Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Everybody knows the famous Monty Hall problem http://en.wikipedia.org/wiki/Monty_Hall_problem

I will try again and I am sorry if I was not clear enough but please understand that I am here and asking because I dont know the answer and also because my english is not my mother language it is not that easy for me to formulate the question clear enough.

Suppose the host are asking 2 people to come on the stage to chose a door out of 3 and is asking the younger one to chose first. The participants are not allowed to chose the same door. The Host will open the 3rd door. In case the price is behind this opened 3rd door the 2 participants will go home without a price. In case the goat is behind the 3rd door the participants wil be asked if they want to stay or to switch. What is the better solution? To stay or to switch.

Now 2nd question/option. The Host will ask 2 people to come on the stage and are asked to chose one door but each one needs to write it on a piece of paper without the other one can see which door he is choosing. When choosing this way it can happen that both will chose the same door or in case of different doors the Host could open the 3rd door and the price will be behind this door. Then both Participants will go home without the price. In case the goat is behind door 3, what will be the best to do? Switch or stay? Regarding this question it looks to me that Henry gave the correct answer IMHO. thanks

share|improve this question
    
I don't think this is well enough specified - how do the two people choose, and which door if any is opened? –  Mark Bennet Oct 8 '13 at 7:17
    
thank You for Your advice. in first situation both people are asked to chose one of three doors and not to chose the same door. the younger of the two people may chose first. in second situation 2 people will give their choice of the door for example with a written note so no one of the 2 can see the other one's choice –  Franz Oct 8 '13 at 7:22
    
@Franz: And what does the host do then, especially in case the third door actually hides the desirable choice (the host is assumed to know whether this is the case)? (It is always assumed the people prefer a car over a goat; I cannot see why this should necessarily be the case.) –  Marc van Leeuwen Oct 8 '13 at 7:26
    
@Marc sorry for not being explicit enough (nit easy this problem), that means in case of 1st example the 3rd door is opened no matter what is behind. there could be a goat or the car. in case the car is there the 2 people go home with a goat=no price. in case of example 2 if both are chosing the same door by chance one door with the goat wil be openen and they will be asked to switch or to stay.in case that each one chose a different door it will be handled like above in first example. –  Franz Oct 8 '13 at 7:32
    
Mate in the first example - if the door is opened to show a goat, what does it mean for them to 'switch'? Which door is opened if they do switch? Which door is opened if they do not switch? –  Donkey_2009 Oct 8 '13 at 11:23

1 Answer 1

In the first case, it makes no difference whether they switch or not. The chance for each that they win the prize is $\frac{1}{3}$.

In the second case, they should always switch. The chance for each that they win the prize is $\frac{4}{9}$ if they switch but $\frac{3}{9}$ if they do not switch.

share|improve this answer
    
please let me think loud. 1st case: taking the MH version if it is one participant he always needs to switch to have 2/3 instead 1/3 chance to win the car. You are saying that if there are 2 participants and the host opens the the 3rd door with the goat it makes no difference if they switch or not. sorry but I just dont understand this because I see it as the original MH problem in case the 3rd door has goat. IMO they should always switch. what did I miss here? regarding second case I understand that there are 9 possibilities and the one who switches will win 4 times out of nine? –  Franz Oct 8 '13 at 9:47
    
@Franz: In the classic MH problem, the host always opens a goat door. In your first problem the host opens a door which might have a goat or might not, so the problems are different: you get no information from the host deciding which door to open, since the host has no choice to avoid the car. –  Henry Oct 8 '13 at 10:34
    
exactly the host has no choice if the car is behind 3rd door, he has to open it. then the participants get no price. but in case the host opens the 3rd door and the goat is behind the door each one has the original MH problem and should IMO switch. –  Franz Oct 8 '13 at 10:51
    
@Franz: There are $18$ equally likely possibilities the car being behind a particular door, the younger choosing, the elder choosing, and the remaining door being opened. So what do you think the probabilities are of the younger winning the car if the younger always switches after seeing a goat and the probabilities of the younger winning the car if the younger never switches after seeing a goat? –  Henry Oct 8 '13 at 11:06
    
I still dont understand it. maybe I need to explain that in case both people have chosen the same door,and in case both will switch they will switch to the other door (which is still closed) that would mean that if the price or the goat is behind 2nd door they will go home without any price.in case only one will switch so one will be the winner of the price. now if each of the 2 participants have a different door and in case only one decides to switch they both will have the same door and will go home without a price. but if both decides to switch they change doors and one will win the price. –  Franz Oct 8 '13 at 15:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.