Monday, 14 April 2014

The Monty Hall Problem

I enjoy problems like this. They remind us that probability can be counter-intuitive.


Mark Wadsworth said...

I first read about this in 1990, and even though I know the answer and logic off by heart, I still struggle to explain it to somebody who does not want to understand it.

A K Haart said...

Mark - I don't find such problems easy to explain either. I'm not sure why, but I suspect there is an interesting issue in there somewhere.

Sam Vega said...

There is another issue in there. How did anything that is true ever get to be counter-intuitive? Why should our brains or thought-patterns develop in such a way that there is a class of facts which regularly eludes us?

A K Haart said...

Sam - I think the problem may be something to do with assumptions. In a sense, the problem is designed to encourage an obvious but mistaken assumption. A trick in other words.

Who plays similar tricks?

Scrobs... said...

It's not 66%/33% at the second chance, it's still 50/50.

It was the former, but isn't now.

Woodsy42 said...

I find this logic difficult, and I am not fully convinced because it depends on how you define the problem - and I think this explanation may be wrong.
Basically it uses the chances (1 in 3) of the first choice to influence the second choice by defining the second choice (change or not to change) as a continuation of the same problem. This is surely rather like saying that a succession of heads when tossing a coin makes a tail more likely.
But previous occurrences do not influence future probability (do they?) so I think of the second decision not as change/no change but as a totally new problem with a situation of 2 doors, one with a goat one with a car, and thus a 50/50 chance - you don't 'change' - effectively you have to once again pick a door - so the chance is 50/50 irrespective?

A K Haart said...

Scrobs and Woodsy - no it's 66/33. By seeing it as 50/50 you are ignoring information you already have about your first choice - that it only has a 33% probability of being right.

The trick is to get you to ignore that information and treat your second choice as a new situation, which it isn't.

At the second choice, the opened door shows a goat, so the remaining closed door must have a 66% probability of hiding the car as your original door still only has a 33% probability. Overall the two must add up to 100%.

Michael said...

Having just read the Wiki article, I'm of the opinion that the laws of probability were dreamed up by exponents of Pythonism!

This could run and run...

James Higham said...

I'm with Woodsy on this. On the first run, the chance of the car was 33%.

The host opening a goat door makes the chance now 50% of picking the car.

And therefore it's 50/50 whether to swap or not. The trick was to conflate both probabilities as a "changing" probability, calling it 66%.

In fact it's not - it's 50% the second time.

Demetrius said...

On the other hand you can milk or even eat a goat.

A K Haart said...

Michael - it seems so!

James - try emulating it using the RAND function in Excel. It may take a few iterations but you will see 66/33 not 50/50.

Demetrius - and goat's cheese can be excellent.

Woodsy42 said...

How about we breed a lot of goats, build a lot of doors and run an experiment?

A K Haart said...

Woodsy - but you'll need the cars too. Sounds expensive so we'll need a government grant which means it will have to be related to climate change.

Woodsy42 said...

"66% probability of hiding the car as your original door still only has a 33% probability."

No AK it doesn't - at least the way I see it. The new information you have been given (the location of one goat) means it is a new situation. the probabilities have changed so that your original choice now has a 50% probability just like the other closed door.
Consider: If a new contestant (having not seen the preceeeding openning) now arrives and is also asked to choose a door (with one goat door already open) he would have a 50/50 chance. So you are suggesting that changing contestants changes the probablity of the car being behind a particular door. How can one door have two different probabilities?

A K Haart said...

Woodsy - there are only three situations.
1 2 3

If you originally choose door 1 and don't change, you win the car in 1/3 of the situations.

If you originally choose door 1 and do change, you win the car in 2/3 of the situations.

Woodsy42 said...

Think of it this way then.
There are 100 doors. Contestant picks a door (1%chance yes?). The host then opens 98 doors displaying goats.
If the contestant changes his choice does he how have a 99% chance of a car?
Honestly I am not being deliberately thick or difficult. I think there is a logical falacy in the explanation - even if 97% of climate (sorry maths)scientists claim otherwise.

A K Haart said...

Woodsy - yes, in your 100 door scenario, if the contestant changes his choice he now has a 99% chance of a car.

This is what Monty Hall is doing when he opens 98 door to show goats. He is showing you which door hides the car - assuming you didn't pick the car first time round.

This is why you can't treat it as two distinct situations. Monty Hall is showing you something about your first choice, he is not creating a new situation with different probabilities.