While it is true that the 99th shot MIGHT have pushed the results up to 66% or even beyond,
it is far from likely to do so… Law of large numbers… the odds were already set by shot 3,4,5,6,7,8… an unknown… last known . 50/50… by symmetry, it remains at 50/50, with some mathematical statement of confidence … 95% sure … something like that.
He had a 50% rate when the coach left the building, so all we can assume is that it was still at 50% when he came back. He hits shot 99, so now his rate is 50/99 or 50.5%. I’m guessing I’m missing something, but that seems logical to me.
I believe you’re failing to take conditional probability into account. For instance, here’s a much less interesting version of the same puzzle: the guy’s third free throw is a tossup, exactly 50/50, but then every further free throw he shoots that session will have exactly the same result as the third.
So the coach watches two, then goes away, then comes back, and the guy is about to shoot #99. From the coaches perspective, what’s the probability of it being made? 50/50, the coach has no information and no way to guess what has happened.
Then he shoots #99, and it goes in. Now the coach knows with 100% certainty that #100 will also go in.
The coach sees #99, and that gives him additional information which lets him assign more accurate probabilities to the likelihood of #100 going in. But in the actual riddle, it’s far more subtle precisely HOW he should go about doing that. But he still has gained additional information.
For instance, here’s a somewhat-in-between version of the riddle. Again, shot #3 is 50/50. And again shot #3 determines all future shots. But not quite perfectly. So if he makes #3, then he’ll make 90% for the rest of the session. If he misses #3, he’ll miss 90% for the rest of the session. Rest of puzzle the same. Again, coach comes back in, sees a make of #99. From his perspective, it’s now far more likely that #3 was made than that #3 was missed, and he can adjust his probabilities accordingly, but it won’t be 100% either way.
(Not going to spoiler my response since it isn’t really related to the probability from the coach’s perspective.)
I initially thought the same as you - the actual % would converge toward 0, 50, or 100%. If you miss a bunch, you’ll keep missing more and more. If you make a bunch, you’ll make more and more. And if you’re 50/50, it will stay there.
But what actually happens is it converges to some number within roughly 50 throws with little variation after that, and the numbers it converges to are anywhere between 0 and 100. Numbers from 0-10, 40-60, and 90-100 are a little more likely, but you still see plenty of outcomes in the 10-40 and 60-90 range.
Which made perfect sense once I thought about it. If I hit 7 of 10 initially, each subsequent throw has a smaller impact on the percentage, so it gets harder and harder to move away from 70%.
This is wrong.
After every n throws, each of the n-1 possible outcomes has exactly equal probability. So it’s just as likely that he hit 64/80 as 2/80 as 30/80.
Without going too much into the math (which Indistinguishable covered already), the reason is because the closer-to-the-middle outcomes have more paths to get there, but each path is individually less likely because it has to “fight” the pull towards the extremes. Those two tendencies end up cancelling each other out precisely.
I guess there probably is a mathematical answer to how much you can tell about what went on while the coach was away from seeing the player hit shot 99, but it’s got to be pretty small, right? No matter what, he won’t be at zero or 100 percent. If he hits shot 99, it’s a tiny bit more likely that he hit shot 3.
Since the answer is posed as ‘from the coach’s perspective’ then the answer should be 100%.
The memory of past trajectories matters here. At 50% after 2 shots, the player will either move to 33% or 66% after the 3rd shot. There are many trajectories that end up very close to 0% or very close to 100%. In particular, there is plenty of downward or upward pressure. By this I mean, if the player makes his 3rd shot (50/50 chance) then he has a 66% chance of making the next shot. So there’s a 2/3rds chance his NEXT shot will be with 75% likelihood of being made. Each shot is a bifurcation point in the trajectory and the probabilities accumulate over time.
The coach is acting on limited information. If he sees the player make a shot quite late into the process, it has to mean that he’s converged on a relatively high overall percentage. The coach is safe to assume that if he’s made the 99th, he’s likely to make the 100th.
That might seem intuitively likely, or the exact opposite might seem intuitively likely, which was my initial inclination, that if #99 was hit then the probability for #100 goes way way way up. But in fact, intuition is wrong, and the answer is 2/3, due to math explained by Indisinguishable, and verified by a computer simulation I wrote.
But the player couldn’t possibly have a 100% chance of making the shot, since we know he’s already missed one.
That’s interesting if that’s the answer, since that’s the percentage he’s hit of the shots that we’ve seen.
Yup.
Hmmm, my simulation shows otherwise. It’s possible I didn’t run it enough times. I’ll have to ponder on the math a little more.
That said, my simulation also shows that for scenarios where he made shot #99, the average made percentage converges to 67%, which is confirmation of your answer.
I stopped reading about half way through the thread. Apologies if someone else has already pointed out the following. In my opinion, the answer given by most is incorrect**. To see this most clearly, suppose that coach has witnessed only one toss and it was successful. Should he estimate 100% for next success?
Answer: The appropriate estimator is
p = (x + a) / (x + y + a + b)
where x,y are the number of observed successes and failures respectively and a,b are arbitrary constants related to any a priori probability. The answer 2/3 is just the special case where a = b = 0 (or a = 2b).
I won’t look for a cite, but a = b = 1 are a choice of constants which solve some well-posed problem in probability estimation with no prior information. In the example, p = (x + 1) / (x+y + 1+1) would be 3/5.
The problem doesn’t really function at all without knowing that the student starts out 1-1. Because if you simply state it as “his probability of making a shot is always equal to his current record”, then whatever he does first, he will continue to do forever. That’s why there has to be a special case that his weird behavior only kicks in after at least one make and at lesat one miss.
If the problem was simply stated as “at each point, his probability of making the next shot equals the percentage of shots he’s already made”, then the only two possibilities are MMMMMMMMMMMMMMMM and HHHHHHHHHHHHHHH, and in fact the coach having seen one shot and knowing that it was an H would mean that 100% of shots would be H forever.
The stipulation of the problem is that we have some probabilistic machine which works as follows: it is guaranteed to make the first throw, and it is guaranteed to miss the second throw. After that, on each throw, conditional on the outcomes of the previous throws, the probability that the machine makes (or misses) the next throw is equal to the proportion of times so far it has made (or missed) throws so far.
The machine is defined to operate with this particular probability distribution. There is no room to quibble with this; it is part of the intended problem statement.
The question, then, is, conditional on the machine making the 99th throw, what is the probability that the machine makes the 100th throw?
Incidentally, we are essentially using a = 1 and b = 1 in your terminology, when the hits and misses we count for x and y, respectively, are those past the first two throws (the ones whose outcomes are variable). And, thus, the answer we get will be (1 + 1)/(1 + 0 + 1 + 1) = 2/3; the numerator includes the one non-guaranteed hit (throw #99, which we have conditionalized upon the success of) and one guaranteed hit (throw #1), and the denominator includes as well the one guaranteed miss (throw #2). But we get this formula not because of any particular usefulness as an estimator for a situation with no information about the relevant probabilities, but, rather, because we know everything about the probability distribution of the machine, and they are stipulated in such a way as that this formula accurately describes the result.
OK, I misunderstood. Sorry. For some reason I thought the machine behavior was known to us, but not to the coach. That would be a silly stipulation.
Blame lack of sleep and ingestion from Cannabaceae.
Strictly speaking, the machine is not known to be guaranteed to make the first shot and miss the second. In fact, we know almost nothing about the machine’s behavior, in general, for its initial shots. We just know that, in this particular run, the machine made the first and missed the second. It could be that, in general, it has a 50% chance on each shot until it’s gotten at least one of each. It could even be that it has a 1% chance of making the first shot and a 99% chance of making the second, and we just witnessed a really rare fluke. But once we know the sequence that established the machine, it no longer matters how it actually arrived at that sequence.
Fair enough. We are conditionalizing on (i.e., restricting attention to) the circumstances under which the machine makes the first shot and misses the second, and thereafter operates with probability distribution as described. (And then further conditionalizing on making the 99th shot, and asking about the conditional probability distribution of the 100th shot).
Of course, all the stuff about basketball players, or machines as I put it, and the actual vs. counterfactual behavior of those machines, etc., is just flavorful story on top of an ultimately intended underlying purely arithmetic problem.
The underlying purely arithmetic question is “Consider the ‘weighting’ function on finite sequences drawn from {make, miss} starting ‘make, miss’ which assigns to any such sequence the product, over each value in the sequence past those first two, of the proportion of previous values which match. Over all such sequences of length 100 whose 99th value is ‘make’, what proportion of the total ‘weight’ is contributed by those sequences whose 100th value is also ‘make’?”.
But people mightn’t be all that interested in this question if it was stated so drily. 
I’m not a statistician. I just want to put down my answer before looking back to find out what it actually was.
Based on the way it works, my expectation would be that the player would either quickly get really really good or really really bad. The way the math works, it applies positive feedback into the system.
Starting at 1 in 2, he will either go up to 2 in 3 or down to 1 in 3. If he goes up to 2 in 3, then the next time he’s more likely to make it, bumping him to 3 in 4, which makes him even more likely to make it, again bumping him to 4 in 5.
While it’s not guaranteed, the general trend will be to approach 1 or 0, the longer it goes on.
So by the 99th shot, if he made it, it’s far more likely that he will make the next one than that he will miss it, because it’s a strong indication that his aim went the positive direction rather than negative.
I could probably figure out a way to do the math, to come up with the most probable value, but that would take too long, so I’m just going to check my general theory.