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1. This website plays fast and gives information about the problem.

http://www.amstat.org/publications/jse/v6n3/applets/LetsMakeaDeal.html

Please discuss your thoughts regarding this information and the problem.

Try playing 20 times, switching curtains each time, then report back to the class the number of times you won the prize. 

So, should you switch curtains after Monty opens the non-winning curtain? What are your initial thoughts? (Be careful!)

Class, suppose you switch curtains. What is the ONLY way you will lose the game?

Now, just for fun, search the Internet for "Monty Hall Problem" (or something similar) and see how many websites you can find devoted to this question.

--Tom

Note: If the above link does not work for you its due to incompatibilities with your JAVA settings or an older JAVA install; either correct or select another Let's Make a Deal simulation from the web as there are more than one could ever count...

Understanding Why Switching Works

Here’s an easier way:

If I pick a door and hold, I have a 1/3 chance of winning.

My first guess is 1 in 3 — there are 3 random options, right?

If I rigidly stick with my first choice no matter what, I can’t improve my chances. Monty could add 50 doors, blow the other ones up, do a voodoo rain dance — it doesn’t matter. The best I can do with my original choice is 1 in 3. The other door must have the rest of the chances, or 2/3.

The explanation may make sense, but doesn’t explain why the odds “get better” on the other side.

Understanding The Game Filter

Let’s see why removing doors makes switching attractive. Instead of the regular game, imagine this variant:

There are 100 doors to pick from in the beginning

You pick one door

Monty looks at the 99 others, finds the goats, and opens all but 1

Do you stick with your original door (1/100), or the other door, which was filtered from 99? (Try this in the simulator game; use 10 doors instead of 100).

It’s a bit clearer: Monty is taking a set of 99 choices and improving them by removing 98 goats. When he’s done, he has the top door out of 99 for you to pick.

Your decision: Do you want a random door out of 100 (initial guess) or the best door out of 99? Said another way, do you want 1 random chance or the best of 99 random chances?

We’re starting to see why Monty’s actions help us. He’s letting us choose between a generic, random choice and a filtered choice. Filtered is better.

But… but… shouldn’t two choices mean a 50-50 chance?

Overcoming Our Misconceptions

Assuming that “two choices means 50-50 chances” is our biggest hurdle.

Yes, two choices are equally likely when you know nothing about either choice. If I picked two random Japanese pitchers and asked “Who is ranked higher?” you’d have no guess. You pick the name that sounds cooler, and 50-50 is the best you can do. You know nothing about the situation.

Now, let’s say Pitcher A is a rookie, never been tested, and Pitcher B won the “Most Valuable Player” award the last 10 years in a row. Would this change your guess? Sure thing: you’ll pick Pitcher B (with near-certainty). Your uninformed friend would still call it a 50-50 situation.

Information matters.

The more you know…

Here’s the general idea: The more you know, the better your decision.

With the Japanese baseball players, you know more than your friend and have better chances. Yes, yes, there’s a chance the new rookie is the best player in the league, but we’re talking probabilities here. The more you test the old standard, the less likely the new choice beats it.

This is what happens with the 100 door game. Your first pick is a random door (1/100) and your other choice is the champion that beat out 99 other doors (aka the MVP of the league). The odds are the champ is better than the new door, too.

http://betterexplained.com/articles/understanding-the-monty-hall-problem/