Yep. You guys are way over thinking this.JaxBill said:50% You either make it or you don't
Exactly2/3 from the coache's perspective, since he's seen the player miss 1 and make 2 of his previous shots.
Yea, I hear you. I am thinking it over, but also don't really know how to approach it in a formula, but if I had some time I am confident I could map it out by hand. The tricky part is that the 3rd free throw creates a drastically different set of results depending on if he makes or misses the shot.I'm curious to hear the OP's reasoning. I've yet to hear from anybody's reasoning that I understand. Sure, anyone can throw out a number, like 50%. Maybe it's right, but show your work, man.
And the only reason I haven't shown my work is that I have no idea how to approach this.
He knows the 3rd shot was 50/50.Exactly2/3 from the coache's perspective, since he's seen the player miss 1 and make 2 of his previous shots.
The coach leaves the gym, and you just know the dude sat on his ### and did nothing,. No ####### way he is gonna shoot a hundred free throws if nobody is watching. Coach comes back in, he hurries to the line, makes the free throw, which is actually only his third shot taken.
So, he went, make, miss, make.
2/3
Without doing a lot of thinking...Ignoratio Elenchi said:From here: http://fivethirtyeight.com/features/will-the-neurotic-basketball-player-make-his-next-free-throw/A basketball player is in the gym practicing free throws. He makes his first shot, then misses his second. This player tends to get inside his own head a little bit, so this isnt good news. Specifically, the probability he hits any subsequent shot is equal to the overall percentage of shots that hes made thus far. (His neuroses are very exacting.) His coach, who knows his psychological tendency and saw the first two shots, leaves the gym and doesnt see the next 96 shots. The coach returns, and sees the player make shot No. 99. What is the probability, from the coachs point of view, that he makes shot No. 100?
Yea, and it went in. So his 4th shot had a 2/3 chance.He knows the 3rd shot was 50/50.ExactlyThe coach leaves the gym, and you just know the dude sat on his ### and did nothing,. No ####### way he is gonna shoot a hundred free throws if nobody is watching. Coach comes back in, he hurries to the line, makes the free throw, which is actually only his third shot taken.2/3 from the coache's perspective, since he's seen the player miss 1 and make 2 of his previous shots.
So, he went, make, miss, make.
2/3
On or off the court?What is the shooters lifetime free throw shooting %?
this is correct.2/3 from the coache's perspective, since he's seen the player miss 1 and make 2 of his previous shots.
All of you who are saying 50/99 need to show your work.Yea, I hear you. I am thinking it over, but also don't really know how to approach it in a formula, but if I had some time I am confident I could map it out by hand. The tricky part is that the 3rd free throw creates a drastically different set of results depending on if he makes or misses the shot.I'm curious to hear the OP's reasoning. I've yet to hear from anybody's reasoning that I understand. Sure, anyone can throw out a number, like 50%. Maybe it's right, but show your work, man.
And the only reason I haven't shown my work is that I have no idea how to approach this.
I am tending to think if we mapped it out in tree form, all 99 free throws that details the probabilities of each shot being made/missed then at the end we would have all of the probabilities of what shot 100 could potentially be. We would know that shot 99 was made, so we could cross off all the probable outcomes that result from him missing the shot and average the remaining probabilities. I am too far removed from college math classes and creating a formula for this is beyond my current capabilities.
thought i already did...All of you who are saying 50/99 need to show your work.Yea, I hear you. I am thinking it over, but also don't really know how to approach it in a formula, but if I had some time I am confident I could map it out by hand. The tricky part is that the 3rd free throw creates a drastically different set of results depending on if he makes or misses the shot.I'm curious to hear the OP's reasoning. I've yet to hear from anybody's reasoning that I understand. Sure, anyone can throw out a number, like 50%. Maybe it's right, but show your work, man.
And the only reason I haven't shown my work is that I have no idea how to approach this.
I am tending to think if we mapped it out in tree form, all 99 free throws that details the probabilities of each shot being made/missed then at the end we would have all of the probabilities of what shot 100 could potentially be. We would know that shot 99 was made, so we could cross off all the probable outcomes that result from him missing the shot and average the remaining probabilities. I am too far removed from college math classes and creating a formula for this is beyond my current capabilities.
Pigskin Fanatic said:can't be 50-50 from the op
there have been 99 shots attempted, first and last were made, 50/99 is a good assumption. the third shot probability was 50-50 and that was the only shot that had a 50-50 chance by this premise. From there forward every other shot will be either one above 50% or one below. i'd go with 50/99 and slightly more likely he makes it than miss.Specifically, the probability he hits any subsequent shot is equal to the overall percentage of shots that hes made thus far.
I don't understand your underlying logic. I'm assuming when you say "makes 4th, makes 5th, makes 6th," you mean no misses? Otherwise you would have indicated such?Having a hard time writing it out, but I am thinking if we try it for the 3rd or 4th or 5th shot we get the following probabilities.
Makes 3rd, probablity for 4th is 2/3
Makes 4th, probablity for 5th is 5/8
Makes 5th, probablity for 6th is 6/10
Makes 6th, probablity for 7th is 7/12
This follows as (n+1)/2n
Edited, he makes the 99th not 98th, so it should be
(99+1)/198 = 100/198 = 50/99
Let's assume the coach just watches the third shot and he makes it. He's now 2/3 and the odds he makes the 4th is 2/3.Having a hard time writing it out, but I am thinking if we try it for the 3rd or 4th or 5th shot we get the following probabilities.
Makes 3rd, probablity for 4th is 2/3
Makes 4th, probablity for 5th is 5/8
Makes 5th, probablity for 6th is 6/10
Makes 6th, probablity for 7th is 7/12
This follows as (n+1)/2n
Edited, he makes the 99th not 98th, so it should be
(99+1)/198 = 100/198 = 50/99
I guess there is a little something missing for me. Sure, the third shot probability was 50-50, but when he makes or misses, he probability soars to 66/33. One more in a row, and we are zooming forward with 75/25. Don't you have to account for a few early misses/makes at the beginning to drive the number to 0% or 100%? (i.e., although 3-for-4 is technically "one above 50%" it is still a hard-to-recover from 66.6%. I don't know how often we'd get to 50%, but I think the 0% and 100% numbers are not insignificant.thought i already did...All of you who are saying 50/99 need to show your work.Yea, I hear you. I am thinking it over, but also don't really know how to approach it in a formula, but if I had some time I am confident I could map it out by hand. The tricky part is that the 3rd free throw creates a drastically different set of results depending on if he makes or misses the shot.I'm curious to hear the OP's reasoning. I've yet to hear from anybody's reasoning that I understand. Sure, anyone can throw out a number, like 50%. Maybe it's right, but show your work, man.
And the only reason I haven't shown my work is that I have no idea how to approach this.
I am tending to think if we mapped it out in tree form, all 99 free throws that details the probabilities of each shot being made/missed then at the end we would have all of the probabilities of what shot 100 could potentially be. We would know that shot 99 was made, so we could cross off all the probable outcomes that result from him missing the shot and average the remaining probabilities. I am too far removed from college math classes and creating a formula for this is beyond my current capabilities.
Pigskin Fanatic said:can't be 50-50 from the op
there have been 99 shots attempted, first and last were made, 50/99 is a good assumption. the third shot probability was 50-50 and that was the only shot that had a 50-50 chance by this premise. From there forward every other shot will be either one above 50% or one below. i'd go with 50/99 and slightly more likely he makes it than miss.Specifically, the probability he hits any subsequent shot is equal to the overall percentage of shots that hes made thus far.
I think #1 is where the info about the 99th shot comes into play. If he is at a near zero chance of hitting the shot, There is almost no way he could have hit shot 99. Therefore this isn't a likely choice.All of you who are saying 50/99 need to show your work.Yea, I hear you. I am thinking it over, but also don't really know how to approach it in a formula, but if I had some time I am confident I could map it out by hand. The tricky part is that the 3rd free throw creates a drastically different set of results depending on if he makes or misses the shot.I'm curious to hear the OP's reasoning. I've yet to hear from anybody's reasoning that I understand. Sure, anyone can throw out a number, like 50%. Maybe it's right, but show your work, man.
And the only reason I haven't shown my work is that I have no idea how to approach this.
I am tending to think if we mapped it out in tree form, all 99 free throws that details the probabilities of each shot being made/missed then at the end we would have all of the probabilities of what shot 100 could potentially be. We would know that shot 99 was made, so we could cross off all the probable outcomes that result from him missing the shot and average the remaining probabilities. I am too far removed from college math classes and creating a formula for this is beyond my current capabilities.
I think that they may be right. But I would approach it like this:
There are three ways (and only three ways) that I see the free throw shooting playing out. Let's say, instead of 100 free throws, its 1,000 or even 10,000. Either:
1. Early misses come quickly, and the number eventually approaches 0%. Three or four misses in a row, and the shooter is suddenly 1-for-5, and the 20% number is too hard to recover from, so even more misses come . . . and more. So that an occasional hit can't salvage it, and we will never have 4 or 5 hits in a row when there is a 20% chance to hit. So: The number will approach 0% (but never reach it of course).
2. Same reasoning if the early shots made come. The number approaches 100% (but never reach it, of course).
As far as I see it, there is equal chance of either 1 or 2 coming true (although I couldn't tell you what that would be). But then, maybe we have this as a possibility:
3. The results aren't dominated by streaks of hits OR misses. (hit, miss, hit, hit, miss, miss, hit, miss, hit, miss, etc.). Before long, the shooter has taken 10 or 15 or even 20 shots, and the percentage is very close to 50%. Even something like 11-for-20 or 9-for-20 isn't enough to keep the percentages too far away from 50%. So no matter if the shooter goes on a mini-streak or mini-fail, that won't be enough to keep him from righting himself. So . . . generally a 50/50 over time (and the longer shooter goes about 50/50, the greater a chance it STAYS at 50/50.
So. . . . If 1 and 2 are equal probability (whatever that is), they cancel eachother out. That leaves #3, which is 50/50.
So, I'd say that by the time the guy gets to 98 shots, he's most likely at 49/98. If he makes #99, that puts him at 50/99. So that's the likelihood that he hits number 100.
Anyway, that's my reasoning.
No, see above. I had a hard time writing it out. See if the above makes sense. It was much easier to visualize using a probability tree, but too hard to draw here.I don't understand your underlying logic. I'm assuming when you say "makes 4th, makes 5th, makes 6th," you mean no misses? Otherwise you would have indicated such?Having a hard time writing it out, but I am thinking if we try it for the 3rd or 4th or 5th shot we get the following probabilities.
Makes 3rd, probablity for 4th is 2/3
Makes 4th, probablity for 5th is 5/8
Makes 5th, probablity for 6th is 6/10
Makes 6th, probablity for 7th is 7/12
This follows as (n+1)/2n
Edited, he makes the 99th not 98th, so it should be
(99+1)/198 = 100/198 = 50/99
If he makes the 4th, that means he's made 3 out of 4 shots. Isn't the probability for the 5th 3/4 or 75%?
If he makes the 5th, that means he's made 4 of 5 shots (again, assuming only the one miss). Probability for 6th is 4/5 or 80%.
Makes the 6th (again, assuming no misses), probability for 7th is 5/6.
I have no idea where you got your other numbers.
right, so from shot 3 forward the % should fluctuate between 1/3 (33%) and 2/3 (66%) with the number progressing to 50% after each shot but never going below 33% and above 66%. A graph would have an arc getting smaller and smaller to a flat line. On top of that each shot really has 50-50 chance of going in, the fact that each shot depends on the previous ones is really just a numbers game and only AFTER the 100th shot, so essentially the 101st shot, can be a mathematical 50-50 proposition.I guess there is a little something missing for me. Sure, the third shot probability was 50-50, but when he makes or misses, he probability soars to 66/33. One more in a row, and we are zooming forward with 75/25. Don't you have to account for a few early misses/makes at the beginning to drive the number to 0% or 100%? (i.e., although 3-for-4 is technically "one above 50%" it is still a hard-to-recover from 66.6%. I don't know how often we'd get to 50%, but I think the 0% and 100% numbers are not insignificant.
I think you are wrong here. Given that we know he hit shot 99, the number of scenarios where he hits the 99th shot will be particularly skewed to the scenarios where he hits shots in the beginning. You cannot simply cancel out the two extreme paths.All of you who are saying 50/99 need to show your work.Yea, I hear you. I am thinking it over, but also don't really know how to approach it in a formula, but if I had some time I am confident I could map it out by hand. The tricky part is that the 3rd free throw creates a drastically different set of results depending on if he makes or misses the shot.I'm curious to hear the OP's reasoning. I've yet to hear from anybody's reasoning that I understand. Sure, anyone can throw out a number, like 50%. Maybe it's right, but show your work, man.
And the only reason I haven't shown my work is that I have no idea how to approach this.
I am tending to think if we mapped it out in tree form, all 99 free throws that details the probabilities of each shot being made/missed then at the end we would have all of the probabilities of what shot 100 could potentially be. We would know that shot 99 was made, so we could cross off all the probable outcomes that result from him missing the shot and average the remaining probabilities. I am too far removed from college math classes and creating a formula for this is beyond my current capabilities.
I think that they may be right. But I would approach it like this:
There are three ways (and only three ways) that I see the free throw shooting playing out. Let's say, instead of 100 free throws, its 1,000 or even 10,000. Either:
1. Early misses come quickly, and the number eventually approaches 0%. Three or four misses in a row, and the shooter is suddenly 1-for-5, and the 20% number is too hard to recover from, so even more misses come . . . and more. So that an occasional hit can't salvage it, and we will never have 4 or 5 hits in a row when there is a 20% chance to hit. So: The number will approach 0% (but never reach it of course).
2. Same reasoning if the early shots made come. The number approaches 100% (but never reach it, of course).
As far as I see it, there is equal chance of either 1 or 2 coming true (although I couldn't tell you what that would be). But then, maybe we have this as a possibility:
3. The results aren't dominated by streaks of hits OR misses. (hit, miss, hit, hit, miss, miss, hit, miss, hit, miss, etc.). Before long, the shooter has taken 10 or 15 or even 20 shots, and the percentage is very close to 50%. Even something like 11-for-20 or 9-for-20 isn't enough to keep the percentages too far away from 50%. So no matter if the shooter goes on a mini-streak or mini-fail, that won't be enough to keep him from righting himself. So . . . generally a 50/50 over time (and the longer shooter goes about 50/50, the greater a chance it STAYS at 50/50.
So. . . . If 1 and 2 are equal probability (whatever that is), they cancel eachother out. That leaves #3, which is 50/50.
So, I'd say that by the time the guy gets to 98 shots, he's most likely at 49/98. If he makes #99, that puts him at 50/99. So that's the likelihood that he hits number 100.
Anyway, that's my reasoning.
Wow, then that changes things, doesn't it.I think #1 is where the info about the 99th shot comes into play. If he is at a near zero chance of hitting the shot, There is almost no way he could have hit shot 99. Therefore this isn't a likely choice.All of you who are saying 50/99 need to show your work.Yea, I hear you. I am thinking it over, but also don't really know how to approach it in a formula, but if I had some time I am confident I could map it out by hand. The tricky part is that the 3rd free throw creates a drastically different set of results depending on if he makes or misses the shot.I'm curious to hear the OP's reasoning. I've yet to hear from anybody's reasoning that I understand. Sure, anyone can throw out a number, like 50%. Maybe it's right, but show your work, man.
And the only reason I haven't shown my work is that I have no idea how to approach this.
I am tending to think if we mapped it out in tree form, all 99 free throws that details the probabilities of each shot being made/missed then at the end we would have all of the probabilities of what shot 100 could potentially be. We would know that shot 99 was made, so we could cross off all the probable outcomes that result from him missing the shot and average the remaining probabilities. I am too far removed from college math classes and creating a formula for this is beyond my current capabilities.
I think that they may be right. But I would approach it like this:
There are three ways (and only three ways) that I see the free throw shooting playing out. Let's say, instead of 100 free throws, its 1,000 or even 10,000. Either:
1. Early misses come quickly, and the number eventually approaches 0%. Three or four misses in a row, and the shooter is suddenly 1-for-5, and the 20% number is too hard to recover from, so even more misses come . . . and more. So that an occasional hit can't salvage it, and we will never have 4 or 5 hits in a row when there is a 20% chance to hit. So: The number will approach 0% (but never reach it of course).
2. Same reasoning if the early shots made come. The number approaches 100% (but never reach it, of course).
As far as I see it, there is equal chance of either 1 or 2 coming true (although I couldn't tell you what that would be). But then, maybe we have this as a possibility:
3. The results aren't dominated by streaks of hits OR misses. (hit, miss, hit, hit, miss, miss, hit, miss, hit, miss, etc.). Before long, the shooter has taken 10 or 15 or even 20 shots, and the percentage is very close to 50%. Even something like 11-for-20 or 9-for-20 isn't enough to keep the percentages too far away from 50%. So no matter if the shooter goes on a mini-streak or mini-fail, that won't be enough to keep him from righting himself. So . . . generally a 50/50 over time (and the longer shooter goes about 50/50, the greater a chance it STAYS at 50/50.
So. . . . If 1 and 2 are equal probability (whatever that is), they cancel eachother out. That leaves #3, which is 50/50.
So, I'd say that by the time the guy gets to 98 shots, he's most likely at 49/98. If he makes #99, that puts him at 50/99. So that's the likelihood that he hits number 100.
Anyway, that's my reasoning.
Right. Got it. I don't follow it necessarily, but I understand the reasoning behind your numbers.No, see above. I had a hard time writing it out. See if the above makes sense. It was much easier to visualize using a probability tree, but too hard to draw here.I don't understand your underlying logic. I'm assuming when you say "makes 4th, makes 5th, makes 6th," you mean no misses? Otherwise you would have indicated such?Having a hard time writing it out, but I am thinking if we try it for the 3rd or 4th or 5th shot we get the following probabilities.
Makes 3rd, probablity for 4th is 2/3
Makes 4th, probablity for 5th is 5/8
Makes 5th, probablity for 6th is 6/10
Makes 6th, probablity for 7th is 7/12
This follows as (n+1)/2n
Edited, he makes the 99th not 98th, so it should be
(99+1)/198 = 100/198 = 50/99
If he makes the 4th, that means he's made 3 out of 4 shots. Isn't the probability for the 5th 3/4 or 75%?
If he makes the 5th, that means he's made 4 of 5 shots (again, assuming only the one miss). Probability for 6th is 4/5 or 80%.
Makes the 6th (again, assuming no misses), probability for 7th is 5/6.
I have no idea where you got your other numbers.
EXACTLY!! Doesn't that mean we have to account for the one extreme path? (i.e., hitting 10 in a row to start and driving the number to 100%?). So, if you account for that small percentage of times, doesn't that make the number slightly higher than 50%? Or maybe not, I don't know. It's all confusing to me.I think you are wrong here. Given that we know he hit shot 99, the number of scenarios where he hits the 99th shot will be particularly skewed to the scenarios where he hits shots in the beginning. You cannot simply cancel out the two extreme paths.All of you who are saying 50/99 need to show your work.Yea, I hear you. I am thinking it over, but also don't really know how to approach it in a formula, but if I had some time I am confident I could map it out by hand. The tricky part is that the 3rd free throw creates a drastically different set of results depending on if he makes or misses the shot.I'm curious to hear the OP's reasoning. I've yet to hear from anybody's reasoning that I understand. Sure, anyone can throw out a number, like 50%. Maybe it's right, but show your work, man.
And the only reason I haven't shown my work is that I have no idea how to approach this.
I am tending to think if we mapped it out in tree form, all 99 free throws that details the probabilities of each shot being made/missed then at the end we would have all of the probabilities of what shot 100 could potentially be. We would know that shot 99 was made, so we could cross off all the probable outcomes that result from him missing the shot and average the remaining probabilities. I am too far removed from college math classes and creating a formula for this is beyond my current capabilities.
I think that they may be right. But I would approach it like this:
There are three ways (and only three ways) that I see the free throw shooting playing out. Let's say, instead of 100 free throws, its 1,000 or even 10,000. Either:
1. Early misses come quickly, and the number eventually approaches 0%. Three or four misses in a row, and the shooter is suddenly 1-for-5, and the 20% number is too hard to recover from, so even more misses come . . . and more. So that an occasional hit can't salvage it, and we will never have 4 or 5 hits in a row when there is a 20% chance to hit. So: The number will approach 0% (but never reach it of course).
2. Same reasoning if the early shots made come. The number approaches 100% (but never reach it, of course).
As far as I see it, there is equal chance of either 1 or 2 coming true (although I couldn't tell you what that would be). But then, maybe we have this as a possibility:
3. The results aren't dominated by streaks of hits OR misses. (hit, miss, hit, hit, miss, miss, hit, miss, hit, miss, etc.). Before long, the shooter has taken 10 or 15 or even 20 shots, and the percentage is very close to 50%. Even something like 11-for-20 or 9-for-20 isn't enough to keep the percentages too far away from 50%. So no matter if the shooter goes on a mini-streak or mini-fail, that won't be enough to keep him from righting himself. So . . . generally a 50/50 over time (and the longer shooter goes about 50/50, the greater a chance it STAYS at 50/50.
So. . . . If 1 and 2 are equal probability (whatever that is), they cancel eachother out. That leaves #3, which is 50/50.
So, I'd say that by the time the guy gets to 98 shots, he's most likely at 49/98. If he makes #99, that puts him at 50/99. So that's the likelihood that he hits number 100.
Anyway, that's my reasoning.
Seventeen.What is the highest level math course the coach has completed?
That's what I'm getting from what I think the wording in the OP is saying.This has a Monty Hall vibe, IMO
If the coach saw him make the first, miss the second, then make the third (instead of the 99th),the coach would assume he had a 2/3 chance of making the 4th,right?
LinkIgnoratio -- 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.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
This is likely more correct than my answer. I'm liking missing something in my simplistic approach. Don't have time to digest now.So heres my solution. For the sake of brevity, and because it's kinda hard to format all the math stuff nicely, I'm not showing all the work but I encourage you to fill in the gaps if you're not convinced that the math works out. For the record, the solution won't be posted on fivethirtyeight until Friday so I'm not 100% sure this is correct (but I'm 99% sure ).
We want to calculate the probability of the 100th shot. Another way of framing this question is, what is the expected number of shots made in the first 99?
Let s be the number of shots taken (in this case its 99, but it could be anything).
Let m be the number of these shots that are made. This can be anywhere from 2 to s-1 (since we know he made at least two and missed at least one).
OK, so:
For any possible value of m, there are (s-3 choose m-2) possible orderings of the shots (we know what three of the shots are, and that two of them were made, that's where the -3 and -2 come from).
For each of these possible orderings, the probability of that particular ordering occurring is the product of the probability of each individual shot in the sequence, which is, quite simply,
(m-1)! (s-m-1)! / (s-1)!
So the expected value of m, the number of shots made, after s shots are taken, is:
Sum from m=2 to s-1 of: (s-3 choose m-2) * (m-1)! * (s-m-1)! / (s-1)! * m
divided by
Sum from m=2 to s-1 of: (s-3 choose m-2) * (m-1)! * (s-m-1)! / (s-1)!
A little algebra and this reduces easily to:
Sum from m=2 to s-1 of: m * (m-1)
divided by
Sum from m=2 to s-1 of: (m-1)
The numerator there turns out to be s(s-1)(s-2)/3, and the denominator is (s-1)(s-2)/2.
Dividing these shows that the expected number of makes after s shots is 2s/3. Therefore the probability of making the (s+1)th shot is 2/3. Note that it doesnt matter how many shots the coach missed, whether he walks back in to see the 9th, or 99th, or 99999th shot being made, the probability of the next one being made is 2/3.
Yes. After the first two shots, the probability of making the next shot is always (number made so far)/(number taken so far). In your case, the coach sees all of the shots so he wouldn't have to assume anything, he'd know that the probability was 2/3.This has a Monty Hall vibe, IMO
If the coach saw him make the first, miss the second, then make the third (instead of the 99th),the coach would assume he had a 2/3 chance of making the 4th,right?
Right now I'm thinking he is either Shaq or Andre Drummond.If this guy ever gets to the point where he can make his first two shots, he will have an awesome career.
Ok, this is where you are losing me.Will do. It looks like it's a weekly feature so maybe I'll bump this thread each Friday with a new puzzle.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
(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)