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The Monty Hall Problem
by vivek gupta - Tuesday, 17 April 2007, 10:59 AM
 

CAT MBA ProbabilityRemember the game show "Khul Ja Sim Sim" where the host, Amen Verma, would show you three doors and you had to select one of them? Actually the game was a direct lift from an English game show "Let's Make A Deal." The show was based on a classic riddle and is now known as the "Monty Hall Problem" named after the host, Monty Hall.

 

THE PROBLEM

 

In the game show, you are given the opportunity to select one closed door of three, behind one of which there is a prize. The other two doors are either empty or hide some "nonprize". Once you have made your selection, Monty Hall will open one of the remaining doors, revealing that it does not contain the prize. He then asks you if you would like to switch your selection to the other unopened door, or stay with your original choice.

 

                                                           CAT MBA Probability                                      

 

Decide your answer before reading further…

 

BACKGROUND OF THE PROBLEM

 

The problem first appeared in Martin Gardner's 1961 book, More Mathematical Puzzles and Diversions, as The Three Prisoner Problem. Although the problem is simple, it is interesting because the solution is counter-intuitive. Most people chose their answers immediately and believe that it is the correct one. But trusting your instincts in this problem and choosing the "obvious" answers will lead you to incorrect solution.


CAT MBA Probabilty

 

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Re: The Monty Hall Problem
by RAJAT SAGAR - Wednesday, 18 April 2007, 04:35 PM
  Thanx Vivek Sir, great article! Me too a 'DCEite', 2006 -Mech. hoping to crack CAT this year. will need your help. and ofcourse, Thanx TG for such a wonderful site!!
Re: The Monty Hall Problem
by Total Gadha - Wednesday, 18 April 2007, 09:12 PM
  Hi Rajat,

Glad to know you're a mechanical engineer. Vivek and I are also both mech engineers. smile

It's good that you are liking the site. So far, we have managed to skim out all the useless jazz and keep only what is useful and necessary. I hope we are able to do that in future also.

Cheers!

Total Gadha
Re: The Monty Hall Problem
by priyanka sharma - Wednesday, 18 April 2007, 11:36 PM
  INTERESTING
Re: The Monty Hall Problem
by Total Gadha - Thursday, 19 April 2007, 12:20 PM
  Priyanka smile
Re: The Monty Hall Problem
by vivek gupta - Friday, 20 April 2007, 09:40 AM
 

Thanks Rajat,

The idea here is to share knowledge with like minded people.

Re: The Monty Hall Problem
by shankar i - Thursday, 3 May 2007, 02:37 PM
 

A very good explanation....

But there is some thing that I want to point out here...

The probability of the door selected initially to be incorrect was 2/3 but once the other incorrect door has been opened, that probability should also go down.

It has been mentioned that, "'you can think of the probabilities in aggregate-2/3-but now eliminate one possible choice". As I understand when a option gets eliminated its possibility of happening gets equally divided into other options, provided they all are mutually dependent, that is if one door has prize other 2 wont.

So the probability of having prize for opened incorrect door(1/3 that is) should be divided among rest two equally that is 1/6 each....and the then the probability of prize for each of them becomes 1/3+1/6=1/2. That is both will have equal chances.

There could be some part that I have misunderstood ...but it seems like a well explained fallacy. Would be more then happy to get some clarification.

Re: The Monty Hall Problem
by Total Gadha - Friday, 4 May 2007, 07:44 PM
  Hi Shankar,

It is a hard problem to understand. Let me ask you two questions:

When you selected a door, what was the probability that the prize was behind it? 1/3 or 33.33% right? Now no matter what I do in this world, this probability will stay 33.33% no?

After Monty opens a door, would the probability of your door holding the prize change? No. Nothing has been done to your door. Therefore, it still has the probability of 33.33% for holding the prize. Therefore, the other door had probability of 66.66% for holding the prize.

I understood the problem this way only. If the probability of my door holding the prize is 33.33%, then it will stay 33.33% no matter what Monty does. Suppose in place, of opening a door, he adds one more door. Would the probability of my door holding the prize become 25% now? No. It does not matter what Monty does.

Total Gadha
Re: The Monty Hall Problem
by shankar i - Monday, 7 May 2007, 03:24 PM
 

well... i again say the same thing ....finding prize behind door are mutually dependent events... 

if u find gift behind one door, u cant find it behind other....

hence the probabilites are dependent on each other for them...

also the way u said....''When you selected a door, what was the probability that the prize was behind it? 1/3 or 33.33% right? Now no matter what I do in this world, this probability will stay 33.33% no? "

the same way the prob of finding prize behind the 2nd left out door was 33.33%...n according to above logic should remain 33.33%...

i agree this is not right and it will increase as u have mentioned in your 2nd para(though not 66.66%)...but so would increase the prob of selected door....

ny ways ....seems this is some thin to do with how we r percievin this problem.. but nice discussion though..will wait 4 reply

Re: The Monty Hall Problem
by Sourabh jain - Monday, 9 July 2007, 06:10 AM
  Hi,

I agree with shankar about the dependency of the events.
As we are left with only 2 doors , the propability of the door containing the prize now gets divided between 2 doors i.e. 50% each.

Regards

Re: The Monty Hall Problem
by Eshika Aggarwal - Monday, 10 September 2007, 12:37 PM
 

Hi TG

I am not able to do the copy paste the monty hall problem as when i try to do that it comes till point1.please help me with that.smile

Thanks and Regards

Eshika

Re: The Monty Hall Problem
by Abhishek Kumar - Tuesday, 16 October 2007, 10:58 PM
 

Well a very fact we all would appreciate about probability is that its totally dependent on your line of thinking.. and how you percieve the situation..

well the situation is itself a conjecture..and a very difficult think to understand unless a deep study is made...

well what has been explained has been done by people who master the science of probaility..

 

but the queries raised ny SHANKAR and Saurab are good.. considering the fact that we are here by encountering one problem but at two different instances with different situations... so the answers about allocation of probability might change...

In case 1... assigning probability to each door ws 1/3

incase two the scenario has changed as one of the doors been opened and you are asked whether to switch or not... this is a different problem in itself... weighing the probable response in terms of probabilities :-

my thought process says either of the two may happen:-

a> as shankar and saurab said.. 50% coz two options are there

b> considering the two problems as a part of same query... sticking to my choice of 1/3 .. so the other door will have a probability of 2/3.

but this 2/3 probability (as a matter of fact also my choise of 1/3) is associatted with a probability of (1/2) of being correct... so this has to be incorporated to in making my finall judgement....

what you all say...

 http://math.ucsd.edu/~crypto/Monty/montybg.html

http://montyhallproblem.com/

 

Re: The Monty Hall Problem
by franklin brown - Wednesday, 30 April 2008, 06:03 PM
  aye Vivek... superb thinkin....approve
Re: The Monty Hall Problem
by Nishant Sharma - Monday, 8 June 2009, 09:30 PM
 

yes, that is correct.

i appreciate it!

but hadn't u took it from the movie 21 ?really good movie.smile

Re: The Monty Hall Problem
by Ghanshyam Chandak - Saturday, 8 August 2009, 02:56 AM
  i don't think monty hall is right while calculating probability.....i have understood but i,m not getting why this is always work.....
Re: The Monty Hall Problem
by Anil ,, - Friday, 11 September 2009, 11:08 PM
  I would like to request you very humbly to review the solution that you have provided once again as I completely disagree with the logic you have mentioned in the solution saying-"since you will be right one-third of time that means that if you stay with your first choice you will get prize one -third of time."
now I tell you why so.
probability of getting prize is 1/3 implies that if you open the door that you have 33.33% chance to get the prize. It has NOTHING to take with the the chances of getting the prize for any other time. a very explicit explanation supporting my argument is that
'the laws of probability says that it is calculated every time independently".

I can assert this with full confidence that it woud not matter at all weather the choice is switched or not.

I welcome your comment in this regard.
Re: The Monty Hall Problem
by Siddharth Arora - Tuesday, 1 June 2010, 09:54 AM
  Hi Anil,
I shall try to give you an insight of the problem which is totally based on my comprehension.See if that helps.

Firstly , here , wish to find the probability of winning in the 2 cases(since our concern is only to get the gift)..

1) stick to your choice - In that case winning is only possible if you select the right door at first shot and stick to it..which happens with a probability of 1/3.

2) Swap the choice. So it starts with selecting a wrong door with a probability of 2/3 and then swapping.

The problem becomes easy if  we are able to see that selection of a wrong door is a more probable event than selecting a right door.
The chance of selecting a wrong door is twice the probability of getting the correct door.I hope you agree to this.
If the wrong door is selected swapping gives a 100% chances to win since Monty has opened the door(which does not contain the gift).The only left out door is the one with the gift.

I would really appreciate other's viewpoint on this.
Re: The Monty Hall Problem
by pramod srini - Tuesday, 22 June 2010, 04:36 PM
  This is a tricky problem. but a systematic analysis will leave no doubts in our minds.
Let Monty hall name the goats as goat1(G1) and goat 2(G2) and the car as "car" .well Monty can arrange the the three things  behind doors A,B,C in the following ways.
DoorA  DoorB  DoorC
  G1      G2    car
  G1      car    G2
  car      G1    G2
  car      G2    G1
  G2      G1    car
  G2      car    G1

Now when given the first choice,let us select door 'B'. According to the problem , Monty can only open the door which does not have a car(the logic is obvious). Hence if we select the door having goat G1 , he can only open a door containing goat'G2' but Not the door having the car.
Monty now offers us the choice of either switching from our choice of door'B' or staying with doorB.  Based on above logic, the different possible combinations are as follows.

let: original choice=C,  monty's choice of opening the door=M
    dont switch=DS      , switch=S

  O choice        Mont choice        DontS        Switch
               
  G2              only door A            G2        car
  car              door A or C            car        G1 or G2
  G1              only door C            G1        car
  G2              only door C            G2        car
  G1              only door A            G1        car
  car              door A or C            car        G1 or G2 

As we can see, the probability of finding a CAR behind those doors if we accept Monty's choice of switching is 4/6=0.6666.
The result is same if we choose door A or door C initially.We can  surely try and verify!
So,we can conclude that irrespective of what our sixth sense says, it is always more favorable to us to SWITCH if we want to find a Car behind those doors :)