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.
Yesterday’s discussion helped the boys understand the problem that Ellenberg is discussion in chapter 10 of his book a bit better (hopefully anyway!). Today we took a crack at replicating the calculations in the book relating to the roulette wheel example.
First we revisited the example from the book to make sure we had a good handle on the problem:
Next we talked through the details of the process that we’ll have to follow to replicate the calculations that Ellenberg does. Following the discussion here the boys did the calculations off camera:
Here we talk through the numbers that the boys found off camera – happily we agreed with the numbers in the book.
At the end of this video I introduce a slight variation on the problem – instead of getting R, R, R, R, R in a test of 5 rolls, we get an alternating sequence of R and B for 20 rolls:
Here are their answers – and a discussion of why they think the answers make sense – for the new case I introduced in part 3 of the project:
This two project combination was really fun. My younger son said that he was confused by the roulette wheel example, but I think after these two projects he understands it. I think it is a challenging example for a 9th grader to understand, but with a little discussion it is an accessible example. It certainly makes for a nice way to share some introductory ideas about Bayesian inference.
Today I had my son explore a little further. He was interested to see if different starting positions led to different distributions of endings. He looked at five different starting positions. Here’s the first (with a quick review of the problem) when the urn starts with 5 black and 5 white balls and we play the game 1,000 times:
Next he looked at how the starting position with 1 black ball and 5 white balls evolved. The way the distribution of the number of white balls at the end changes is pretty amazing:
Now for the most surprising one of all – the starting position with 1 white ball and 1 black ball – it seems that ending with 1 white ball or 1001 white balls (or any amount in between!) is equally likely:
Finally he looked at the starting position with 1 black ball and 10 white balls. This one is a little less surprsing having already seen the 1 black ball and 5 white ball game, but still it was neat to see:
This is a fun little game for kids to study. It is also a nice introductory programming exercise, too. Thanks so much to Ole and Marcos for sharing their ideas!
My younger son is working through an introductory probability and stats book this school year. This week he came across a problem in the Bayes’ Theorem section that really gave him a lot of trouble. I was a little caught off guard by how much difficulty he was having with the problem (and how little my help was helping!) but then we got a great bit of luck when Julia Anker posted her solution to the problem:
First I used a Venn diagram with just A and B. Once I had that one figured out I used that and the rest of the info to figure out the overlapping C parts. pic.twitter.com/83MoPQpeNU
I had my son work through Anker’s solution yesterday and today we talked about the problem. Here’s the introduction to the problem and to his thoughts about Anker’s work:
At the end of the last video I had my son pick two of the regions in Anker’s diagram to see if we could verify the calculation. Here’s the calculation for the first region:
Here’s the calculation for the second region and an overall wrap up on the problem:
I’m extremely graful to Julia Anker for sharing her solution to the problem. It is an odd problem to have spent 5 days on for sure, but thanks to the extra help my son was really able to turn the corner and understand the problem. Math twitter is the best!
Yesterday we did a neat geometry project inspired by an amazing thread from Freya Holmér:
given three points, you can always find a circle passing through them all
1. draw lines from a to b to c 2. draw perpendicular bisectors (dashed) 3. the circle center is their intersection point 4. the circle radius is the distance from the center to any of the three points pic.twitter.com/9VceD4N8vG
Today we are extending that project by trying to find the expected area of the circle when the three points are inside of a unit square.
To start the project we talked through a bit of the geometry that we need to answer the question about the expected area of the circle:
Before jumping into the computer simulation we had to check a few more geometric details – here we talk about using Heron’s formula:
Now my son took 15 min off camera to write a simulation to find the expected value of the area of the circle. Here he walks through the program and we look at several sets of 1,000 trials:
Finally, we finish up with a bit of a surprise – switching to 10,000 trials, we find that the mean still doesn’t seem to converge!
Turns out the expected area of the circle is infinite – that’s why we aren’t seeing the mean in our simulations converge. I think this is a great way to show kids an example where the Central Limit Theorem doesn’t apply.
My younger son wants to learn more about statistics. I’m excited to come up with some projects, though our journey here is probably not going to look like a typical statistics class.
My first idea was to have him investigate coin flips and look at similarities and differences with 10, 100, and 1000 coin flips. But really before we even got started he had a really interesting question.
So, here’s how he described the program I had him write ahead of time, and then we discuss the question he had -> In 100 flips, why did it seem that the chance of getting less than or equal to 50 heads was more than 50%?
To start diving in to his question, we first looked at sequences of 100 flips to get a better sense of what was going on. Interestingly, on a test look at 10 sequences, we did find that 6/10 had less than 50 heads:
Now we looked at the distribution of heads in 100 flips using Mathematica’s Histogram function – it was really interesting to hear him describe the different distributions that we saw:
Finally we talked about why my son was seeing what he was seeing in his coin flip program.
From just a little bit of time on the site, I thought having my younger son read and play with some of the ideas would make a great project. So I asked him to spend 20 min reading and exploring, and then we talked.
Here are his initial thoughts:
One of the things he thought was interesting was the idea of 3 and 6 degrees of separation when you have a few connections and how much the network changes when you just add one connection (on average) per person:
Another thing he thought was interesting was the companion site that allowed you to modify connections in the network. Here he looked at the size of the largest group when you made the change from connections with only essential works to again adding 1 connection on average for everyone:
I really like how the ideas of network connections are explained on this site. Their work makes a fairly complex idea accessible to everyone – including kids. Thanks to Bill Hanage for sharing this site!
I thought that talking through these problems would make a nice project for the boys today, so we started in on Alex’s problem. The nice thing right from the start is that the boys had different guesses at what the expected value of one die would be when the combined roll was 8:
Now that we had a good discussion of the case where the sum was 8, we looked at a few other cases to get a sense of whether or not the intuition we developed from that discussion was correct:
Next I introduced the problem on Gil Kalai’s blog – again the boys had different guesses for the answer:
I had the boys write computer programs off screen to see if we could find the answer to the problem on Kalai’s blog via simulation. The interesting thing was that the boys approached the problem in two different ways.
First, my younger son started looking at dice roll sequences and he stopped when he found a 6 and always started over when he saw an odd number. He found the expected length of the sequence of rolls was roughly 1.5:
My older son looked at dice roll sequences and he stopped when he found a 6 but instead of starting over when he found an odd number, he just ignored the odd number. He found the expected length of the sequences looking at them this way was 3:
This turned out to be a great project. I’m glad that the boys had different ideas that we got to talk through. These conditional expectation puzzles can be tricky and subtle, but they are always fun!