Okay, so, today I messed around with something that I thought was simple but turned out to be a bit of a head-scratcher. It all started when I got curious about basketball free throws, yeah, those shots players take after a foul. I mean, how hard could it be to figure out the chances of making a few shots in a row, right?
So, I started with this idea that a player has a 60% chance of making a free throw. Sounds reasonable, a good player should be able to sink more than half their shots. I figured, let’s see what happens if they take, say, three shots. What are the odds they make two out of three? That’s where things got interesting.
Digging into the Problem
First, I tried to visualize it. I made a little list of all the possible outcomes, like this:
- Make, Make, Miss (MM)
- Make, Miss, Make (MM)
- Miss, Make, Make (MM)
- And so on…
There are actually eight different ways three shots can go down. But only three of those ways end up with two makes and one miss. I thought, “Cool, so it’s just 3 out of 8, right?” Wrong.
Then, it hit me. Each shot is independent. That means the outcome of one shot doesn’t affect the next. So, I needed to think about the probability of each specific outcome. A 60% chance of making a shot is the same as a 0.6 probability, and a 40% chance of missing is 0.4. For example, the probability of a player making the first two shots and missing the third (Make, Make, Miss) is 0.6 times 0.6 times 0.4, which equals 0.144.
Putting It All Together
I did this for all three scenarios where the player makes two out of three shots:
- Make, Make, Miss: 0.6 0.6 0.4 = 0.144
- Make, Miss, Make: 0.6 0.4 0.6 = 0.144
- Miss, Make, Make: 0.4 0.6 0.6 = 0.144
Turns out, each of these scenarios has the same probability, 0.144. So, to get the total probability of making exactly two out of three shots, I just added these up: 0.144 + 0.144 + 0.144, which gives us 0.432. That’s a 43.2% chance.
It was a bit more complicated than I initially thought, but it was pretty cool to figure it out. This little experiment reminded me that even seemingly simple things can have layers of complexity. Plus, it got me thinking about all the other scenarios, like what if the player took five shots, or ten? The possibilities are endless, and each one is a new puzzle to solve. Also, I realize that I was actually trying to use the binomial probability formula without knowing it, lol. Maybe I should use the formula next time.
