My reaction was that it would be fun to turn this into a project for kids, but this one would need a little introduction since conditional probability can be incredibly non-intuitive. During the week I came up with a plan, and we began to look at the problem this morning.

Here’s the introduction – I asked the boys to give their initial reaction to the seeming paradox:

Next we looked at an example that is slightly easier to digest -> rolling two dice and asking “do you have at least one 6?”

My younger son had a little trouble with the conditional probability, so I’m happy that we took this introductory path:

Next we moved to a slightly more difficult problem -> rolling 3 distinct dice. I used a 6x6x6 Rubik’s cube to represent the 216 states. To start, I asked the boys to count the number of states that had at least one six. Their approach to counting those 91 states was really fascinating:

Finally, we looked at the analogy to the 2nd ace paradox in our setting. So, if you have “at least one 6” what is the chance that you have more than one six, and if you have “a six on a specific die” what is the chance that you have more than one six?

Again, my younger son had a little trouble understanding how the cube represented the various rolls, but being able to hold the cube and see the states helped him get past that trouble:

Tomorrow we’ll move on to studying the paradox with the playing cards. Hopefully today’s introduction helped the boys understand