James Tanton’s counting problem part 2

Yesterday we looked at a really neat problem James Tanton posted last week:

That project is here:

Working through a challenging counting problem from James Tanton

Our first look at the question involved some dice rolling and a computer simulation. Today we are going to look at an exact solution to the problem. That solution involves studying all of the different things that can happen when you roll 5 dice. It turns out that there are 7 different patterns that can happen, and these patterns related to the ways you can write 5 as the sum of positive integers.

(1) 5 different numbers, which I’ll represent as 1 + 1 + 1 + 1 + 1

(2) 3 different and 2 the same -> 1 + 1 + 1 + 2

(3) 2 different and 3 the same -> 1 + 1 + 3

(4) 1 different and 4 the same -> 1 + 4

(5) 1 different and 2 pairs -> 1 + 2 + 2

(6) 1 pair and 1 triple -> 2 + 3

(7) All numbers the same -> 5

For today’s project we’ll count the number of ways that each of these 7 patterns can occur. We know that the total number of arrangements is 7,776, so that’s going to help us make sure we have counted correctly.

Here’s the introduction to the problem and to the approach we are going to take today:

Now we began to count some of the arrangements. In this video we count the number of dice rolls in (1), (4), and (7) above:

Now we moved on to some of the more challenging arrangements. Here we looked at (6) and (2) above:

Now we looked at case (5). This case proved challenging because dealing with the 2 pairs caused a little confusion between over counting and under counting. But, after looking at the cases carefully we did manage to get to the answer.

At this point we had only one case left -> (3) from above. But, the counting practice that we’d had up to this point helped this case go pretty quickly.

Finally, we added up our numbers and checked that we’d found all 7,776 cases. We did!

The one thing left to do was to count the different numbers that we saw in each case and find the average. I’d done that ahead of time just to save a bit of time in the movie. Our final answer was (27,906) / (7,776) or about 3.588. The exact answer was (happily!) very close to the two estimates that we had found in our simulations yesterday.

I love Tanton’s problem. It is a great estimation problem as well as a great counting problem. We might do one more project tonight on yet a different way to solve the problem using Markov chains:

Looks like a fun idea – I’ll be thinking about how to talk through this approach with the kids during the day today.

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