This afternoon my son got a pack of rockwool for a science project:
I managed to squeeze in a path counting project before he used it!
I started with a question about counting paths that he could solve by listing all of the possibilities:
In the last video my son was getting a little stuck trying to find a way to count the paths, but he did have a few ideas. Here we looked at a different set of paths and were able to find a good counting method:
Finally, with our counting formula in hand, we calculated the number of paths from the bottom left corner to the top right one:
Obviously this wasn’t a planned project, but it was fun to share this neat counting idea!
On New Year’s eve my older son was playing poker with his friends using 2 decks of cards – see if you can spot the oddity in the picture 🙂
For today’s project we decided to explore some of the probability ideas around playing poker with 2 decks of cards. First we just looked at the possible hands and talked about some potential questions to ask:
For an introductory problem, we looked at the number of ways of getting a Royal Flush and then all types of flushes with a 5 card hand dealt from a single deck of cards:
Now we looked at how regular flushes could happen when dealt from a two decks of cards shuffled together. We also had a good discussion about whether or not it was more likely or less likely to get a flush in the 2 deck situation:
Finally, we went back to Mathematica to take a look at the numbers for a general flush with one and two decks. Here we are lumping all kings of flushes together – regular ones, straight flushes, and Royal flushes are all the same.
Compute the chances of various hands with 2-deck poker is a pretty fun math exercise for kids.
A couple of years ago Jim Propp suggested a neat counting exercise for the boys – counting tilings of 2xN rectangles by 2×1 dominos. We’ve played with this idea twice before, but thought it would be fun to revisit it today.
The old projects are here:
A fun counting exercise for kids suggested by Jim Propp
Revisiting counting 2xN domino tilings
I started by reminding the boys of the problem and we checked out a few of the simple cases:
Next we looked at the 2×4 case and found that there were 5 possible arrangements:
Now we moved on to the 2×5 case – we had a long discussion about how to determine if we had found all of the possibilities:
Finally, we discussed why the Fibonacci pattern we were finding was correct. The argument here is a slightly sophisticated one for kids, but they were able to find it!
Yesterday we looked at the famous Birthday problem – how many people do you need to have in a room to have a 50/50 chance of two people having the same birthday? That project is here:
Diving into the Birthday problem with kids
Today we continued the project (with just my older son as my younger son was hiking) and studied the problem that originally motivated this project -> If you have 24 students in a class, what is the chance that exactly 3 pairs of students will share a birthday? This is the surprisingly fun situation in my son’s English class.
We will – as I think it standard for the introductory version of this problem – be making the assumption that all birthdays are equally likely. If you want to see a really neat discussion – though not really a math for kids paper! – see the paper in this tweet:
So, to start the project today we first reviewed the main ideas from yesterday:
Next we took a step towards solving the problem by looking at the chance of having exactly 2 pairs. Once piece of the counting here is tricky, so we used the computer to help see what the problem was.
Now we tackled the “exactly 3 pairs problem”:
Finally, I had my son make up a problem to solve – he decided to find the chance of all 24 students pairing up. This problem wasn’t too hard given the prior work. It was also a fun challenge to try to estimate the chance of this happening.
Earlier in the week I learned that my older son’s high school English class has 24 students and 3 pairs of students who share the same birthday. None are twins, so no tricks or anything like that, just a fun fact for this particular class.
I thought it would be fun to figure out how rare something like this would be – assuming, of course, that all of the birthdays are randomly distributed amount the 366 possible birthdays (366 because many of the kids were born in 2004).
It turns out the chance of having exactly three pairs of kids with the same birthday (and no other shared birthdays) in a class of 24 kids is roughly 2.3%, or if you prefer the exact answer:
Instead of continuing with Mosteller’s book this weekend, I thought it would be fun to dive into the birthday problem. I started today with the standard problem – how many people do you need in a room for a 50% chance of two people sharing the same birthday. This is not an easy problem and the answer is not intuitive.
Here’s how we got started – not surprisingly, down a path that wasn’t quite right:
After coming up with a formula in the last video, we went to Mathematica to see what it said. Here we discovered that the formula was giving answers that were not correct:
Now we returned to the whiteboard and the boys found a new formula – this one calculated the chance of having exactly 1 pair with the same birthday. I was happy that they were able to derive this formula and even happier for the chance to show them it didn’t agree with our computer modelling!
Now we went back to the computer to see the surprise that our new – and much closer to correct – formula actually didn’t agree with the modelling. What was wrong?
Finally, having figured out why the two approaches didn’t match, the boys were able to find the correct formula to solve the problem. Tomorrow we’ll dive into the more complicated problem of finding the probability of 3 pairs:
Problem #9 from Mosteller’s 50 Challenging Problems in Probability is about the game of craps. The question asks, essentially, does the player of the casino have a better chance of winning the game.
This is both a fun and reasonably difficult problem for kids, but it led to a terrific conversation.
Here’s how I introduced the problem:
Following the introduction, I had the boys solve for the probabilities of an immediate win or loss:
Now we moved on to the harder question – what happens if you roll. say, and 8 on the first roll. How do we find the probability that you win the game in this situation?
One way would be summing an infinite series, but I hoped to introduce the boys to a simpler way of seeing the probability here:
Having solved one of the hard cases exactly in the last video, we moved on to solve the rest of them here:
Finally, we went to Mathematica – not for anything super complicated, just to add up the fractions – so we could find out whether or not the player or the casino had the advantage in the game:
This problem is a great one for kids to explore – it really shows how a systematic approach to problem solving can help you get throw a pretty challenging problem.
We are up to problem #8 in Mosteller’s 50 Challenging Problems in Probability.
The problem today is a classic -> What is the probability that you would be dealt a bridge hand with 13 cards of the same suit from a well-shuffled deck of cards?
We started off by taking a quick look at the problem and getting a few ideas from the boys about how to solve it. One question that came up was whether or not the method used to deal the cards would matter:
The first cut at solving the problem involved dealing the cards in a circle – so what I’d think of as the standard way to deal cards:
Next up we took a little detour into choosing numbers because some of the details of how those numbers worked were a little fuzzy. It was a nice review and I was happy that the boys had recognized that the choosing numbers were somehow related to problem:
Finally, we wrapped up by checking to see if the probability changed when we used a different method of dealing.
I think this is a great question for kids to think through. The thoughts from the boys here are probably representative of some of both the struggles and connections that kids will have thinking through this problem.