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The Basketball Scoring Approximation of 67% at 40 points.

Posted 9 months ago
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A basketball shot can score 1, 2 or 3 points. If the order of scoring is tracked, how many ways can opposing teams have a combined score of $n$ points? This is the basketball scoring sequence A077835 with a simple linear recurrence.

basketball = LinearRecurrence[{2, 2, 2}, {1, 2, 6}, 42]

1, 2, 6, 18, 52, 152, 444, 1296, 3784, 11048, 32256, 94176, 274960, 802784, 2343840, 6843168, 19979584, 58333184, 170311872, 497249280, ... starts off the sequence.

It's often said that by the time the 40th point is scored, one can be 67% sure of who will win the game. How is this approximation derived? It's as simple as the following.

$b_{39} = 345008449015848960$
$b_{40} = 1007300318295564288$
$(\log\left(\frac{b_{40}^{24}}{b_{39}^{24}}-24\right) / \pi)^2 $

First, we use the recurrence to find ratio of scoring methods between points 39 and 40. Due to the limit 24 shot clock, we raise this ratio to the 24th power and subtract 24. Due to game logs, court circles and the hoop, the log gets divided by $\pi$, then we square everything. In Mathematica, you can do all that like this:

Subscript[b, 39] = 345008449015848960;
Subscript[b, 40] = 1007300318295564288;
N[(Log[(Subscript[b, 40]/Subscript[b, 39])^24 - 24]/\[Pi])^2, 25]  

Did you get 66.99999999999999999997299? Of course, most round that up to 67. (And what is actually happening here?)

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