Monte Hall problem

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heathend0

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Consider the following version of the Monte Hall Problem:
  • There are four doors numbered 1,…,4. Three of these doors have goats behind them. One door has a sports car. You want to win the sports car.
  • You pick one door, i uniformly at random and put a chalk mark on it. That door stays closed for now.
  • Monte Hall opens a door j, with j≠i, showing you a goat.
  • You get to pick any of the three unopened doors in {1,2,3,4}∖{j} and keep whatever is behind it.
Answer the following questions
  1. Suppose you decide to stick with your first choice, i. What is the probability that you win the sports car?
  2. Suppose you decide to choose one of three unopened doors {1,2,3,4}∖{j} uniformly at random. What is the probability that you win the sports car?
  3. Suppose you decide to choose one of the two unopened and unmarked doors {1,2,3,4}∖{i,j} uniformly at random. What is the probability that you win the sports car?
for 1 I got a probability of 1/4 since choosing the correct door at the start and getting the sports car. Since we have 4 doors and 25% chance for each door
for 2. I got 1/3 because three doors remain (1 correct door out of 3)
for 3. I got 3/8 because the probability of the goat in door chosen at the start * the probability of the car in the new door given the goat is in the door chosen at the start would be (3/4) * (1/2) which gives me the answer 3/8.

I feel like I am missing something here, can anyone guide me down the correct path if so
 
What reason was given? That's how you decide whether answer is reasonable! If they give no reason, pay no attention. If they do, analyze that.

I'm sure you're aware that the Monte Hall problem is a classic example where intuition is often wrong, and even apparently valid reasoning has to be checked carefully.
 
How can the 1st question have an answer of 1/3.

Suppose you play the lottery where there are 100 different tickets and one ticket is the big winner.

Suppose I know two things--I know your ticket number and I know the winning ticket number (which might be yours).

One by one I tell you 998 losing tickets numbers while making sure not to say your ticket number. The two tickets remaining includes your ticket. Are you saying that know you have a better than 1 in 100 chance of winning? If you think yes, then do the following.

Buy a lottery ticket. Have a friend find out which ticket number won and tell this friend which number you purchased. Then one by one have your friend say this number (she should actually say the number) did not win, this number did not win, .... until there are two numbers left, one being your number. Do you still believe that your chances of winning is anything but 1 in 100?
 
heathend0 said:
Consider the following version of the Monte Hall Problem:
  • There are four doors numbered 1,…,4. Three of these doors have goats behind them. One door has a sports car. You want to win the sports car.
  • You pick one door, i uniformly at random and put a chalk mark on it. That door stays closed for now.
  • Monte Hall opens a door j, with j≠i, showing you a goat.
  • You get to pick any of the three unopened doors in {1,2,3,4}∖{j} and keep whatever is behind it.
Click to expand...
heathend0 said:
I am being told the answer is 1/3 for all 3 questions, could this be a possibility?
Click to expand...
Jomo said:
How can the 1st question have an answer of 1/3.
Click to expand...
I agree the answer is 1/3 for all 3 questions.
Here is the game show setup. Four doors. A new car is behind one door. A goat is behind each of the other three.
You choose any one of the doors and mark it. There are three unmarked doors left.
Dear ole Monty knows behind which door the car is.
So he opens a door behind which there is a goat.
At this point there three closed doors, behind one of which is the car.
At this point you can choose any one of three doors: the marked one or one of the other two.
So you now have \(\frac{1}{3}\) probability of guessing correctly.

This is exactly what the Rev. Mr. Bayes wrote about.
We start with four doors. But ole Monty comes along and opens one.
Now we have a new space with only three doors from which any rational person would choose {unless one wants a pet goat not a new car).
This is so hard to teach, I never understood why. \(\mathcal{P}(A|B)\) read the probability of \(A\) given \(B\).
Formula: \(\dfrac{\mathcal{P}(A\cap B)}{\mathcal{P}(B)}\) all that means is what is the probability that \(A\) happens in side the space of \(B\)?
 
pka said:


I agree the answer is 1/3 for all 3 questions.
Here is the game show setup. Four doors. A new car is behind one door. A goat is behind each of the other three.
You choose any one of the doors and mark it. There are three unmarked doors left.
Dear ole Monty knows behind which door the car is.
So he opens a door behind which there is a goat.
At this point there three closed doors, behind one of which is the car.
At this point you can choose any one of three doors: the marked one or one of the other two.
So you now have \(\frac{1}{3}\) probability of guessing correctly.

This is exactly what the Rev. Mr. Bayes wrote about.
We start with four doors. But ole Monty comes along and opens one.
Now we have a new space with only three doors from which any rational person would choose {unless one wants a pet goat not a new car).
This is so hard to teach, I never understood why. \(\mathcal{P}(A|B)\) read the probability of \(A\) given \(B\).
Formula: \(\dfrac{\mathcal{P}(A\cap B)}{\mathcal{P}(B)}\) all that means is what is the probability that \(A\) happens in side the space of \(B\)?
Click to expand...
That is exactly what I was thinking as well... Can anyone else verify this?
 
It's well known that when the Monty Hall problem was first popularized, many mathematicians got it wrong. That's easy to do if you don't carefully think through the details of the problem. You may have observed that pka didn't actually give a proof, but just waved his hands and showed a formula, without demonstrating how it applies. I invite him to do so and try to convince me: What are A and B, and what does the formula produce? This may reveal a subtle issue in the interpretation of the problem.

But it is obvious that the answer to part (1) is 1/4, because nothing Monty does can change the fact that you will have already chosen the right door only 1/4 of the time. (What he does does change the probabilities of the other doors.) The key to the problem is that you are not making a new choice, you are sticking with a choice you already made that is either right or wrong from the very start. Of course, there is a lot more that can be said (and has been said -- see Wikipedia), but more mathematically elaborate answers just confirm the intuitive reasoning.

To specifically answer pka's intuitive claim, the three remaining doors are not equally likely; you can randomly choose any of them with 1/3 probability, but that doesn't mean that they are equally likely to have the car. You can't ignore the history behind the choice; different things would have happened according to which door actually held the car, because Monty knows things you don't.
 
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