I’m teaching AP Statistics for the first time this year, and my first week just ended. I’ve taught statistics as portions of other secondary math courses and as a semester-long community college class, but never under the “AP” moniker. The first week was a blast.

To connect even the very beginning of the course to previous knowledge of all of my students, I decided to start the year with a probability unit. For an early class activity, I played the classic Monte Hall game with the classes. Some readers will recall the rules, but here they are just in case you don’t know them.

- A contestant faces three closed doors. Behind one is a new car. There is a goat behind each of the other two.
- The contestant chooses one of the doors and announces her choice.
- The game show host then opens one of the other two doors to reveal a goat.
- Now the contestant has a choice to make. Should she
- Always stay with the door she initially chose, or
- Always change to the remaining unopened door, or
- Flip a coin to choose which door because the problem essentially has become a 50-50 chance of pure luck.

Historically, many people (including many very highly educated, degree flaunting PhDs) intuit the solution to be “pure luck”. After all, don’t you have just two doors to choose from at the end?

In one class this week, I tried a few simulations before I posed the question about strategy. In the other, I posed the question of strategy before any simulations. In the end, very few students intuitively believed that staying was a good strategy, with the remainder more or less equally split between the “switch” and “pure luck” options. I suspect the greater number of “switch” believers (and dearth of stays) may have been because of earlier exposure to the problem.

I ran my class simulation this way:

- Students split into pairs (one class had a single group of 3).
- One student was the host and secretly recorded a door number.
- The class decided in advance to always follow the “shift strategy”. [Ultimately, following either stay or switch is irrelevant, but having all groups follow the same strategy gives you the same data in the end.]
- The contestant then chose a door, the host announced an open door, and the contestant switched doors.
- The host then declared a win or loss bast on his initial door choice in step two.
- Each group repeated this 10 times and reported their final number of wins to the entire class.
- This accomplished a reasonably large number of trials from the entire class in a very short time via division of labor. Because they chose the shift strategy, my two classes ultimately reported 58% and 68% winning percentages.

Curiously, the class that had the 58% percentage had one group with just 1 win out of 10 and another winning only 4 of 10. It also had a group that reported winning 10 of 10. Strange, but even with the low, unexpected probabilities, the long-run behavior from all groups still led to a plurality winning percentage for switching.

Here’s a verbatim explanation from one of my students written after class for why switching is the winning strategy. It’s perhaps the cleanest reason I’ve ever heard.

The faster, logical explanation would be: if your strategy is staying, what’s your chance of winning? You’d have to miraculously pick the money on the first shot, which is a 1/3 chance. But if your strategy is switching, you’d have to pick a goat on the first shot. Then that’s a 2/3 chance of winning. In a sense, the fact that there are TWO goats actually can

helpyou, which is counterintuitive on first glance.

Engaging students hands-on in the experiment made for a phenomenal pair of classes and discussions. While many left still a bit disturbed that the answer wasn’t 50-50, this was a spectacular introduction to simulations, conditional probability, and cool conversations about the inevitability of streaks in chance events.

For those who are interested, here’s another good YouTube demonstration & explanation.

There are some holes in your statement of the problem (unsurprising, given how often those holes appear in statements of the problem I’ve seen over the last 25 years or so) that make it ill-posed. It is vital for the probability of winning 2/3 of the time in the long haul by switching to actually make sense for some additional constraints to be in place.

1) the host always know which door contains the prize. (You’ve implied that but it should be explicit.

2) the host ALWAYS offers the contestant the option of switching. If this isn’t explicit, then the host can skew the results capriciously or intentionally by randomly deciding when to offer that option or, more slyly, only when the contestant has in fact picked correctly to begin with, in which case the chances of winning with “Switch” are 0.

I read an excellent, detailed analysis of this problem in the early ’90s in one of the magazines of either AMS or MAA, and I know that since then, there have been follow-up articles by both the same authors and others. There is subtlety here that may not appear to be the case to even those of us who know how things are supposed to work out, and were we simulating with a real game show host, minus those constraints, things could go amiss for the knowledgeable, clear-thinking player.

Your student’s analysis, by the way, is really good, one I’ve not heard before. More typically, I’ve argued that if you change the game to 100 doors with 99 goats and one desirable prize (assuming you don’t want a goat), play the game with the same basic rules, but have the host open 98 of the 99 remaining doors to show a goat behind each after the contestant makes a pick. That tends to hammer home to many skeptics that switching is a vastly better strategy that only gets better with an increase in doors and boobie prizes.

Stipulated on both points. You are correct; I had always assumed your points to be obvious, but are much better stated.

Also funny that you mention 100 doors and 99 goats. I think another student this week mentioned something about more goats making the switching option more viable. Funny coincidence.

Years ago when this first came known (when Marilyn vos Savant was called many nasty names) I found the best way to talk about it was to ask the students what the chances of them choosing the car on their first guess. Hence 1/3 of the time Monty can choose between two doors to show a goat, leaving a goat behind the unopened unchosen door. However, 2/3 of the time the contestant will choose a goat door and Monty has only one choice to show a goat leaving the car behind the last door. This is a problem that never dies–so it must be a good problem.

Doug,

I did share the vos Savant story after the class had grappled with the outcome of the game. Note my side comment about the “degree flaunting PhDs”!

I agree that this really is quite an excellent problem. Much better, IMO, because of its superb counter-intuitive finale.

Thanks, much, for the continued commentary.

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