Ignoratio -- I just don't understand the math. Once the 538 answer is posted, please post here. That 2/3 number is really surprising to me
Will do. It looks like it's a weekly feature so maybe I'll bump this thread each Friday with a new puzzle.(clip)
The insight to have here is that the denominators will always go 2, 3, 4, ... , (s-1). And the numerators will always include 1, 2, 3, ..., (m-1) (for each of the made shots, whatever order they come in) and 1, 2, 3, ... , (s-m-1) for each of the misses. That is, the probability of making exactly 3 of the 5 shots is the same, no matter what order they're made in.
The same is true for larger cases - there are an absurdly large number of ways that you can make, say, 50 out of 99 shots, but the probability of every single one of those sequences is the same.
(clip)
Ok, this is where you are losing me.
I don't understand how the bolded can be true. The probability that he makes 49 shots in a row (after his 1/2 performance) followed by missing the next 49 shots has got to be really miniscule, right? Because that would mean he would be at 50 made shots out of 51 shots attempted. The likelihood that he'd make the next one is well over 95%, right? So the likelihood that he'd go on a run of 49 missed would be really really slim, right? I don't see how that would be "just as likely" as where he goes miss, hit, miss, miss, hit, hit, miss, hit, miss, etc." until he's at 50 out of 99.
In short, because the next basket made is dependent on the percentages made before it, I don't understand how "the probability of every single one of the sequences is the same."
I kept the example above small so it might not be apparent what's happening. Here's the way to think about it. Remember that the probability of making a shot at any point is simply (number of shots made so far)/(number of shots taken so far). It's not stated explicitly, but you can also see that this means the probability of missing a shot at any point is simply (number of shots missed so far)/(number of shots taken so far). For example, if after 10 shots he's made 7 of them (and thus missed 3), the probability of making the next one is 7/10 (and thus the probability of missing the next one is 3/10).
Ok, so let's say he makes 50 of the 99 shots. He makes the first and misses the second, we know that already. At some point, he'll make a second shot (this might come on his third try, or his fourth, or his 20th, whatever). When he does, we know that the probability of him making that shot was 1/(however many shots he's taken so far), because up to that point he'd only made 1 shot. At some point after that, he'll make another one. And again, even though we don't know when he'll make it, we know the probability of making the shot at that point is 2/(however many shots he's taken so far). And then at some point after that he'll make another one, with probability 3/(however many shots he's taken so far)... all the way up until he makes his 50th shot, with probability 49/(however many shots he's taken so far).
Similarly, at some point he'll miss. We don't know when, but whenever it happens we know that the probability of missing at that point was 1/(however many shots he's taken so far). At some later point, he'll miss again with probability 2/(however many shots he's taken so far)... all the way up until he misses his 49th shot with probability 48/(however many shots he's taken so far).
To calculate the total probability of the whole sequence occurring, we just multiply all of these individual probabilities together, and multiplication of fractions is associative in that we can multiply the numerators together in any order, and then multiply all the denominators together in any order, and then take the quotient. Like 4/9 * 6/7 is the same as 6/9 * 4/7. I can switch around the tops and bottoms and the answer is the same.
So, we've got a bunch of fractions where the numerators are 1, 2, 3, ..., (m-1) (for all the shots that were made) and 1, 2, 3, ..., (s-m-1) (for all the shots that were missed. If we multiply them all together, that first thing is just (m-1)! and the second thing is (s-m-1)!
And in these fractions, the denominators are just 1, 2, 3, ..., (s-1). Each one is just the number of shots that have come before. Multiplying all of these together is (s-1)!
So putting them together, we get (m-1)! * (s-m-1)! / (s-1)! If we have a sequence of s shots where exactly m of them are made, this is the probability of it occurring, regardless of the order in which they're made.
Not sure if that's clear and/or convincing enough so maybe a longer example will help. Pretend he took 10 shots and made 5 of them. Let's look at three ways this could happen (1 is make, 0 is miss):
1011110000 (he gets hot early then cools off)
1010101010 (he alternates making and missing shots)
1001000111 (he starts pretty cold then gets hot)
They all start with 10, we know that. Starting with the third shot then, let's look at the probabilities (and try to notice the pattern):
First sequence = 1/2 * 2/3 * 3/4 * 4/5 * 1/6 * 2/7 * 3/8 * 4/9.
Second sequence = 1/2 * 1/3 * 2/4 * 2/5 * 3/6 * 3/7 * 4/8 * 4/9.
Third sequence = 1/2 * 1/3 * 2/4 * 3/5 * 4/6 * 2/7 * 3/8 * 4/9.
The individual shot probabilities vary, but when you multiply them all together they're all the same answer. The bottom just goes from 2 up to 9 each time. And the tops go from 1 to 4 (for the 5 makes) and 1 to 4 (for the 5 misses) in some order.