The Monty Hall Problem – Extended!

A brief summary of the Monty Hall problem. 

There are 3 doors.  Behind 2 doors are goats and behind 1 door is a car.  You choose a door at random.  The host then opens another door to reveal a goat.  Should you stick with your original choice or swap to the other unopened door?

A good IA?

This is a reasonably predictable IA topic – which whilst not especially original, can offer some opportunities for investigation if the student alters the original premise and does some personal modeling (e.g. with Python etc.).  It’s also a very well known counter-intuitive puzzle which shows how ill equipped we are to understand probabilities!

Understanding this with tree diagrams

There are a number of ways to explain this, I will use tree diagrams as they allow some nice extension ideas.  We can make the following tree diagram:

This is a tree diagram to show the possible outcomes when a person starts by choosing Door 1.  (By symmetry this then allows us to understand the general problem).  The tree diagram then maps the possibilities of the car, and the door that the host will open.  For example when the contestant chooses Door 1 and the car is behind Door 3, then there is a 100% that the host will open Door 2 (the bottom branch).

This allows us to see that the probability of winning a car when swapping is:

Extension ideas

So far, so ordinary – there’s not much originality so far.  So we can make things more interesting by changing the original problem.  Let’s say the following:

There are 4 doors.  Behind 3 doors are goats and behind 1 door is a car.  You choose a door at random.  The host then opens another door to reveal a goat.  Should you stick with your original choice or swap to another unopened door?

Tree diagram

Once again this is a tree diagram to explain the possible doors the host opens when a contestant starts by opening Door 1.  We can see that if the car is behind Door 4 that the host has a choice of 2 doors to open (Door 2 and Door 3).  Say the host opens Door 3, this then gives me a 50% of winning if I swap to another door (as I can swap to either Door 4 or Door 2).  This is shown in the bottom branch.

This allows us to see the following probabilities for swapping:

and not swapping:

So we can see that the probability of winning has changed from 1/4 to 3/8 when swapping.

Finding a general formula:

We can then notice a pattern in our tree diagram to see that for 5 doors with 4 goats and 1 car we have a probability of swapping given by:

and then the general form for swapping with n doors with n-1 goats is:

We can see that when n=3 (our original case of 3 doors) this does indeed give 2/3.

More insights?

We can also use this general formula to prove why we should always swap.  The probability of winning by sticking is 1/n, so we can see that the following inequality always holds for n at least 3:

More generality

There is also a general equation for when there are n doors and the host opens p other goat doors.  This also follows from our tree diagram ideas:

So – even though it’s a well known topic, there is still a lot of explore.  You could try and make your own gameshow and work out the probability of someone winning.