The Revised Monty Hall Problem

Probability Level 1

Many people have probably heard of the Monty Hall problem. Here is a different version of it.

You are playing on a game show and have made it to the last round. You have a chance to win a car! The game show host shows you three doors. He claims that behind one of the doors is the car, and behind the other two are goats. The game show host asks you to pick a door and explains what will happen after you pick your door. Before your door is opened, the host will open one of the other two doors at random . If that door reveals the car, you automatically win it! If it turns out to reveal a goat, you will have the option to swap your chosen door and open the door you didn't choose in the beginning. If the door you open reveals the car, you win the car.

So here is what happens: You pick the door on the right. The host chooses to open the door in the middle. As it turns out, that door was holding a goat. Now you have the option to switch. Should you switch, stick, or does it not matter what you choose to do?

Which action would give you the highest chance of winning the car?

Stick Switch Doesn't matter

6 solutions

Brian Moehring
Aug 16, 2018

Initially, you choose the car with probability $\frac{1}{3}$ and a goat with probability $\frac{2}{3}.$

If you chose the car initially, then the host will open a door with a goat with probability $1$ .
If you chose a goat initially, then the host will open a door with a goat with probability $\frac{1}{2}$ (In the classical Monty Hall problem, this probability is $1$ )

Together, this means the probability that the host chooses a goat is $\frac{2}{3},$ so by Bayes, the probability that you chose the car initially given the fact the host has revealed a goat is $P\left(\text{you chose car} \Big| \text{host chooses goat}\right) = \frac{P(\text{you chose car})}{P(\text{host chooses goat})}\times P\left(\text{host chooses goat}\Big|\text{you chose car}\right) = \frac{1/3}{2/3} \times 1 = \frac{1}{2}$

I have a question. Why don't you have 2/3 a chance of winning at the beginning since you either win if the host randomly picks the winner or you pick the winner?

James Tourkistas - 2 years, 1 month ago

You do have a 2/3 chance of winning. The important matter for this problem is that regardless of which strategy you choose to implement (swap or not swap, given the option), you'll still have a 2/3 chance of winning.

Brian Moehring - 2 years, 1 month ago

Oh ok. I think I get it. So, since it is a random process, you also kind of have 2/3 a chance of losing. Does it make sense to think of it as 4 possibilities since we enter a random process in it with two possible outcomes? So, at the beginning, there are really 4 possibilities. You pick right, the host removes goat1, you pick right, the host removes goat2, you pick wrong/the host removes a goat1, you pick wrong/the host removes goat2. So, it ends up being 2/4 and 2/4. I'm looking for an intuitive bright line way to think about the host purposefully picking something and us getting additional information and him doing so through a random process?

Also, why isn't it important that technically there are a couple more possibilities? You pick goat 1, the host removes goat 2. You pick goat 1, the host removes car. You pick goat 2, the host removes car. You pick You pick goat 2, and the host removes goat 1. That would leave us with 6 possible outcomes. Only two of which we picked the car. At the risk of answering my own question, is it because we now have additional information and know the car wasn't picked so we no longer consider that in determining our odds of winning.

James Tourkistas - 2 years, 1 month ago

"you also kind of have 2/3 a chance of losing" - No. In the beginning, you have a 2/3 chance of winning and a 1/3 chance of losing. If the host shows the car behind his door, you win automatically, with probability 1. If the host shows a goat behind his door, you will then win with probability 1/2, so it doesn't matter if you switch doors or not. With a little prep-work, this actually would provide an indirect method to calculate the conditional probability we want.

For the rest of your comment: Listing outcomes and calculating the number of favorable outcomes divided by the total number of outcomes only works if each outcome is equally probable .

In your first situation (with 4 possibilities), you used "you pick right" and "you pick wrong" to create the 4 possibilities, but these are not equally probable, so merely counting outcomes is insufficient to find the probability.

In your second situation, you correctly identify the six equally probable possibilities (can you show that they are equally probable? How does this fail for the classical Monty Hall problem?). However, as you mention, you must remove the cases where the host had chosen a car since we are told that the host chooses a goat. Since we started with equally probable possibilities, this removal of two cases results in 4 equally probable possibilities, of which 2 had you choose the car originally (and $\frac{2}{4} = \frac{1}{2},$ as we showed originally)

Brian Moehring - 2 years, 1 month ago

Nope, it's really simple. You have a 1/3 chance of picking the car and winning if you stick. Your chance of the host picking the car is 2/3 x 1/2 = 1/3. So your total chance of winning is 1/3 + 1/3 = 2/3. However, the one chance of winning by the host picking the car is lost, and that was a 1/6 probability. So now your chance of winning is 2/3 - 1/6 = 1/2 if you just sit tight. But that means that you also have a 1/2 chance of losing. So it doesn't matter if you switch , the odds are the same either way... 1/2!

James Schuller - 1 year, 1 month ago

If the probability that I chose a car given that the host chose a goat is 1/2 as illustrated in the answer, there's no need to switch, right? The answer should be "Doesn't matter"

Divya Pattisapu - 8 months ago

If the probability that I chose a car given that the host chose a goat is 1/2 as illustrated in the answer, there's no need to switch, right? The answer should be "Doesn't matter"

Divya Pattisapu - 8 months ago

Mohammad Farhat
Aug 16, 2018

This is a case of randomness vs purposeness. If It was purposeness, then the probability is 2/3 according to Bayes' Theorem and if it is randomness then the probability is 1/2

Bayes' to the rescue!

Roger Marchant - 6 months, 3 weeks ago

Joshua Lowrance
Aug 16, 2018

Because the host is opening at random, the probability is 1/2.

Björn Holmgren
Mar 31, 2021

its random, and Monty still have 1/3 to pick the car. and the last door must have 1/3 as well then (so sum is 1).

B D
Sep 16, 2018

When you pick the door on the right you have 1/3 chance of winning the car. After the host opens the door in the middle and reveals a goat you have 1/2 chance of winning the car so it doesn't matter.

That this idea is incorrect is very easily demonstrated. Imagine that MH opens a door without saying whether his action is random or not, and a goat is revealed. What is the probability of winning by switching then? it is, as before, 2/3 irrespective of MH's knowledge.

Roger Bodey - 10 months, 3 weeks ago

Nguyen Nghi
Jul 31, 2019

Anyone, please help with this question. ' WHY( PURPOSE) the host need to open 1 door in random or normal case?"

The Revised Monty Hall Problem

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