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.
We are working through Mosteller’s 50 Challenging Problems in Probability this fall. Today we tackled problem #3 which is an problem that is definitely accessible to kids and has a fun and surprising result.
The problem can be summarized like this:
You have two people who each have probability p of making a correct decision and a third person who just flips a coin. If this group of three reaches a decision by majority rule, what is the probability that they make the correct decision?
I started the project today by sharing this problem with the boys and asking them what they thought:
They had a pretty good idea about how to approach the problem, so for the 2nd part of the project we dove in to the calculation and found the surprising result:
To wrap up, we wrote a short computer program to simulate the problem and see if the results of that program matched what we’d found in the second part:
I really like this problem – easy for kids to understand, not too difficult to compute the answer, the computation allows a few different approaches and lessons about probability, and the result has a nice surprise! Fun project 🙂
About a week ago I saw this neat tweet:
So, I bought the book (plus one other from the thread!):
My plan for this fall is to go through as many of the 50 challenging probability problems as we can. I don’t know how many are accessible to the boys, but hopefully most of them are with some help.
Today we tackled problem #1 – You have some red socks and some black socks in a drawer. When you pick two socks at random the probability of a red pair is 1/2. What is the smallest number of socks that could be in the drawer?
Here’s how I introduced the problem to the boys and their initial solution:
The second part of the problem asks what the minimum number of socks is assuming that the number of black socks is even. This problem gave the boys a bit more trouble, but was a great learning opportunity for them.
In the last video they showed that the number of black socks couldn’t be 2 or 4 with direct computation. Here they showed that it could be 6 using algebra. This was a nice opportunity for some factoring practice.
Finally, we went to the computer and wrote a short (and obviously inefficient) program to test out some solutions. It was fun to find a few more (sadly we were super rushed for time here, but still had a nice conversation).
I saw a neat tweet from Annie Perkins last week:
Today I thought it would be fun to play around with this idea with my younger son. First I introduced the 4-person problem and let him think through it. His thought process is a great example of what a kid learning math can look like:
At the end of the last video he’d determined that there were 3 different arrangements of the 4 people sitting around the table. In this video I asked him to find those arrangements:
Next we moved to the 5 person problem:
Finally, having decided that there were 12 different arrangements with the 5 person problem, I asked him to try to write down all 12. This is a good exercise in using counting techniques to make an organized list:
Definitely a fun problem for students, and also a really nice introduction to counting and symmetry. Thanks to Annie for sharing!