Monty Hall Problem - $4$ doors where you pick one by one

Question

I encountered the problem of Monty Hall Problem with $4$ doors ($3$ doors $1$ goat). However in this variation you pick an initial door. Monty reveals one goat. Thereafter you either switch or stay - whatever you pick, Monty reveals a second goat and you get to choose again. What scenario is most optimal? I have seen this problem before on this site but the solution is wrong in my opinion. Can anyone see If I have done it right down below.

Scenario: Stay at first choice and Stay at second choice. (Stay1, Stay2)

If you are to win by staying in general, the first choice must be the right one and to win, the probability of staying for the second and third pick will be $1$. This means that

$\frac{1}{4} \cdot 1 \cdot 1 = \frac{1}{4}$

The probability of winning by staying is equal to the probability of choosing the car at first, $\frac{1}{n}$, and in this case $\frac{1}{4}$.

Scenario: Stay1 & Switch2

If you have a switch move included in the strategy then the first pick must not be the car as it will be impossible to win. So the first pick must be a goat hence $\frac{3}{4}$.

$\frac{3}{4} \cdot 1 \cdot 1 = \frac{3}{4}$

There is a $1$ probability of staying as the contestant must stay with the goat even though Monty reveals one if the next move is to switch.

Scenario: Switch1 & Stay2 If the contestant’s strategy includes switching as a first step and staying as second then the right pick must not be the first.

$\frac{3}{4} \cdot \frac{1}{2} \cdot 1 = \frac{3}{8}$

Scenario: Switch1 & Switch2

???? - do not know how to calculate here

I have seen that it is $\frac{5}{8}$ but I can not reflect which stops from calculating.

Open for critics and any further help.

Is it possible to write a general formula for this with $n$ doors? If so how?

Answer 1

If you use Switch1 & Stay2 strategy, you win with probability $3/8$; but after the 2nd goat is revealed, only 2 doors remain, and certainly the car is behind one of the 2 doors; consequently, if you switch (Switch1 & Switch2 strategy) you win with probability $1-3/8=5/8$

Answer 2

The probability that your original choice is correct is $\frac14$ and if this has occurred then by following the double switch strategy it is for sure that you will win the car.

The probability that your original choice is wrong is $\frac34$ and if this has occurred then the double switch strategy gives you a chance of $\frac12$ to win. You will win if your first new choice (which is a choice out of $2$ options and one of them positions you at the door that hides the car) is wrong and you will loose otherwise.

This gives in total a probability of: $$\frac14\cdot1+\frac34\cdot\frac12=\frac58$$ to win by applying the double switch strategy.

Answer 3

Regarding the generalization question you made, you can generalize it as follows.

Suppose you have $n$ doors and you're given the choice to swap $n-1$ times. In the original Monty Hall problem, you are asked to devise the best strategy to maximize your chances of winning.

For any $n$, your optimal strategy is to stay $n-2$ times, and at the final prompt to switch. Then, your probability of getting the prize is $\frac{n-1}{n}$; interestingly, as the number of empty prizes increases, so does your probability of finding the prize. The reason for this is that the optimal strategy (staying with your original choice for $n-2$ prompts and swapping at the final one) turns your chance of winning from randomly choosing amongst $n$ items to winning if you choose any of the $n-1$ losing items.

Similarly, you can see that by refusing to swap all $n-1$ prompts your chance of finding the prize is $\frac{1}{n}$ (getting it right in the first try).

You probably can generalize to find the probabilities for the other $2^{(n-1)} - 2$ strategies; I'm not going to do that here, as I'm confident it will be pretty messy, but the values of the other strategies will be in the range $[\frac{1}{n}, \frac{n-1}{n}]$ (between the probability of the worst and best strategies).

As a clarification, staying on all prompts is the worst strategy as you are ignoring all the information you are amassing from the reveals of the empty boxes.