Dorothy Revisited -- Another View...

Mathmom contributed some insightful thoughts about how most middle school students might feel about the probability investigation from the other day. I agree with her that some would be able to compute the results or even devise a general formula but "proving" it in the general case might be too ambitious. In my reply, I suggested there might be another way of deriving the formula 1/N for the probability of losing the game. Here's what I came up with. It still requires some careful development to show that the outcomes are equally likely but I will indicate how it could be done in the particular case where N = 10.

Brief Explanation of Method:
There are N equally likely (to be shown) ways for the game to end (i.e., when the red card is selected). Of these, only one will result in a loss -- when the red is the last card chosen. Therefore, the probability of losing is 1/N, hence the probability of winning is 1 - 1/N or (N-1)/N.

Demonstrating "Equally Likely" for N = 10:
P(game ending after one card) = 1/10
P(game ending after 2 cards) = P(black selected followed by red) = (9/10)(1/9) = 1/10
P(ending after 3 cards) = P(black,black,red) = (9/10)(8/9)(1/8) = 1/10
etc...

The general case is similar using N in place of 10. I do think that students with some understanding of algebra could follow it but deriving it on their own is another story!

I also indicated that I might provide a program for the TI-83 or -84 which could be used to simulate the game. The programming skills needed are not that advanced and some high schoolers or even middle schoolers can pick up on the code and begin writing their own programs - I've seen it happen! Here it is...











T represents the number of times the game is played with 3 cards. I entered 100 for the number of trials. K stores the number of times Dorothy won when playing 100 times. Can you make sense of the rest of the code?

The experimental probability of 0.68 is reasonably close to the theoretical probability of 2/3. I often feel more confident of my reasoning in difficult probability problems when my simulation approximates my answer. This doesn't prove anything but it does have value IMO. There is also the opportunity to demonstrate some important stat concepts by running the program several times and having students plot the experimental probabilities and observing their distribution.

This Logic Challenge is 'Par for the Course'!

Don't forget our MathAnagram for Aug-Sept. Thus far we have received a couple of correct responses. You are encouraged to make a conjecture!
Look here for directions. Here is the anagram again:

PRINCE? NAH! E-ROI!

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A former student sent me a wonderful reasoning problem involving mean, median, and mode, so it is accessible to middle schoolers. The question came from his teacher so I decided to revise it, put it in a different context but preserve the essence of the logic. The student will need to know some basics of scoring in golf but most of it should be clear. If not, this may help.

This kind of question will frustrate some but reasonable frustration can often lead to 'pearls of wisdom.' Clear thinking and careful attention to detail is necessary. Certainly basic knowledge of measures of central tendency is a prerequisite, but this question can also serve to review these ideas.

Have fun with it yourself and, if you can, try it as a 5-minute warmup in class, preferably with students working in pairs. Let us know if they make a 'hole in one'! Again, thanks to my student and his teacher for the original source of this challenge.

Alex played 18 holes of golf and we know the following information:
His maximum score on any hole was a '5' and he shot this on six holes.
His median score on the 18 holes was 4.
The mode was 3.
What was the lowest possible mean score he could have achieved on the 18 holes?
Express your answer to 2 places (rounded).