Yesterday we did not get to the optimal solution, but rather looked at the strategy of stopping when you get a 6 on the first or second roll, and then at stopping when you get a 4 or higher on the first or second roll. I asked my son to think about the problem a bit more this morning while I was out and he was able to find the optimal solution.

Here’s what he did in his own words:

Next he showed how he used Mathematica to help him find the best solution:

Finally, I showed him an alternation approach to finding the optimal solution that comes from working backwards. This is the approach that Cirillo takes in his discussion of the problem: