I love this problem – even though it is just an exercise – it’s still kind of fun to work out.

It is also known as The Monty Hall and was floating around a while back in the early 1990’s on Marilyn vos Savant’s web page. Most of my current students have little or no idea who Monty Hall is and so I have fun explaining to them the whole idea behind the game show, Let’s Make a Deal.

Anyway, readers of this blog are probably familiar with the show and the problem, but for completeness sake, here it is in a nutshell.

You are on a game show where there are 3 doors. Behind two of the doors is a mule, and behind the third door is a Cadillac. You choose a door. The host then gives you an opportunity to switch your choice after opening one of the three doors – not the one you chose, of course – and revealing what is behind it. The question is, should you switch or keep to your original choice?

You have no doubt already clicked on the link above and checked out the solution, so you know that in the long run it is actually to your advantage to switch after the host reveals what is behind one of the unopened doors. But I have another question for you: What if there are 4 doors, and after you select one, the host reveals what is behind one of the remaining 3 doors – not the one you selected, of course – and gives you the option of switching. What should you do? And what is the exact probability of winning or losing in any case?

I’ll post an answer soon – unless you beat me to it! No Googling, now!

### Like this:

Like Loading...

*Related*

Tags: cadillac, doors, let's make a deal, marilyn vos savant, monty hall, mule, probability, switch or stay

This entry was posted on April 21, 2011 at 11:38 am and is filed under Uncategorized. You can follow any responses to this entry through the RSS 2.0 feed.
You can leave a response, or trackback from your own site.

April 25, 2011 at 3:28 pm |

In your new scenario, are there prizes behind 3 doors? or only behind 2 doors?

The original problem is a lovely one, but for perhaps a different reason than many problems. It’s lovely because the solution defies our intuition. It SEEMS like it should be a 50-50 proposition, but it’s not.

So then here’s the interesting teaching and learning question for me…what sorts of evidence will students accept (really accept, not just nod their heads to) that their intuition is wrong? What sorts of questions do they need to work on in order to develop new intuitions?

April 26, 2011 at 12:52 pm |

The new scenario still has only a single prize – so fours doors, one Cadillac and three mules.

And it is true that in fact most of my students at first believe that it is a 50-50 proposition when the situation is first presented. We simulate the game using cards – students working in pairs are given 2 kings and 1 ace from a deck of playing cards. The kings represent the mules and the ace represents the Caddy. The cards are shuffled and placed face down on the table. One student, the Host, peeks to see where the ace is. The other student, the Contestant, selects a card. Then the Host reveals a king from the two unselected cards and offers an opportunity to switch. They repeat this process 20 times switching, and another 20 times not switching and record whether the prize was chosen. All of the results for the entire class are compiled and that is when the students believe that it is actually better to switch than to stay. It’s remarkable how close the experimental values come to the actual probabilities.

Are you making any progress on the 4-door version? Or do you want to see my take on it?

May 12, 2011 at 5:06 pm |

[…] so in my regular Monty post I promised an answer to the 4 door Monty question and here it […]