For our project today, I though it would be fun to talk about the Poisson distribution. For me it is one of the most interesting and important ideas in probability. This question, for instance, is fascinating -> If a random event happens on average once per time period, what is the probability that it happens twice?
I started the introduction with a version of the idea I mentioned above and asked my son for some estimates of what he thought the answer would be:
Then we looked at some simulations. Here I’m looking at the idea of a random event that happens on average once per year and chopping the year up into 52 weeks:
Next I chopped the year up into 365 days – would we get different answers?
This project turned out to be a little more interesting to my son than I was expecting – I’m looking forward to exploring Poisson distributions a bit more next week.
Last week I learned about an really interesting probability problem from Pasquale Cirillo:
Today I asked him to think about the problem while I was out and when I got back home he walked me through his solution:
The last video shows his general approach – now he calculated the asnwer:
To wrap up I showed him how to modify his original argument just a bit to avoid the infinite series calculation. This is a much shorter way to solve the problem, but does require a bit more mathematical sophistication:
I really love the problem and think that it is a great one to share with kids. Even if kids can’t quite solve it, it would be really fun to hear their thought process and how they might estimate the probability of winning.
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:
Yesterday we did a project on this fun problem from Futility Closet:
Today we finished the project by talking about the 2nd part of the problem and then having a discussion about why the answers to the two questions were different. Unfortunately there were two camera goofs by me filming this project – forgetting to zoom out in part 1 and running out of memory in part 4 – but if you go through all 4 videos you’ll still get the main idea.
Here’s the introduction to the problem and my son’s solution to the 2nd part of the problem. Again, sorry for the poor camera work.
Next we went to the computer to verify that the calculations were correct – happily, we agreed with the answer given by Futility Closet.
In the last video my son was struggling to see why the answers to the two questions were so different. I’d written two simulations to show the difference. In this part we talked about the difference, but he was still confused.
Here we try to finish the conversation about the difference, and we did get most of the way to the end. Probably just needed 30 extra seconds of recording time 😦 But, at least my son was able to see why the answers to the two questions are different and the outputs from the simulations finally made sense to him.
So, not the best project from the technical side, but still a fun problem and a really interesting idea to talk through with kids.
This problem teaches a couple of good counting lessons. Today we focused on the first part – if you have at least one ace, what is the probability that you have more than one. First, though, we talked through the problem to make sure my son understood it:
Next I asked my son to work through the calculation for the number of hands that have “at least one ace.” He made a pretty common error in that calculation, and we discussed why his calculation wasn’t quite correct:
Now we talked about how to correct the error from the last video via complementary counting:
Now that we had the number of hands that had at least one ace, we wanted to count the number of hands with more than one ace. My son was able to work through this complementary counting problem, which was really nice to see:
Finally, since we had all of our numbers written down as binomial coefficients and these numbers were going to be difficult to compute directly, we went to Mathematica for a final calculation:
Excited to continue this project tomorrow and hear my son’s explanation for the seeming paradox.
James Tanton’s Solve This book is full of incredible math projects to do with kids:
Today we went back to do a project that we’ve looked at a few times before – cutting various versions of a Möbius Strip.
We started with a cylinder just to get going with an easy shape:
Next we moved on to a Möbius strip – but when we made this one we didn’t notice that we’d made a full twist rather than a half twist. So we sort of got two surprises. I’ve included this mistake to show that you do have to be a little careful when you do this project – it is easy to accidentally make the wrong shape:
Here’s the correction and the actual Möbius Strip:
Next was a shape where we had a 2nd little error – this time the tape held the final shape together a bit too much. In our video you can sort of guess what the actual shape will be, but I definitely encourage everyone to try this one out for yourself!
The 4th shape is where things really get interesting. For this and the next two shapes we are making a circular cut through a cylinder with a hole cut out of it. Each of the shapes has a interest set of twists in it – the shape in this video has a single half twist on one of the “arms”:
The 5th shape has half twists (going the same direction) in each of the arms:
The 6th shape has half twists going in different directions in each of the arms:
This really is an amazing project for kids. As our version of the project here shows (accidentally!), you do have to be careful with the preparation, but even with the couple of errors we had, it was still an incredibly fun morning.
I left two copies of the puzzle for my son to work through while I was out this morning. For the first run through I asked him to solve the puzzle as it was stated. Here’s his work and his explanation:
For the second run through I asked him to solve the problem assuming that the radius of the circle was X rather than 5. This was first step in what I was hoping would be an interesting algebra exercise. Here he was successfully able to use the quadratic formula even though the equation he found had 2 variables:
For the last part of the project I wanted to see if he could factor the equation he found in the last video. This turned out to be a significantly more difficult challenge, but he figured out how to do it just as we ran out of space on the memory card!
I suspected that the factoring challenge would be more difficult than simply using the quadratic formula, though I didn’t realize how much different it would be. I might try to find some more challenges that involve multiple variables just to get a bit more practice with these ideas.
Last night I asked my younger son what he wanted to do today for a project and he said that he wanted to talk about the permutahedron a bit more. In yesterday’s project we talked about the permutations of the set (1, 2, 3, 4), so today we started by going down to some simpler sets of permutations:
Next we looked at the shape made by the permutations of the set (1, 2, 3). The way my son thinks through this problem shows why I love sharing ideas from math research to my kids.
To wrap up today we dove a little deeper into one of the ideas we talked about yesterday – in the permutations of the set (1, 2, 3, 4) is there a permutation that requires 4 or more flips to get back to the starting point of (1, 2, 3, 4)?
The permutahedron is a really neat shape to explore with kids, and hearing them talk about and think through the shape itself is incredibly fun.
Today we explored the and idea that Ardila discussed in the podcast – finding paths on the permutahedron.
We started by just reviewing what the shape is and what it represents:
Next we tried an easy example of finding a path on the permutahedron going from a random permutation back to the correct order:
For the last part of the project we tried a more complicated scramble of the cards and found that walking back to the unscrambled state would take a minimum of 3 steps:
I love playing with this shape with kids. It is a great way to get them talking about fairly advanced mathematical ideas and also allows them to see a really neat 3d shape that research mathematicians find interesting!