By looking at how many shots would land in a given location (and thus make the goal), I can know how hard this would be. Here are my assumptions.

• The launch position is essentially constant – meaning I am not changing this.
• The variation in the left-right launch angle for a basketball is similar to the data for me throwing a small ball. Oh, I know you will complain – I am ok with that.
• Same for the up-down launch angle. I am also going to assume the standard deviation of distribution does not change with angle (same variations for all chosen launch angles).
• The ratio of standard deviation to launch speed for the basketball is similar to that for the small ball that I threw (again – this is just an assumption)
• For both angles and launch speed, I am going to assume each throw is independent of the previous (no learning).
• Finally, I am going to assume that the distributions of angles and speeds are normal distributions.

The Plan

I am going to numerically simulate the shot of a basketball off the huge statue as shown in this post. My default value for the initial velocity will be the same as what I ended with in that post. It may not be the exactly correct conditions – but that is ok. I am looking at the variation in the landing locations, not the actual landing site. How do these launch parameters vary? Here are the launch parameters I will start with (assuming normal distributions the the +/- representing the standard deviation of that distribution – oh, and I modified these values a little bit from my previous experiment assuming that these basketball guys can throw better than I can):

Here ? is the angle left or right of the target and ? is the angle the ball is thrown above the horizontal. Just as a sample, here is the distribution of the x-component (towards the target) of the launch velocities for 1,000 throws.

Looks normal, right?

The Data

Ok, now what about the landings? First, I have one more assumption. I am going to assume that the ball at the end of its trajectory is basically going straight down (which isn’t a bad assumption). This means that I don’t have to worry about the angle that the ball approaches the goal. So, how far off could the ball be and still make it? Here is a diagram.

By looking at the difference between the size of the goal and the ball, the ball can be as far off as 10.9 cm from the center and still go through. Let me call it an even 11 cm (even if it hits a little bit on the rim, it will still go through). Note that I am not considering backboard goals or any other type of rolling around the rim thing.

What is the distribution of the landing locations of the balls in the simulations? Instead of looking at both the x and z coordinates of the landing position, I can just look at the distance from the center of the goal. For 1,000 shots, this is what I get:

How many of these are within the 11 cm? It is sort of difficult to tell from that plot, but from the data I can tell you the answer. One. Just one of those shots made it within 11 cm of the center. That is 1 out of a thousand. Oh, sure – maybe my parameters are off. Maybe these guys are better than that. Maybe they are super good. I will give you that. Lets say they make 3 out of 1000 shots.

How many shots?

If I use the above, then I can say that the chance of making this shot is 3 out of 1000 or 0.3%. Well, how many times would they have to do this to get it to work? There is no answer to that question. It is possible that they could go up to the top of the statue and throw it – BOOM. Basket. I know that is not the answer you are looking for, so let me start with something else. Rolling dice.

If I roll a six-sided die, what is the probability I will roll a 1? For an unloaded die, this should be 1/6. How many times would I have to roll to expect a 1? That question is more complicated. How about instead I look at a the probability of rolling a 1 as a function of the number of rolls. What if I roll the die twice? What is the probability that out of those two rolls, neither is a 1?

There are two possible things that can happen when I roll the die twice. Either I can get a 1 or I can not get a 1. I just calculated the probability of not getting a 1, so the probability of getting a 1 would be the rest of the probability:

This can be generalized to n rolls so that the probability of rolling a 1 once would be:

Maybe it would be nice to see this graphically:

After 25 rolls, you can see the probability of getting a 1 is very close to 1 (100%) – actually it is 98.7%. Now, I can do the same thing with this basketball shot. The only difference is that instead of having 1/6 has the probability of success, I have 3/1000. Graphically, this would look like:

After 200 throws, there is a 45% chance that they will have made the shot. How many throws to get to a 70% chance of success? About 400.

How long would it take to shoot 300 times?

Could these guys even take 300 shots in one day (about 60% chance)? How long would it take to just make one shot? Well, you would need to carry the ball up to the top of the statue and then throw it. A little time would be needed to say “hi” to the camera (just in case you make it). The time for the ball to be thrown would be small (around 3 seconds). You could make it easier by carrying multiple balls to the top. Let me estimate some stuff:

• The viewing platform is about 5 stories high (120 foot pedestal)
• Two guys could carry 8 balls total (per trip)
• Climbing 5 stories would take about 1 minute – just a guess
• Setup (including hiding missed balls and yet to be thrown balls) = 10 seconds.

This would give an effective per shot time of 17.5 seconds. Let me just put this at 20 seconds per shot. This means that it would take 1 hour and 40 minutes (with no bathroom breaks).

It could be done. Even if you change the parameters around a bit, you are still going to be in the same ballpark.

Authors: Rhett Allain

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