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.

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