Monty Hall: What's wrong with my approach

Question

Let us say we have n boxes, and one car.

The probability that one of them might have a car = p

and each box is equally likely to have a car, therefore the change that a single box has car = p/n

Now let us consider the chance that box 1 has car, given other boxes don't have car

P(Box 1 contains car | other n-1 boxes don't contain car)

$$=\frac{P(Box\space1\space contains\space car \cap other\space n-1\space boxes\space don't\space contain\space car)}{P(other\space n-1\space boxes\space don't\space contain\space car)}$$

$$=\frac{P(Box\space1\space contains\space car)}{P(other\space n-1\space boxes\space don't\space contain\space car)}$$

$$=\frac{\frac{p}{n}}{ \left( 1 - \frac{p}{n} \right)^{(n-1)} }$$ [1]

But if we apply this result to classical monty hall problem where we have 3 boxes. Let us say contestant chose box 1.

Other two boxes have p = 2/3 to have car, n = 2,

each box will have probability of p/n = $\frac{2/3}{2} = 1/3$, of having a car

If monty opens up say box 3(which doesn't have car),

the probability of box 2 having car

$$=\frac{\frac{p}{n}}{ \left( 1 - \frac{p}{n} \right)^{(n-1)} }$$

here p = 2/3

n = 2 $$=\frac{\frac{2/3}{2}}{ \left( 1 - \frac{2/3}{2} \right)^{(2-1)} }$$ $$= 1/2$$

which is not the correct answer. I have done something wrong and [1] is not valid result.

What is wrong with it?

PS: I know how to solve monty hall problem using other approach. I want to know what is wrong with this approach?

Answer 1

First off, your are not actually computing the Monty Hall Problem. You are computing the probability: $$ P(\text{box 2}\mid\text{not box 3})=\frac12 $$ Whereas the Monty Hall Problem is concerned with the probability: $$ P(\text{box 2 OR box 3}\mid\text{(not box 3) OR (not box 2)})=\frac23 $$

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To comment on your derivation, by $n$ you do not mean the total number of boxes since otherwise specifying that $n-1$ boxes do not contain a car would be redundant. This was unclear when I read it at first.

Instead suppose we have $N>n$ boxes in total. Then you are asking:

What is the probability that the car is found in the first of the $n<N$ boxes given that it is NOT found in the other $(n-1)$ boxes.

To be clear, this entails the possibility that the car may in fact be in one of the remaining $N-n$ boxes.

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Then most of your derivation is correct, except for your denominator. It considers the $(n-1)$ boxes as $(n-1)$ independent trials, which it is not. We are only considering a single trial here. Rather the denominator should be: $$ P(\text{not the other $n-1$ boxes})=1-(n-1)\cdot\tfrac pn $$

Answer 2

Firstly, your definition of $p$ is rather vague.
Try something in the lines of: Assuming there are $n$ boxes, each box has the probability of containing the car equal to $p = \frac{1}{n}$.
Now, the probability you propose $$P(\text{Box 1 contains car} | \text{other n-1 boxes don't contain car}) = 1$$ since the statement "Box 1 contains car" is equivalent to "other n-1 boxes don't contain car" because there is only 1 car.
The probability you are looking for is $$P(\text{Box 1 contains car} | \text{1 other box don't contain car}).$$ Other than that, I think you're on the right track.