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Sunday, May 22, 2005

How to approach a probability problem

If you are dealt a hand of five cards, what is the probability that you have three diamonds and two clubs?

There are (at least) two different approaches to this kind of problem:
  1. The chance the first card is a diamond is 13/52, then the next is 12/51, then 11/50; there are still 13 clubs, so we have 13/49 and then 12/48. Multiply these all together. But since the problem didn’t say that it had to be the first three that were diamonds, we have to multiply by 5C3 since there are that many triples of slots for the diamonds, and then by 5C2 since there are that many pairs of slots for the clubs. I get 0.0085834.

  2. The definition of probability says that I should count the number of ways to get the 5 specified cards and divide it by the number of combinations of 5 cards in the whole deck. So, I need 3 of the 13 diamonds (13C3) and 2 of the 3 clubs (13C2). Divide them by 52C5 and I get 0.0085834.

Years ago I approached problems like this with the first method. Now I use the second, which seems simpler to me. But one of my esteemed colleagues claims that the first method is actually simpler, and that students understand it better.

Now I suppose the truth must be that some students understand one method better, and some understand the other one better. You could argue that that means that we should teach both, but that’s just going to confuse all the weaker students. What should we do?

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