LSV's Free Throws

Earlier today, LSV posted a tweet which I found quite intriguing:

It's an interesting stats kind of question, and unsurprisingly a follower quickly tweeted a correct response along with an explanation, which can be found (with some formality) here. The answer, by the way, is ~18.68 free throws (expected value)

Further conversation with Patrick Chapin,however, brings up some more interesting points and topics. In lots of cases, probably most cases really, expected value just isn't that useful of a concept. There's lots of reasons for this, but the biggest one is probably diminishing returns. At a certain point, you are going to die before you can use all the benefits of your winnings; or conversely, the penalties may kill you (or stop you from continuing the game) after a certain point. In short, the abstraction of the scenario breaks down after a while. In any event, there are definitely lots of reasons why you might be interested in knowing when the threshold of success or failure is met in terms of precise probabilities rather than just a smoothed-out average.

To that end, let's go through this toy scenario, figuring out the chance you'll be done after any given shot. Trivially, you can't get there before 10 shots (so the probability on 0-9 is 0). And on the 10th shot, the chance is p^10. On the 11th shot, you necessarily need the 1st shot to have been a miss (else we would have ended with 10 successes on shot 10), and then getting 10 consecutive hits. So this is p^10*(1-p).

An interesting thing happens on shot 12, when you need the second shot to have been a miss (this is, once again, the only way that shot 12 will be precisely the 10th hit in a row), as well as 3-12 all being makes, but you don't care at all what happened on shot 1. Either way, 12 will be the final shot. So you get the exact same p^10*(1-p). And this trend continues for 13, 14, 15, etc.

But not all the way ad infinitum. The pattern shifts again once we get to shot 21. Now, not only do we need 12-21 to all be hits, and 11 to be a miss, but we need at least one of 1-10 to be a miss as well, or we would have stopped on shot 10. This could look like a daunting calculation at first, but thankfully, recursion gives us a helpful shortcut: we already calculated the chance of one of shots 1-10 being a miss - or more precisely, we calculated its compliment when we checked the chance of it being the final shot.

So for shot 21 to be the last one, we get the following: p^10*(1-p)*(Chance the game wasn't over by shot 10 or earlier). And we can extend this formula on ad infinitum, so long as we keep a running count for each number of the chance that the game would have ended at that point or earlier.

You're left with the following:

If you're interested, you cross 95% after 42 shots, 99% on your 60th shot, and 99.9% on your 87th shot.

Anyway, I hope you found this interesting - I know I did.