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Math Puzzles (from FiveThirtyEight - new puzzle every Friday) (1 Viewer)

This just shows you've gotta come out hot or else it's not gonna be a good night for you take that to the bank bro-clever-hans

 
For me, the clue was in the way the problem was written. With the limited information the coach has, the answer is pretty clearly 2/3, but he leaves the gym, so his limited-information conclusion has to be wrong, right? Turns out that no, it's right whether he left or not.
If he stayed at the gym, and watched every single shot up to shot 99, the chances that it would be exactly 2/3 at that point is miniscule.
No. See this from Greg:

"I think you keep getting lost in the fact that though the odds will eventually converge towards either 0% or 100% over enough trials, our coach is given no information which way it is diverging with just knowing the first 2 shots. The coach might agree it is likely the player made most or missed most rather than making 50%. But based on the info he can work from, those are equally likely so leave him at 50% odds of making the next basket... until such time as he knows the outcome of additional baskets such as learning the result of shot 99."
I disagree on the convergenceThe odds of making n shots in a row (after shot 2) is 1/(n-1). Which means the odds of converging to 100% decrease over time

Similarly the odds of converging to 0 decrease over time
Agreed, but I think the solution to the problem is the same no matter how many shots the coach misses seeing. But now with your post and moops' thoughts I'm getting hung up on whether or not the solution is the same if the coach sees all of the shots.

 
For me, the clue was in the way the problem was written. With the limited information the coach has, the answer is pretty clearly 2/3, but he leaves the gym, so his limited-information conclusion has to be wrong, right? Turns out that no, it's right whether he left or not.
If he stayed at the gym, and watched every single shot up to shot 99, the chances that it would be exactly 2/3 at that point is miniscule.
No. See this from Greg:

"I think you keep getting lost in the fact that though the odds will eventually converge towards either 0% or 100% over enough trials, our coach is given no information which way it is diverging with just knowing the first 2 shots. The coach might agree it is likely the player made most or missed most rather than making 50%. But based on the info he can work from, those are equally likely so leave him at 50% odds of making the next basket... until such time as he knows the outcome of additional baskets such as learning the result of shot 99."
I disagree on the convergenceThe odds of making n shots in a row (after shot 2) is 1/(n-1). Which means the odds of converging to 100% decrease over time

Similarly the odds of converging to 0 decrease over time
Agreed, but I think the solution to the problem is the same no matter how many shots the coach misses seeing. But now with your post and moops' thoughts I'm getting hung up on whether or not the solution is the same if the coach sees all of the shots.
For every shot the coach sees the player take, the percentages are going to change. I think.

If he saw the guy make, miss, make, then took a break, and again came back at shot 99 for another make, the odds are greater than 2/3 that he makes shot 100. I don't know if it is as simple as it would now be a 75% chance, but intuitively that is what I am going with

 
For me, the clue was in the way the problem was written. With the limited information the coach has, the answer is pretty clearly 2/3, but he leaves the gym, so his limited-information conclusion has to be wrong, right? Turns out that no, it's right whether he left or not.
If he stayed at the gym, and watched every single shot up to shot 99, the chances that it would be exactly 2/3 at that point is miniscule.
No. See this from Greg:

"I think you keep getting lost in the fact that though the odds will eventually converge towards either 0% or 100% over enough trials, our coach is given no information which way it is diverging with just knowing the first 2 shots. The coach might agree it is likely the player made most or missed most rather than making 50%. But based on the info he can work from, those are equally likely so leave him at 50% odds of making the next basket... until such time as he knows the outcome of additional baskets such as learning the result of shot 99."
I disagree on the convergenceThe odds of making n shots in a row (after shot 2) is 1/(n-1). Which means the odds of converging to 100% decrease over time

Similarly the odds of converging to 0 decrease over time
Agreed, but I think the solution to the problem is the same no matter how many shots the coach misses seeing. But now with your post and moops' thoughts I'm getting hung up on whether or not the solution is the same if the coach sees all of the shots.
Depends how you word the question

If he sees all of the shots then he knows the exact number which could be any number from 2/99 to 98/99 and if he is paying attention he will know that number and the odds of it being exactly 66/99 would be small.

However, If we take the weighted average of the binary tree of every possible result, we would get 2/3 as the expected value

 
It's interesting, that no matter how many shots the coach misses, once he comes back and watches the player take a shot, the odds of the player making the next is 2/3 if he makes it and 1/3 if he doesn't according to the coach.
Not just that - in the original problem, the answer is 2/3 no matter how many shots he didn't see while he was out of the room. If we change it slightly so that he sees make-miss-make, then leaves, and comes back to see a third make, the probability is 3/4. What's really interesting to me is if we make basically everything unknown:

He sees the player make the first and miss the second. Then he leaves the room, during which time the player takes some unknown number of shots. He returns and sees the result of another shot. Then he leaves again for some unknown number of shots, comes back and sees another one, and does this some random number of times at random intervals, and then finally returns at the end and sees the player make his last shot (this last part is important). What's the probability the player makes his next shot?
The answer (I think) is just (however many makes the coach saw)/(however many total shots he saw). Basically the probability from the coach's perspective depends entirely on the shots he's seen (no matter how many he's seen, how many of those seen shots were made, in what order he's seen them, etc.) and not at all on any shots he doesn't see. On the one hand that seems exceedingly obvious, and on the other hand it seems really hard to believe it actually boils down to something so simple.

 
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Not just that - in the original problem, the answer is 2/3 no matter how many shots he didn't see while he was out of the room. If we change it slightly so that he sees make-miss-make, then leaves, and comes back to see a third make, the probability is 3/4. What's really interesting to me is if we make basically everything unknown:
hey!


If he saw the guy make, miss, make, then took a break, and again came back at shot 99 for another make, the odds are greater than 2/3 that he makes shot 100. I don't know if it is as simple as it would now be a 75% chance, but intuitively that is what I am going with
 
It's interesting, that no matter how many shots the coach misses, once he comes back and watches the player take a shot, the odds of the player making the next is 2/3 if he makes it and 1/3 if he doesn't according to the coach.
Not just that - in the original problem, the answer is 2/3 no matter how many shots he didn't see while he was out of the room.
In the original problem, is the answer 2/3 if the coach does not leave the room at all and sees every shot?

 
It's interesting, that no matter how many shots the coach misses, once he comes back and watches the player take a shot, the odds of the player making the next is 2/3 if he makes it and 1/3 if he doesn't according to the coach.
Not just that - in the original problem, the answer is 2/3 no matter how many shots he didn't see while he was out of the room.
In the original problem, is the answer 2/3 if the coach does not leave the room at all and sees every shot?
No. Well maybe. The answer is exactly related to what actually happened, not what is likely to happen

 
It's interesting, that no matter how many shots the coach misses, once he comes back and watches the player take a shot, the odds of the player making the next is 2/3 if he makes it and 1/3 if he doesn't according to the coach.
Not just that - in the original problem, the answer is 2/3 no matter how many shots he didn't see while he was out of the room.
In the original problem, is the answer 2/3 if the coach does not leave the room at all and sees every shot?
No. Well maybe. The answer is exactly related to what actually happened, not what is likely to happen
Well, I guess that's right; this is a probability problem after all. So I think for there to be any mystery (as Rove has been saying), the coach has to leave the room and miss seeing at least one shot. (?)

 
It's interesting, that no matter how many shots the coach misses, once he comes back and watches the player take a shot, the odds of the player making the next is 2/3 if he makes it and 1/3 if he doesn't according to the coach.
Not just that - in the original problem, the answer is 2/3 no matter how many shots he didn't see while he was out of the room.
In the original problem, is the answer 2/3 if the coach does not leave the room at all and sees every shot?
No. Well maybe. The answer is exactly related to what actually happened, not what is likely to happen
Well, I guess that's right; this is a probability problem after all. So I think for there to be any mystery (as Rove has been saying), the coach has to leave the room and miss seeing at least one shot. (?)
I think he needs to leave and miss seeing all but one shot, for this to work out the way it does.

Think about this. If he stayed, and watched every single shot, and by some small, super small chance, the guy made every single free throw from 3-99, the odds of him making that 100th shot are far greater than 2/3. The guy has just made 98/99 shot. The probability of him making the next is just that, 98/99.

 
It's interesting, that no matter how many shots the coach misses, once he comes back and watches the player take a shot, the odds of the player making the next is 2/3 if he makes it and 1/3 if he doesn't according to the coach.
Not just that - in the original problem, the answer is 2/3 no matter how many shots he didn't see while he was out of the room.
In the original problem, is the answer 2/3 if the coach does not leave the room at all and sees every shot?
I don't think that's really a properly framed question. If he sees every shot, he knows exactly what the probability is of making the next shot. He's not making any kind of weighted estimation. The probability of making the next shot might sometimes be 2/3, but in any particular scenario it probably won't be exactly 2/3.

But if we consider all the scenarios where the coach watches every shot, and he sees the last shot go in, and average out what the probabilities are in each of them, then yes, I think it averages to 2/3. That is, all else being equal, half of the time a sequence of shots will end with a make. If we just consider all those sequences that end in a make, then the average probability of the next shot being made is 2/3.

 
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It's interesting, that no matter how many shots the coach misses, once he comes back and watches the player take a shot, the odds of the player making the next is 2/3 if he makes it and 1/3 if he doesn't according to the coach.
Not just that - in the original problem, the answer is 2/3 no matter how many shots he didn't see while he was out of the room.
In the original problem, is the answer 2/3 if the coach does not leave the room at all and sees every shot?
I don't think that's really a properly framed question. If he sees every shot, he knows exactly what the probability is of making the next shot. He's not making any kind of weighted estimation. The probability of making the next shot might sometimes be 2/3, but in any particular scenario it probably won't be exactly 2/3.

But if we consider all the scenarios where the coach watches every shot, and he sees the last shot go in, and average out what the probabilities are in each of them, then yes, I think it averages to 2/3. That is, all else being equal, half of the time a sequence of shots will end with a make. If we just consider all those sequences that end in a make, then the average probability of the next shot being made is 2/3.
Thanks. So is the coach leaving the room just a metaphor for probabilistic estimation? That's what I figured the puzzle was about since I first read it.

 
I don't have time to work it out now, but I'm thinking that one potentially surprising result is that the odds of hitting any number of shots out of 99 is the same. I.e a complete lack of convergence

I.e in the binary tree there is one path to 98/99 and the probability of that path is 1/97

But I'm saying that if we take all of the paths that lead to say 17/99, add those all up, we would get 1/97

For each x/99 from 2 to 98, I think they would all total the same....

 
I think you guys are missing the point. The coach knows he made shot 99. The coach knows his tendencies. So, from his point of view, he knows the player is either on heater and making them all, or somewhere around 50/50. He's ruling out that he's on the cold streak because he saw him hit shot 99. So, the coach is probably putting it @ 75% that he hits shot 100.
Right, but he doesn't know anything about shots 3-98. He could have missed every one of those shots and there would still be a probability > 0 of hitting 99. It would be highly improbable, but not impossible.
Yes, but the coach is likely not a stats nerd, right? He probably figures it's either a heater or a middling run and excludes the outside chance of a cold streak. So, the coach figures 75%.

 
No disagreement on that - I was only pointing out the shaky foundation of hulk's comment that he saw him hit 99 meant that the coach could rule out that he was on a cold streak. It's almost unfortunate that the actual binary tree calculations work out to the intuitive result that is based on faulty logic.
You are missing my point entirely.

 
Ignoratio Elenchi said:
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?
From here: http://fivethirtyeight.com/features/will-the-neurotic-basketball-player-make-his-next-free-throw/
The coach isn't some epic math nerd on a quest to pinpoint exactly what the probability of the next shot going in is.

 
I don't have time to work it out now, but I'm thinking that one potentially surprising result is that the odds of hitting any number of shots out of 99 is the same. I.e a complete lack of convergence

I.e in the binary tree there is one path to 98/99 and the probability of that path is 1/97

But I'm saying that if we take all of the paths that lead to say 17/99, add those all up, we would get 1/97

For each x/99 from 2 to 98, I think they would all total the same....
This is only true if we consider every possible sequence of shots, including those that end with the player missing the 99th shot.

In the given problem, where we know the player makes the 99th shot, then the path that leads to 98/99 is much more likely than the path that leads to 2/99. They're not all equally likely - that's precisely why, when we average them all out, we get 2/3 instead of 1/2.

 
I don't have time to work it out now, but I'm thinking that one potentially surprising result is that the odds of hitting any number of shots out of 99 is the same. I.e a complete lack of convergence

I.e in the binary tree there is one path to 98/99 and the probability of that path is 1/97

But I'm saying that if we take all of the paths that lead to say 17/99, add those all up, we would get 1/97

For each x/99 from 2 to 98, I think they would all total the same....
Nope, it's a bell curve. There is only one path to him having only hit 2 shots, and only one path to him having hit 98. He would have to have missed or hit every shot in between, which is a tremendous outlier either way. There are literally billions of billions of ways he could have hit 49 shots.

In fact there are billions of billions of billions of says that he could have started out with hit miss hit miss hit and end with him having hit 49 shots.

 
I don't have time to work it out now, but I'm thinking that one potentially surprising result is that the odds of hitting any number of shots out of 99 is the same. I.e a complete lack of convergence

I.e in the binary tree there is one path to 98/99 and the probability of that path is 1/97

But I'm saying that if we take all of the paths that lead to say 17/99, add those all up, we would get 1/97

For each x/99 from 2 to 98, I think they would all total the same....
Nope, it's a bell curve. There is only one path to him having only hit 2 shots, and only one path to him having hit 98. He would have to have missed or hit every shot in between, which is a tremendous outlier either way.There are literally billions of billions of ways he could have hit 49 shots.

In fact there are billions of billions of billions of says that he could have started out with hit miss hit miss hit and end with him having hit 49 shots.
So... the coach's sample size is small and meaningless?

 
I don't have time to work it out now, but I'm thinking that one potentially surprising result is that the odds of hitting any number of shots out of 99 is the same. I.e a complete lack of convergence

I.e in the binary tree there is one path to 98/99 and the probability of that path is 1/97

But I'm saying that if we take all of the paths that lead to say 17/99, add those all up, we would get 1/97

For each x/99 from 2 to 98, I think they would all total the same....
Nope, it's a bell curve. There is only one path to him having only hit 2 shots, and only one path to him having hit 98. He would have to have missed or hit every shot in between, which is a tremendous outlier either way.There are literally billions of billions of ways he could have hit 49 shots.

In fact there are billions of billions of billions of says that he could have started out with hit miss hit miss hit and end with him having hit 49 shots.
So... the coach's sample size is small and meaningless?
I think the sample size is a distractor. Him seeing that last made shot is the key.

 
I don't have time to work it out now, but I'm thinking that one potentially surprising result is that the odds of hitting any number of shots out of 99 is the same. I.e a complete lack of convergence

I.e in the binary tree there is one path to 98/99 and the probability of that path is 1/97

But I'm saying that if we take all of the paths that lead to say 17/99, add those all up, we would get 1/97

For each x/99 from 2 to 98, I think they would all total the same....
Nope, it's a bell curve. There is only one path to him having only hit 2 shots, and only one path to him having hit 98. He would have to have missed or hit every shot in between, which is a tremendous outlier either way.There are literally billions of billions of ways he could have hit 49 shots.

In fact there are billions of billions of billions of says that he could have started out with hit miss hit miss hit and end with him having hit 49 shots.
In terms of number of possible paths, you're right. In terms of probability though, he's kinda right (if we ignore the stipulation that the paths end with a make).

Consider, there's only one path to him hitting 98 shots, but once he gets that streak started, the probability of continuing to make shots keeps increasing. By the time he's made 97 in a row, he's probably going to sink the 98th. On the other hand, there are lots of ways of making exactly 50 shots but the probability of any one of those precise situations occurring is really low. When you add up the probabilities of all the many different ways to make exactly 50 shots, it really does equal the probability of the one way to make 98 shots (again, ignoring the part of the problem where we see him make the 99th shot).

Edited to add the boring stuff:

We know he makes the first and misses the second. Number of ways to make 97 of the remaining 97 shots = (97 choose 97) = 1 (obviously).

Probability of this sequence = 1/2 (for the 3rd shot) * 2/3 (for the 4th shot) * 3/4 * 4/5 * ... * 96/97 * 97/98 = 97!/98! = 0.0102.

Number of ways to make 49 of the remaining 97 shots = (97 choose 49) ~= 1.27 x 10^28. Like, 12.7 octillion.

Probability of a particular sequence = 1/2 * 2/3 * 3/4 * ... * 49/50 (for the 50 makes) * 1/51 (see how the numerators now start over at 1 for the misses) * 2/52 * 3/53 * ... * 48/98 = 49!*48!/98! ~= 0.000000000000000000000000000000080. A ridiculously tiny number which, when multiplied by 12.7 octillion, works out to 0.0102.

 
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I don't have time to work it out now, but I'm thinking that one potentially surprising result is that the odds of hitting any number of shots out of 99 is the same. I.e a complete lack of convergenceI.e in the binary tree there is one path to 98/99 and the probability of that path is 1/97But I'm saying that if we take all of the paths that lead to say 17/99, add those all up, we would get 1/97For each x/99 from 2 to 98, I think they would all total the same....
This is only true if we consider every possible sequence of shots, including those that end with the player missing the 99th shot.In the given problem, where we know the player makes the 99th shot, then the path that leads to 98/99 is much more likely than the path that leads to 2/99. They're not all equally likely - that's precisely why, when we average them all out, we get 2/3 instead of 1/2.
I guess I meant that leading up to shot 99, I goofed and conflated that with the original problem...

The intuitive thought is that he's going to get really hot or really cold

In actuality, before shot 99, I believe that each total of made shots had an equal chance

 
I don't have time to work it out now, but I'm thinking that one potentially surprising result is that the odds of hitting any number of shots out of 99 is the same. I.e a complete lack of convergenceI.e in the binary tree there is one path to 98/99 and the probability of that path is 1/97But I'm saying that if we take all of the paths that lead to say 17/99, add those all up, we would get 1/97For each x/99 from 2 to 98, I think they would all total the same....
This is only true if we consider every possible sequence of shots, including those that end with the player missing the 99th shot.In the given problem, where we know the player makes the 99th shot, then the path that leads to 98/99 is much more likely than the path that leads to 2/99. They're not all equally likely - that's precisely why, when we average them all out, we get 2/3 instead of 1/2.
I guess I meant that leading up to shot 99, I goofed and conflated that with the original problem...

The intuitive thought is that he's going to get really hot or really cold

In actuality, before shot 99, I believe that each total of made shots had an equal chance
I still don't think is correct.

There is only one sequence that results in 1/98 shots being made. Where as, if he was to make 2/98, he could have gotten the 2nd shot on his 3rd, or 4th, or 5th, or 6th, etc. try ....

 
OK, so I'm with the logic of 2/3 so far, as it made sense to me early on and still does that the greater and lesser probabilities for hits and misses along the way with the unseen shots would cancel each other out and leave us with 50-50 before the 99th shot.

I was bored and tried to remember some stats, but I'm really rusty. I tried to use the multiplication principle but forget which of these two ways makes sense (assuming either is the right way to do it, which probably isn't so safe to assume).

Looking at all the possibilities when 5 total shots are taken, with the first two shots known (probability of 1), and then throwing out the ones with a miss on the 4th shot I came up with

10111 = (1)*(1)*(1/2)*(2/3)*(3/4) = 6/2410110 = (1)*(1)*(1/2)*(2/3)*(1/4) = 2/2410101 = (1)*(1)*(1/2)*(1/3)*(2/4) = 2/24 throw out10100 = (1)*(1)*(1/2)*(1/3)*(2/4) = 2/24 throw out10011 = (1)*(1)*(1/2)*(1/3)*(2/4) = 2/2410010 = (1)*(1)*(1/2)*(1/3)*(2/4) = 2/2410001 = (1)*(1)*(1/2)*(2/3)*(1/4) = 2/24 throw out10000 = (1)*(1)*(1/2)*(2/3)*(3/4) = 6/24 throw outProbability of 5th shot being a hit = (6+2)/(6+2+2+2) = 8/12 = 2/3But if we take it as a given (probability of 1) that the 4th shot is a hit, would you do it this way?

10111 = (1)*(1)*(1/2)*(1)*(3/4) = 3/810110 = (1)*(1)*(1/2)*(1)*(1/4) = 1/810101 = (1)*(1)*(1/2)*(0)*(2/4) = 010100 = (1)*(1)*(1/2)*(0)*(2/4) = 010011 = (1)*(1)*(1/2)*(1)*(2/4) = 2/810010 = (1)*(1)*(1/2)*(1)*(2/4) = 2/810001 = (1)*(1)*(1/2)*(0)*(1/4) = 010000 = (1)*(1)*(1/2)*(0)*(3/4) = 0Probability of 5th shot being a hit = (3+2)/(3+1+2+2)= 5/8I guess the 2nd way seems wrong because it's throwing away the weight that first 2 remaining options had over the last 2 from the results of the 3rd shot. And putting the 1 or 0 probability in for the 4th shot would imply that, like the 1st two shots, it existed outside the probability rules of the game.

Thoughts?

 
OK, so I'm with the logic of 2/3 so far, as it made sense to me early on and still does that the greater and lesser probabilities for hits and misses along the way with the unseen shots would cancel each other out and leave us with 50-50 before the 99th shot.

I was bored and tried to remember some stats, but I'm really rusty. I tried to use the multiplication principle but forget which of these two ways makes sense (assuming either is the right way to do it, which probably isn't so safe to assume).

Looking at all the possibilities when 5 total shots are taken, with the first two shots known (probability of 1), and then throwing out the ones with a miss on the 4th shot I came up with

10111 = (1)*(1)*(1/2)*(2/3)*(3/4) = 6/2410110 = (1)*(1)*(1/2)*(2/3)*(1/4) = 2/2410101 = (1)*(1)*(1/2)*(1/3)*(2/4) = 2/24 throw out10100 = (1)*(1)*(1/2)*(1/3)*(2/4) = 2/24 throw out10011 = (1)*(1)*(1/2)*(1/3)*(2/4) = 2/2410010 = (1)*(1)*(1/2)*(1/3)*(2/4) = 2/2410001 = (1)*(1)*(1/2)*(2/3)*(1/4) = 2/24 throw out10000 = (1)*(1)*(1/2)*(2/3)*(3/4) = 6/24 throw outProbability of 5th shot being a hit = (6+2)/(6+2+2+2) = 8/12 = 2/3But if we take it as a given (probability of 1) that the 4th shot is a hit, would you do it this way?

10111 = (1)*(1)*(1/2)*(1)*(3/4) = 3/810110 = (1)*(1)*(1/2)*(1)*(1/4) = 1/810101 = (1)*(1)*(1/2)*(0)*(2/4) = 010100 = (1)*(1)*(1/2)*(0)*(2/4) = 010011 = (1)*(1)*(1/2)*(1)*(2/4) = 2/810010 = (1)*(1)*(1/2)*(1)*(2/4) = 2/810001 = (1)*(1)*(1/2)*(0)*(1/4) = 010000 = (1)*(1)*(1/2)*(0)*(3/4) = 0Probability of 5th shot being a hit = (3+2)/(3+1+2+2)= 5/8I guess the 2nd way seems wrong because it's throwing away the weight that first 2 remaining options had over the last 2 from the results of the 3rd shot. And putting the 1 or 0 probability in for the 4th shot would imply that, like the 1st two shots, it existed outside the probability rules of the game.

Thoughts?
The second way was the way I did it originally. I can't wrap my head around why your second way is wrong other than the first option seems right.

I can't sem to find the flaw but their must be one. Thanks for detailing it out there like you did as this is exactly what I was thinking.

BTW, if you keep going you get 2/3, 5/8, 6/10, 7/12, ... (n+1)/2n

 
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OK, so I'm with the logic of 2/3 so far, as it made sense to me early on and still does that the greater and lesser probabilities for hits and misses along the way with the unseen shots would cancel each other out and leave us with 50-50 before the 99th shot.

I was bored and tried to remember some stats, but I'm really rusty. I tried to use the multiplication principle but forget which of these two ways makes sense (assuming either is the right way to do it, which probably isn't so safe to assume).

Looking at all the possibilities when 5 total shots are taken, with the first two shots known (probability of 1), and then throwing out the ones with a miss on the 4th shot I came up with

10111 = (1)*(1)*(1/2)*(2/3)*(3/4) = 6/2410110 = (1)*(1)*(1/2)*(2/3)*(1/4) = 2/2410101 = (1)*(1)*(1/2)*(1/3)*(2/4) = 2/24 throw out10100 = (1)*(1)*(1/2)*(1/3)*(2/4) = 2/24 throw out10011 = (1)*(1)*(1/2)*(1/3)*(2/4) = 2/2410010 = (1)*(1)*(1/2)*(1/3)*(2/4) = 2/2410001 = (1)*(1)*(1/2)*(2/3)*(1/4) = 2/24 throw out10000 = (1)*(1)*(1/2)*(2/3)*(3/4) = 6/24 throw outProbability of 5th shot being a hit = (6+2)/(6+2+2+2) = 8/12 = 2/3But if we take it as a given (probability of 1) that the 4th shot is a hit, would you do it this way?
Code:
10111 = (1)*(1)*(1/2)*(1)*(3/4) = 3/810110 = (1)*(1)*(1/2)*(1)*(1/4) = 1/810101 = (1)*(1)*(1/2)*(0)*(2/4) = 010100 = (1)*(1)*(1/2)*(0)*(2/4) = 010011 = (1)*(1)*(1/2)*(1)*(2/4) = 2/810010 = (1)*(1)*(1/2)*(1)*(2/4) = 2/810001 = (1)*(1)*(1/2)*(0)*(1/4) = 010000 = (1)*(1)*(1/2)*(0)*(3/4) = 0Probability of 5th shot being a hit = (3+2)/(3+1+2+2)= 5/8
I guess the 2nd way seems wrong because it's throwing away the weight that first 2 remaining options had over the last 2 from the results of the 3rd shot. And putting the 1 or 0 probability in for the 4th shot would imply that, like the 1st two shots, it existed outside the probability rules of the game.Thoughts?
Right, you can't just assign/ignore the probability of the 4th shot. Doing it the second way, you're not properly weighting the path it took to get to each outcome, e.g. you're saying that starting out 1001 has the same probability as starting out 1011, which isn't true. If you start 100, you're more likely than not to miss the 4th shot; if you start 101, you're more likely to make it. You have to account for that difference (which is the crux of the whole problem).

In the original problem if you just assign the 99th shot probability 1, you get the 50/99 answer some people have come up with, which is basically saying that knowing the 99th shot gives you no insight at all into what happened between shots 3 and 98 ("he probably went 48/96 on all the shots we didn't see, plus the 2 we saw him make").

 
I can't sem to find the flaw but their must be one.
The flaw is that by assigning probability 1 to the nth shot, it ignores the actual probabilities it would've taken to get there. It basically results in an answer of, "I have no way of knowing what happened on the shots I didn't see, so I'll just assume he shot 50% on those. Plus I saw him make 2 out of 3, so when I add those in I get an answer a little higher than 50%." That's where the (n+1)/2n is coming from (it's like if a career .500 hitter goes 2/3 one day, it raises his career average just a little bit over .500).

 
I don't have time to work it out now, but I'm thinking that one potentially surprising result is that the odds of hitting any number of shots out of 99 is the same. I.e a complete lack of convergence

I.e in the binary tree there is one path to 98/99 and the probability of that path is 1/97

But I'm saying that if we take all of the paths that lead to say 17/99, add those all up, we would get 1/97

For each x/99 from 2 to 98, I think they would all total the same....
Nope, it's a bell curve. There is only one path to him having only hit 2 shots, and only one path to him having hit 98. He would have to have missed or hit every shot in between, which is a tremendous outlier either way.There are literally billions of billions of ways he could have hit 49 shots.

In fact there are billions of billions of billions of says that he could have started out with hit miss hit miss hit and end with him having hit 49 shots.
So... the coach's sample size is small and meaningless?
I think the sample size is a distractor. Him seeing that last made shot is the key.
I was kidding; just wanted to quote Assani.

 
knowns:

- player has taken 99 shots

- player has made at least 2 shots

- player has missed at least 1 shot

- therefore, player has made between 2 and 98 shots, inclusive

if you take a straight average of those possible outcomes, it comes to 50.5%.

but i do not know if each outcome has an equal probability. even if not, instinct tells me it should have a normal distribution centered around the midpoint of 50 makes (again, 50.5%).

 
the moops said:
Why are you looking at 5 total shots taken?
Just because it's the easiest example # where you have the 2 initial shots, some number of unknowns (in this case, only one), the next shot that's seen, and the shot we are considering.

The # of unknowns shouldn't matter - you should still get 2/3. This allowed me to see another way to do it that was wrong but seemed somewhat legit on its face.

 
oso diablo said:
knowns:

- player has taken 99 shots

- player has made at least 2 shots

- player has missed at least 1 shot

- therefore, player has made between 2 and 98 shots, inclusive

if you take a straight average of those possible outcomes, it comes to 50.5%.

but i do not know if each outcome has an equal probability. even if not, instinct tells me it should have a normal distribution centered around the midpoint of 50 makes (again, 50.5%).
That's basically where I started too, but what you actually know is that the player has taken 3 shots and made two, in this order -- made, miss, made...

And if you look at the conditional probability of the next shot after this sequence (regardless of how many intervening shots there are) I think the folks saying 2/3rd are right.

 
Rove! said:
Aerial Assault said:
the moops said:
Aerial Assault said:
For me, the clue was in the way the problem was written. With the limited information the coach has, the answer is pretty clearly 2/3, but he leaves the gym, so his limited-information conclusion has to be wrong, right? Turns out that no, it's right whether he left or not.
If he stayed at the gym, and watched every single shot up to shot 99, the chances that it would be exactly 2/3 at that point is miniscule.
No. See this from Greg:

"I think you keep getting lost in the fact that though the odds will eventually converge towards either 0% or 100% over enough trials, our coach is given no information which way it is diverging with just knowing the first 2 shots. The coach might agree it is likely the player made most or missed most rather than making 50%. But based on the info he can work from, those are equally likely so leave him at 50% odds of making the next basket... until such time as he knows the outcome of additional baskets such as learning the result of shot 99."
I disagree on the convergenceThe odds of making n shots in a row (after shot 2) is 1/(n-1). Which means the odds of converging to 100% decrease over time

Similarly the odds of converging to 0 decrease over time
I think you're probably right, there will be a point where each new free throw only changes the shooting percentage a negligible amount. Brute forced some tests in Excel with multiple thousands of free throws and the results are a spectrum. There are a few that are very low or very small, but mostly it's whereabouts they were as the number of free throws taken got large. At around 150-300 the percentage mostly settled down, though some moved another 3-4% either direction over the next several thousand free throws.

 
If there were really a coach like this then he pull a player if they missed their first shot of the game. Then the player would stay on the bench until another player went 0-2.

 
the moops said:
Why are you looking at 5 total shots taken?
Just because it's the easiest example # where you have the 2 initial shots, some number of unknowns (in this case, only one), the next shot that's seen, and the shot we are considering.

The # of unknowns shouldn't matter - you should still get 2/3. This allowed me to see another way to do it that was wrong but seemed somewhat legit on its face.
Gotcha

 
Ignoratio Elenchi said:
nice work by the crew to get to that 2/3, especially all the long math guys. :thumbup: new puzzle required viewing the image so probably should just look at the link.
:goodposting:

For the record in case all the math scared off some folks, I'm pretty sure I got the answer to the new puzzle today, and my solution requires basically no math. Just have to think about the problem the right way. We'll see. :shrug:

 

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