Switch or Stay? Monty Hall problem

Question

So this is the Monty Hall problem that became a craze in the US. It attracted the attention of general population and academics alike. Its a fairly simple problem but the average finds it very difficult. Lets see if you can figure out if you want to Switch or Stay. Please state your reasons for either Switching or Staying.

Assume that a room is equipped with three doors. Behind two are goats, and behind the third is a shiny new car. You are asked to pick a door, and will win whatever is behind it. Let's say you pick door 1. Before the door is opened, however, someone who knows what's behind the doors (Monty Hall) opens one of the other two doors, revealing a goat, and asks you if you wish to change your selection to the third door (i.e., the door which neither you picked nor he opened). The Monty Hall problem is deciding whether you do.

souvik101990 · Answer

Switch! Always.

elavendan1 · Answer

Switch. Because when you switch, you get 67% of getting a car. Initially it is 33.33% probability that you'll get a car. Now when you switch, it's added to your initial probability. So it's better to switch

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rhrforever · Answer

Switch.
If you stick with door 1, your probability of winning is 1/3, but if you switch, probability becomes 2/3

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chetan2u · Answer

Hi
I have seen this earlier and answer is to switch.

But I really don't agree with the solution.

Yes, if I am asked before he opens the door that " will you switch after someone opens one of the other door" Yes, PROBABILITY goes upto 67%

But in present scenario..No
I know it is not in the open door, so it has to be one of the remaining two. Probability is simple 50%. Why should it be 67%

My take.. it should not make any difference as PROBABILITY remains 50% on both

rhrforever · Answer

See @chetan2u, it didn't work that way. 
Let me explain this by breaking the solution into 3 parts where the man always switches
1. Assume that the car is behind door 1. The host opens any one of the other two doors and shows you the goat. Now, you switch the door and lose. (Since car is behind door 1)

2. Assume that the car is behind door 2. You've again chosen door 1. Now, the host won't show you door 2 for sure, since it has the car. He shows you the goat behind door 3. You switch the doors and win the car behind door 2.

3. Assume that the car is behind door 3 and you choose door 1 again. Now, like the previous case, host won't show you door 3 since it has the car. He shows you the goat behind door 2. You switch the door and win again.

As we can see, in 2 out of 3 times, you win by switching the door. So, probability is 2/3 for winning when you switch doors.

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chetan2u · Answer

Hi,
You have simplified it by assuming the host will open only the door which has a goat.
NO,....As per the situation, he opens the door and just opens one door which has a goat. ONLY thing given is that the host knows what is behind the door choosen.

If I have to prove 67%, I don't require to take door wise.
It is straight, you are talking of door 2 and 3, one of which is opened by the host and second you choose..
So it's 2 door v's 1 door thus 67%

It may not be exactly the same but similar to .. 2 black ball and one red ball..
You see inside the box and take out a black. Left is two one red and black. 
Ofcourse you can confuse the situation by saying balls are behind door as this puzzle does and the solution given on net.

jkolachi · Answer

Chetan has a great point. Although i think it helps to think of the doors as probabilities. 1/3 + 1/3 + 1/3. So you pick 1, your probability = 1/3. But you have the option to pick (Door 2 + Door 3) which equals 2/3. Since the goat door will be eliminated its as if you get to choose 2 doors by switching....

You could also make the argument that after the goat is revealed, the probability becomes 50/50 i suppose.

You could also make the argument that after the goat is revealed, the probability becomes 50/50 i suppose.

mvictor · Answer

i'm with chetan.
you get 2 doors in front of you.
your probability remains 50%.
if you chose the door that the host opened, thank him, as he saved your butt and gave you a 50% chance of win.

stonecold · Answer

Wait a second.
Hold on.
Hold on.
Hold on.

Why would i switch.
The probability of winning a car or getting a goat is 50 percent.

Don't switch.
Never switch.

I am sticking with my guns on this one

GMATNinja · Answer

Ha! I love that you posted this, jkolachi . I've taught too many statistics courses, so I'm going to keep my mouth shut on this. But it really is one of those things that makes sense for a minute, and then it doesn't make sense anymore, and then it makes sense again, and then it stops making sense... kind of like the GMAT? :shock:

Klaus 74 · Answer

The rules of the MHP are specific and they include that the host knows where the car is and must reveal a goat from a door not chosen by the contestant. This is what gives the contestant twice as much of a chance to win if he switches his guess to the remaining door, from 1/3 to 2/3. The chances to win by switching is calculated by the probabilities of both the contestant and the host picking goats, therefore 2/3x1=2/3.

If the host does NOT know where the car is and the door he opens happens to reveal a goat, then it's 50/50. The chances to win by switching is calculated, again, by the probabilities of the contestant and the host picking goats, but this time it's 2/3x1/2=1/3, same as in staying which is 1/3x1=1/3. The other 1/3 of the games are lost early as the host will reveal the car 1/3 of the time thus there is no chance given to stay or switch.

GMATNinja · Answer

Klaus  74 -- I think you deserve some sort of prize for making your first-ever post on GMAT Club... about Bayesian statistics! I love it.

Welcome to GMAT Club!

Klaus 74 · Answer

Oh my...thank you for your warm reply GMATNinja, it certainly came as nice surprise. I'm not an academic by any means but I do understand the MHP thoroughly and am happy to help out anyone who may have issues understanding it. Again thank-you!!

Klaus 74 · Answer

With regard to your last sentence, the car IS behind the remaining door 2/3 of the time. Since there is one car and two goats there is a 1/3 chance that the contestant selects the correct door. This translates to the HOST having a 1/3 chance of having a goat/goat set of doors and a 2/3 chance of having a goat/car set of doors from which to reveal a goat. Therefore 2/3 of the time he leaves the car remaining for the contestant to switch to.

Switch or Stay? Monty Hall problem

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