Does Monty Hall Problem have any effect on an equation with multiple choices?

Question

I am having a discussion with someone that insists the Monty Hall Problem would improve the probability of guessing the correct door. However, the circumstances are different.

You have four doors. Only one has the right answer. We are not concerned about the probability of guessing the correct answer on the first try, but whether or not using Monty Hall Problem will improve your chances on guessing the correct door earlier. You choose one door. It's wrong. You choose again until you get the correct door. Does using the Monty Hall Problem better your chances of guessing the correct door with less guesses?

My argument is that since there are multiple choices, the Monty Hall Problem probability factor doesn't apply. The equation is reset each time you choose a new door, so the base probability factor is simply evenly split amongst the remaining doors.

The other person's argument is following a 4 door Monty Hall Problem that only has 1 door opened. You choose Door 1, you open Door 2. Find out that Door 2 is wrong. (S)he says there is now a 25% chance it is Door 1, and a 37.5% chance it is Door 3 and a 37.5% chance it is Door 4.

Who is right? I'm not sure if I'm overlooking something. I don't believe I am though.

If I am right, how would I thoroughly explain how the Monty Hall Problem doesn't apply here? I've tried explaining it how I have here to no avail and the other person still believes they're right.

Answer 1

To do the Monty Hall problem you need Monty Hall. Without him you have this.

You pick Door 1 and open Door 2.

1 out of 4 times the car will be in door 1 and Door 2 is a goat.

1 out of 4 times the car will be in door 2 and you see the car.

1 out of 4 times door 1 and door 2 will be wrong and the car is behind door 3.

1 out of 4 times door 1 and door 2 will be wrong and the car is behind door 3.

You are assuming the second case just didn't happen. So given the condition that it didn't happen you have 3 equally likely scenarios.

The car was behind door 1

The car was behind door 3

The car was behind door 4

You are right.

But if you had a monty hall then what would happen is

The door you choose is called A and the door monty shows you is B. The other two are C and D.

1 in 4 the car is behind A.

3 in 4 the car is not. 3 in 8 the car is behind C. ANd 3 in 8 the car is behing D.

Your friend is right.

But this is only if you have a host.

If you don't:

1 in 4 it is behind A.

1 in 4 it is actually behind B and you SEE it (that never happens with the host)

1 in 4 is is behind C.

and 1 in 4 it is behind D.

==== if you actually do the Monty Hall problem====

1 out of 4 cases you choose the right door the first time.

Monty shows you a goat. There are 3 doors left. The one you choose with a care and two with a goat. You switch to a goat.

There are 3 doors left. The one you first chose with the car, the one you now chose with a goat, and another with a goat. Monty shows you that one with a goat.

There are 2 doors left. The one you first chose with a car. And the one you currently chose the one with a goat. You switch and choose the car.

1 out of 4 times this will happen.

3 out of 4 times you will choose a door with a goat.

There will be 3 doors you didn't pick one with a car and two with goats. Monty will show you a door with a goat. There are three doors remaining. The door you currently picked with a goat; a door you didn't pick with a car; a door you didn't pick with a goat. You switch.

A) 3 out of 8 times you will switch to the door with a car. B) 3 out of 8 times you will switch to the door with a goat.

A) There are three doors. The one you picked with the car, and two you didn't pick with a goat. Monty shows you one of the goats and you switch to the other goat. There are two unopened doors. You have picked the one with the goat. You lose.

This happens 3 out of 8 times.

B0 There are three doors. The one you picked with a goat. The one you didn't pick with a car. The one you didn't pick with a goat. Monty shows you the one with the goat. You switch to the car. You win.

That happes 3 out of 8 times.

So the Monty Hall method works $5$ out of $8$ times.

Answer 2

You are right. The Monty Hall effect only shows up when someone else knows the correct answer. In the standard Monty Hall problem, the host knows which door has the prize, and so when he chooses a door to open he gives you useful information.

Answer 3

Using Bayes theorem:

Your scenario WITHOUT a host.

$P(\text{You pick right door}|\text{shown door is wrong}) = \frac {P(\text{shown door is wrong}|\text{You pick right door})\times P(\text{you pick right door})}{P(\text{shown door is wrong})}=$

$\frac {1\times \frac 14}{\frac 34}=\frac 13$.

So probability one of the other two doors is correct. $1 - \frac 13 = \frac 23$. And each is equally likely so probability a specific other door is correct is $\frac 13$

But WITH A host. It's the host job to make sure the shown door is wrong. Sp $P(\text{shown door is wrong}) = 1$

So

$P(\text{You pick right door}|\text{shown door is wrong}) = \frac {P(\text{shown door is wrong}|\text{You pick right door})\times P(\text{you pick right door})}{P(\text{shown door is wrong})}=$

$\frac {1\times \frac 14}{1}=\frac 14$.

So probability one of the other two doors is correct. $1 - \frac 14 = \frac 34$. And each is equally likely so probability a specific other door is correct is $\frac 38$

.... or ... WITHOUT a host these happen.

Car is behind door 1. 1-4 times.

1: Car, 2: Not Car show, 3: Not Car, 4: Not Car (1 in 12)

1: Car, 2: Not Car , 3: Not Carshow, 4: Not Car (1 in 12)

1: Car, 2: Not Car , 3: Not Car, 4: Not Car show (1 in 12)

Car is behind door 2. 1-4 times.

$\color{red}{\text{1: Not Car, 2: Car *show*, 3: Not Car, 4: Not Car (1 in 12)}}$

1: Not Car, 2: Car, 3: Not Car show, 4: Not Car (1 in 12)

1: Not Car, 2: Car, 3: Not Car , 4: Not Car show(1 in 12)

Car is behind door 3. 1-4 times.

1: Not Car, 2: Not Car show, 3: Car , 4: Not Car (1 in 12)

$\color{red}{\text{1: Not Car, 2: Not Car , 3: Car *show*, 4: Not Car (1 in 12)}}$

1: Not Car, 2: Not Car, 3: Car , 4: Not Car show(1 in 12)

Car is behind door 4. 1-4 times.

1: Not Car, 2: Not Car show, 3: Not Car , 4: Car (1 in 12)

1: Not Car, 2: Not Car, 3: Not Car show , 4: Not Car (1 in 12)

$\color{red}{\text{1: Not Car, 2: Not Car , 3: Not Car, 4: Car *show* (1 in 12)}}$

If we know the red ones didn't happen each of the others are equally likely.

And WITH a host:

Car is behind door 1. 1-4 times.

1: Car, 2: Not car shown, 3: Not car, 4: Not car (1 in 12)

1: Car, 2: Not car , 3: Not car shown, 4: Not car (1 in 12)

1: Car, 2: Not car , 3: Not car, 4: Not car shown (1 in 12)

Car is behind door 2. 1-4 times.

1: Not Car, 2: car, 3: not car shown, 4: not car. (1 in 8)

1: Not Car, 2: car, 3: not car, 4: not car shown. (1 in 8)

Car is behind door 3. 1-4 times.

1: Not Car, 2: not car shown, 3: car, 4: not car. (1 in 8)

1: Not Car, 2: not car, 3: car, 4: not car shown. (1 in 8)

Car is behind door 4. 1-4 times.

1: Not Car, 2: not car shown, 3: not car, 4: car. (1 in 8)

1: Not Car, 2: not car, 3: not car shown, 4: car. (1 in 8)

WITH a host we know that a car will never be shown and that changes everything.