At this time, two colleagues came to order Beef rice ...... So there is only one mushroom rice and the Beef rice in my hand left. Should I put down the Beef rice and take the mushrooms? Or take Beef rice?