The Arbitrarily Large Monty Hall Problem

I've just fallen off a plane from the UK, and am a little tired, but here's some interesting mathematics that kept me busy `on the road'.

I've loved the Monty Hall problem since I first heard about it. It is not a difficult problem, but the outcome can seem quite counter intuitive. Before I look at the problem in more detail (more detail than I should?) you should have a look at this little video as a refresher.


So, let's try and represent all we've seen in the movie in this as a picture. 


At the top is the initial situation, with the car, C, and two goats, G, behind the doors. The next level down is what we end up with, either with you choosing the car and one goat door open (with a probability of 1/3), or choosing a goat and a goat door revealed (with a probability of 2/3).

Looking at this , it's clear that if you choose to stick with your original choice of doors, you will win the car with a probability of 1/3 and will win the goat with a probability of 2/3.

However, if you swap doors, the probability of getting a goat is now 1/3 and winning the car is 2/3; the chances of winning a car has increased by a factor of two by simply switching doors.

How cool is that!

But I started thinking, if we change the problem, change the number of doors, cars and goats, but always reveal one door with a goat, will it always boost your chances to swap?

OK, here we go. Now we have 4 doors to start with, but one car and now three goats. So, similar to the picture above, we get to the final state, with you either choosing the car with the first guess, with the probability of 1/4, or a goat with the probability 3/4.



So, looking at this, then of you stick with the original choice, the chance of winning the car is 1/4 and winning a goat is 3/4. But what if you choose to swap?

Clearly, if you chose the car to start with, then when you swap you will get a goat. 

If, however, you chose the goat to start with, then if you swap then you could swap to the car or you could swap to a goat. At this stage, if you swap you have a 50-50 chance of getting the goat or car. What the chance of winning the car now? 


So, the chance of winning the car if you don't swap is 1/4 (or 2/8), but if you swap it is 3/8, so there is an improved chance of 1.5 times in winning the car if swapping doors than keeping your original choice. 

OK, let's spice things up a little. Let's now have two cars and two goats. Same picture as before, but now the initial chance of choosing a car or a goat is 1/2. 


So, if you stick with your original choices, you have a 50-50 chance of winning a car. But what if you swap?

If you originally picked a car, then there is a 50-50 chance that you will swap to a car or a goat, so you can win or lose. If you originally chose a goat, then when swapping you have a 100% chance of getting a car.

So, what's the chance of winning when swapping now?


So, again, the chances of you winning are boosted by a factor of 1.5 times by swapping rather than keeping your original choice.

So, what if you have originally n cars and m goats, hidden behind (n+m) doors. What's the ratio winning when swapping compared to keeping your original choice? I will leave the algebra to the reader, but you can show this ratio is


Woooooh! This result does not depend upon the actual number of cars and goats, only the number of doors (I should point out, there has to be at least one car and two goats for this to work).

Let's just check this works. In the original problem, there are three doors, so ndoor=3, and this ratio is 2. Excellent. What about four doors, with ndoor=4? The ratio becomes 1.5, just as we saw before.

I don't know if this has been derived before, but I think it is a cool result. 

It tells you a few things. Firstly, as we increase the number of doors, the ratio between swapping and sticking approaches unity, irrespective of the number of cars and goats. But more importantly, the limit is approached from above; the numerator of the fraction is always bigger than the denominator, and so the ratio is always greater than one. It might be only a little bit bigger than one, but it always is.


The moral of the story is to always swap, no matter how many doors are presented to you. Good luck!

(note - badly formed maths fixed since original post - kids, don't blog while jet-lagged!)

Comments

  1. I think the correct answer to Monty Hall is "it doesn't matter". Yeah, I said it.

    The "you should switch" solution comes from assuming some extremely unrealistic prior information: that the probability the host's dramatic opening of a door has zero probability of exposing the car.

    Sure, that's usually given in the statement of the problem, but it's a really weird assumption.

    ReplyDelete
    Replies
    1. Huh? The host *knows* what is behind the doors, and so there is a 100% probability of them dramatically opening the door and revealing a goat. He's not guessing.

      Delete
    2. That's the usual scenario. But I think it is unrealistic and would make for bad television :) There should be some chance that the host will open the door with the car, to make the TV show better.

      Delete
    3. Scratches head... "unrealistic" ??? - lost for words....

      Delete
    4. In the Monty Hall problem we are calculating probabilities from the point of view of the contestant, who has the same information as the audience. Even if the host knows he's going to reveal a goat on that particular episode, that's irrelevant. What matters is whether the contestant/audience knows in advance that the host is going to reveal a goat.

      If they do, that's a boring show, and the opening of a door to create tension is pointless because everyone already knew it was just going to be a goat.

      Am I explaining myself poorly?

      Delete
    5. Yes, you're explaining yourself poorly. Yes, the audience knows he will reveal a goat.

      Have you ever watched "Deal or no deal"? What if Monty offered a financial deal for sticking or switching etc? You can make it exciting :)

      Delete
  2. Thanks for the general solution, though. :-)

    ReplyDelete
  3. "Firstly, as we increase the number of doors, the ratio between swapping and sticking approaches unity, irrespective of the number of cars and goats."

    This is a good example of a sentence which would completely befuddle someone not familiar with the context. Who knows what he would think this is about?

    ReplyDelete
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