Saw this gem of a probability problem tweeted out by Christopher Long earlier this week:
This problem is – obviously – not a problem for kids, but I thought it would be fun to talk through it with kids anyway. Many of the ideas involved in solving this problem are actually accessible to kids and I’ll treat the ones that are not really accessible to them as details.
Two things that were interesting to me in our conversation this morning were:
(i) The ideas that kids have about fairly complicated math problems are fascinating. It is really fun to hear how kids process problems like this one.
(ii) You can use discussions of complicated problems like this one to talk about not-so-complicated ideas is math. In our conversation today we spent a lot of time talking about some basic ideas in arithmetic. Those ideas on their own probably would not have led to a conversation that was all that engaging to the boys, but in the context of the more complicated problem they remained engaged.
So, we began with a quick introduction to the problem and a short discussion of what “greatest common divisor” means. Immediately my older son thinks that it will not be all that likely that the GCD’s will be the same, and they also wonder about how likely it will be that two of the numbers will be the same.
At the end of the last video, my older son thought that it would be the case that the GCD of two random positive integers would be equal to 2 about 1/4 of the time. We explored that idea in a little more depth for this section because there’s a subtle mistake in his reasoning, but talking about that mistake is a great way to start thinking about the solution of the original problem.
At the end of the last video we made some progress understanding when two integers would not have 2 as a common divisor. I wanted to explore that idea a little more to make sure it was sinking in. There were a few misconceptions so it was lucky that we talked about the ideas a little more. This part of the project is mostly the boys working through the idea that two numbers selected at random will not share 3 as a common factor.
Because it took a little extra time to work through the ideas about two numbers sharing 3 as a factor, I decided to do one extra talk about numbers sharing 5 as a factor. It seemed as though the ideas were starting to sink in, which was good.
At the end of this talk we begin to talk about how to calculate the probability that two randomly chosen integers will have no common factors. There’s a couple of interesting bit of confusion – (i) do we add or multiply the probabilities that we’ve been calculating to get the final answer, and (ii) what do we expect to happen when we multiply a bunch of numbers that are all between 0 and 1?
Next we moved to the computer to see if we could calculate the probability from the last video. We explore the calculation of that probability using a short program in Mathematica:
Finally, I wrapped up the talk today by showing them that the number we calculated in the prior video was – amazingly – nearly equal to 
So, a fun morning talking about an amazing probability problem. We are about half way to the solution now – we’ll try to finish up the discussion tomorrow morning.
Mike's Math Page 