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:
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