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.
Sunday, June 28, 2009
Dorothy Revisited -- Another View...
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5 comments:
Here's what I think is an easier way to think of it, without the difficulty (I think) of having to prove the possibilities equally likely.
Imagine that the game does not end when the red card is drawn, and that all N cards are drawn. Dorothy wins as long as the red card is not drawn last. If all N cards are drawn, the red card is equally likely to be drawn first, second, third, ... , Nth. So it is drawn last with probability 1/N, in which case Dorothy loses. Otherwise she wins, probability 1 - 1/N.
Middle schoolers could *follow* this, but even given a fair bit of experience with this type of problem, I don't think they'd *come up with* it. Still a great problem to make them think. :)
Nice way to explain it, Mathmom!
I do feel it is still an elusive idea. One can think of it as N different colors, so it's no more or less likely that the red card (or any other color) is drawn each time. It still seems somewhat paradoxical to me since we are dealing with dependent events (cards are not replaced). Very interesting...
The other way I was thinking of it was, like you, to assume the black cards are distinguishable, and think about all the permutations of orders in which all N cards could be chosen. There are N! of these. How many of them have the red card last? (N-1)! permutations of the non-red cards in the first N-1 spaces. So probability of "red card last" is ((N-1)! / N!) = 1/N
Nice!
I don't know if anyone else is compelled by this discussion but this problem has opened up many important avenues of instruction for solving combinatorial problems.
Your explanation applies equally well to demonstrating that the probability of drawing the red card on any other draw is also 1/n. Since there are n-1 ways to win, the winning probability is therefore (n-1)/n directly.
What I find also instructive for students here is that the counting of permutations is equivalent to determining the probability of each separate draw using dependent events, then applying the multiplication principle. This equivalency is subtle for learners.
I almost always have to do a permutations/combinations question more than one way in order to be confident of my answer. ;-) So it's good to know lots of ways to do it. I do still get temporarily stumped from time to time by a kid who will propose a reasonable-seeming method that does not come out with the correct answer -- figuring out what is wrong with an approach can be very tricky -- a possible challenge problem to pose to students sometimes.
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