Having my son talk through an interesting probability problem from Pasquale Cirillo

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.

Talking through a neat introductory probability / expected value problem from Pasquale Cirillo with my younger son.

This problem was posted by Pasquale Cirillo last week and I thought it would be great to talk through with my son:

This is the video where Cirillo talks through the problem if you want to see his full solution:

I’d asked my son to think about it a bit ahead of us sitting down to talk. Here are some of his ideas:

Here we find the expected winnings if we use my son’s first rule – stop in either of the first 2 rounds if you get a 6:

We ended the day today by talking through the strategy of stopping in either of the first 2 rounds if you see a 4, 5, or 6.

At the end of the last video I let my son know that we hadn’t quite found the best strategy, yet. Tomorrow we’ll finish up and find that best strategy.

The paradox of the 2nd ace part 2

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.

Revisiting (again!) the Paradox of the Second Ace

Today my younger son and I are going back to a problem that we’ve looked at before – the “Paradox of the Second Ace.” This is a problem I learned about from Futility Closet:

Here’s the problem as described on their site

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.

Why I love sharing ideas from research mathematicians with my kids

Yesterday we did a project inspired by the great podcast conversation between Steven Strogatz and Federico Ardila here:

That project is here:


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.

Revisiting the permutahedron with my younger son after listening to Steven Strogatz’s interview with Federico Ardila

This week I listened a great conversation between Steven Strogatz and Federico Ardila on the Joy of X podcast:

We have played with the shape a few times before – see these blog posts of you are interested in seeing other ways that kids can explore the shape:



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!

Revisiting Larry Guth’s “no rectangles” problem – one of the best math activities for kids I’ve seen

I was looking for a relatively stress-free project to do with the boys this morning and thought it would be a good day to review one of my all time favorite math projects for kids -> Larry Guth’s “no rectangles” problem. One of my all time favorite moments doing math with kids came when I used this problem for a 2nd and 3rd grade Family Math night at my younger son’s elementary school a few years ago. The problem is accessible to kids of all ages and also of interest to research mathematicians.

We started with a quick review (and lucky clarification!) of the problem and then the boys tackled the 3×3 case:

Next we moved on to the 4×4 case. The thought process the boys went through here I think shows why this is such a great math problem for kids to talk through:

Next I had a film goof up – luckly it was just 30 seconds of introducing the 5×5 case and telling them that answer to the 5×5 case was 12 squares. Don’t know what happened to this piece of the film, but since the plan was for the boys to play with this part off camera anyway I didn’t bother trying to fix it.

In any case, here’s the 12 square covering for the 5×5 they found and then a brief discussion about the surprise that comes when you move to the 6×6 square:

Again, this is one of my favorite math projects for kids – and kids of all ages. It is a really fun problem to play around with.

Revisiting Futility Closet’s “Paradox of the 2nd Ace” with my younger son

This week my younger son told me he wanted to learn a bit more about statistics. By lucky coincidence I happened to stumble on one of our old projects while trying to answer a question on twitter:


This project was about a fun probability problem I learned in this tweet from Jon Cook:

I stared by introducing the problem to my son and asked what he thought the answer was going to be:

Then we started in on the calculations. Finding the probability that someone having “at least one ace” had more than one was a little challenging, but we found the right approach after a few tries:

Next up was calculating the probability that someone who had the ace of spades would have a second ace. After the work we did in the last part, this calculation was easier:

Finally, we went to the Futility Closet page to see the numbers and then I asked my son why he thought the surprising result was true.

Definitely a fun project and a neat probability surprise for kids to see.

Having my younger son work through Graph Theory for Kids by Joel David Hamkins

I’ve been playing around with a big of graph / network theory with my younger son the last two days. This morning I had him take a look at a graph theory project made for kids. This project is actually aimed at kids who are a little younger than my son, but I thought it would still be a good exercise for him.

You can find the pdf for the project on Joel David Hamkins’ website:


My son spent about 20 min working through the project and then we talked through all of the pages. Here are his ideas. If you listen to the conversation we have, you’ll see what a great little exercise Hamkins’ project is for kids:

Looking at a version of a problem originally proposed by Nassim Taleb that is accessible to kids

Last week Nassim Taleb proposed a really nice problem as an extension of a problem proposed by Zhou Xi:

Nassim’s problem really isn’t accessible to kids, but a slight variant is -> how many sequences of 250 coin flips are there where no run of heads or tails is longer than 2 flips?

I decided to go through that problem with the boys this morning. It was just at the right level to really challenge them, but still fit inside of a 30 min project.

We started by looking at Xi’s problem and they both had pretty good intuition for which sequence was which:

After the short introduction we started trying to figure out how to tackle the problem about sequences where the longest run was at most 2. After thinking of a few other ideas first, they decided to take a look at some shorter sequences to see if that would help us get some intuition for how many had runs that were no longer than 2:

Now we took a look at the sequences of length 4. Luckily there are only 16 different coin flip sequences of length four, so we could write them all out. The boys found that there were 10 sequences with runs no longer than 2. That led to an idea about how many would work in general:

Now we had a conjecture – there would be 16 sequences of length 5 that had runs no longer than 2 – so we tried to count those sequences directly to see if the conjecture was right:

Finally, we sketched a general proof of the conjecture (I’m intentionally being vague on what it is to not give it away). This part was also a little difficult for the boys, but they eventually saw the right pattern and that pattern led to the general proof:

This problem made for a really fun project this morning and Nassim’s problem led to some great twitter discussions that lasted all week. I was happy to be able to find a piece of Nassim’s problem that the boys could tackle.