To your question, I suppose I’m a little unclear what technology Riley has used in his lesson. Plastic bowls? Candy? Whiteboard markers?

My appreciation of the lesson is rooted in one particular moment, where Riley takes a student rebuttal and runs with it, drawing fifty marks on the board and asking a pointed question that helps the student strengthen his own conceptual framework for the Monty Hall problem.

That took courage, content knowledge, and empathy; not much technology. But that’s just my interpretation of the whole thing.

I am ultimately curious with why my first interpretation of Dan Meyer’s comment, “And then he kills the probability lesson anyway.” was opposite of his intent. Seems to me the measure of good teaching strategy is clarity of explanation combined with slick technology use.

Am I off? I would have thought good teaching might be measured with something like quantifying (or characterizing in some form) student learning. Engaging kids is clearly a huge part of the battle. But creating mathematical ways of knowing & understanding for learners is the pedagogical hurdle.

Else, it is merely another, slicker, snake oil sale.

No hate intended to all commenters. I am a fan, and truly do dig the multimedia use and the impressive ideas for engaging kids.

If you choose at random the second time, you have a .5 probability of winning, however this probability is not distributed evenly. 1/6 of the time you would win by sticking with your original choice, 1/3 of the time you would win by switching, 1/3 of the time you would lose by sticking with your original choice, and 1/6 of the time you would lose by switching.

Because this probability is not evenly distributed, you should choose to switch. Your probability of winning by staying then is 1/3 and your probability of winning by switching is 2/3.

Now the explanations I’ve seen didn’t do it for me. What did it form me was to think of it as sets of door combinations. Assuming you’ve picked door 1, there are three possibilities for these doors: {W, L, L}, {L, W, L}, and {L, L, W}. Monty is always going to show one of the losing doors that you didn’t pick, reducing our sets to: {W, L}, {L, W}, and {L, W}. If you pick at random, there is one way of winning by sticking with door 1, and two ways of winning by switching to the remaining door.

I’ve read this over three times now and think it makes sense. Please tell me I haven’t botched it again. :)

NetLogo by
http://ccl.northwestern.edu/netlogo/
The Monty-Hall simulation is included in the program.

Take a lot of turtles and make them advance on a football field when they win a game. Configure some of them to always switch door, some of them to always keep the door and some of them to randomly switch.

Now make them play the game until they reach the end of the football field. That should show clearly that switching is a great strategy. I’m not sure how useful it would be pedagogically (I’m a programmer).

I’m also developing a probability website with simulators (it’s free), but it’s in French, it’s beta and it needs the latest Flash Player. It does include a Monty-Hall simulator:
http://www.netmaths.net/FeteForaine/

Have fun and thx for the great post
Carl from BuzzMath.com

Here’s the setup: G1 G2 C (Goat1 Goat2 Car)

Scenario A: Always switch
I chose G1, he showed G2, I switched to C -> I won
I chose G2, he showed G1, I switched to C -> I won
I chose C, he showed G1, I switched to G2 -> I lost

Wins 2, Losses 1

Scenario B: Never switch
I chose G1, he showed G2, I didn’t switch -> I lost
I chose G2, he showed G1, I didn’t switch -> I lost
I chose C, he showed G1, I didn’t switch -> I won

Wins 1, Losses 2

I was getting tripped up on the textual description of “winning,” that switching will cause you to win 2/3 of the time. In fact, the probability of winning is actually 1/2 as we intuitively expect it to be.”

No. This is incorrect.

Here’s another way of understanding this. We’re going to play two variations of the same game, only with cards. I have three cards: two 2’s and one Ace. You’re trying to find the Ace.

In the first variation, I lay them face down in front of you and tell you to point to one but don’t look at it. You point at one of the cards. I then discard one of the other two, showing you that it’s a 2. That means that there’s a 2 and an Ace still on the table. Now, I PICK UP the two cards, shuffle them where you can’t see, then lay them back down and tell you to pick one. Obviously, you have a 50/50 chance of getting the Ace.

Now for the second variation. I lay the three cards face down in front of you and tell you to pull one to yourself and KEEP it there, face down. I then pick up the other two cards, discard one — showing you that it’s a 2 — and lay the remaining one face down and ask you if you want to switch. In this variation, there is only 1 chance in 3 that the card in front of you was the Ace when you chose it. It’s stayed in front of you, and I haven’t done anything to it. Therefore, it’s STILL only 1/3 likely to be the Ace. No matter what I’ve done to the other SET of cards, the Ace is STILL more likely to be on my side of the table. Thus, if I am forced to get rid of a 2, whatever card I have left is 2/3 likely to be the Ace.

Does this help clear things up?