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.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 timeNo. See this from Greg: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.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.
"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."
Similarly the odds of converging to 0 decrease over time
For every shot the coach sees the player take, the percentages are going to change. I think.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.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 timeNo. See this from Greg: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.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.
"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."
Similarly the odds of converging to 0 decrease over time
Depends how you word the questionAgreed, 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.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 timeNo. See this from Greg: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.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.
"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."
Similarly the odds of converging to 0 decrease over time
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: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.
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.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?
hey!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:
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
In the original problem, is the answer 2/3 if the coach does not leave the room at all and sees every shot?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.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.
No. Well maybe. The answer is exactly related to what actually happened, not what is likely to happenIn the original problem, is the answer 2/3 if the coach does not leave the room at all and sees every shot?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.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.
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. (?)No. Well maybe. The answer is exactly related to what actually happened, not what is likely to happenIn the original problem, is the answer 2/3 if the coach does not leave the room at all and sees every shot?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.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.
I think he needs to leave and miss seeing all but one shot, for this to work out the way it does.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. (?)No. Well maybe. The answer is exactly related to what actually happened, not what is likely to happenIn the original problem, is the answer 2/3 if the coach does not leave the room at all and sees every shot?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.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.
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.In the original problem, is the answer 2/3 if the coach does not leave the room at all and sees every shot?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.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.
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 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.In the original problem, is the answer 2/3 if the coach does not leave the room at all and sees every shot?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.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.
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.
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%.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.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.
You are missing my point entirely.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.
The coach isn't some epic math nerd on a quest to pinpoint exactly what the probability of the next shot going in is.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?
So more Byron Scott than Brad StevensThe coach isn't some epic math nerd on a quest to pinpoint exactly what the probability of the next shot going in is.
Don't lie. You came up with 2/3 cause you read the thread.I come up with 2/3 -- since he's seen three shots and two went in.
This is only true if we consider every possible sequence of shots, including those that end with the player missing the 99th shot.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....
haha honestly not lying!Don't lie. You came up with 2/3 cause you read the thread.I come up with 2/3 -- since he's seen three shots and two went in.
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.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....
So... the coach's sample size is small and meaningless?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.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....
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 think the sample size is a distractor. Him seeing that last made shot is the key.So... the coach's sample size is small and meaningless?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.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....
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).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.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....
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 guess I meant that leading up to shot 99, I goofed and conflated that with the original problem...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 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....
I still don't think is correct.I guess I meant that leading up to shot 99, I goofed and conflated that with the original problem...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 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....
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
https://forums.footballguys.com/forum/index.php?/topic/740599-math-puzzle-from-fivethirtyeight/?p=18794268I 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 ....
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.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?
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).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?
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?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
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 can't sem to find the flaw but their must be one.
I was kidding; just wanted to quote Assani.I think the sample size is a distractor. Him seeing that last made shot is the key.So... the coach's sample size is small and meaningless?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.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....
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.
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 moops said:Why are you looking at 5 total shots taken?
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...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%).
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.Rove! said: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 timeAerial Assault said:No. See this from Greg:the moops said: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.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.
"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."
Similarly the odds of converging to 0 decrease over time
No idea, but from my point of view this is a dumb question.What is the probability, from the coachs point of view, that he makes shot No. 100?
GotchaJust 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 moops said:Why are you looking at 5 total shots taken?
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.
From my point of view the Jedi are evilNo idea, but from my point of view this is a dumb question.What is the probability, from the coachs point of view, that he makes shot No. 100?
nice work by the crew to get to that 2/3, especially all the long math guys.Ignoratio Elenchi said:
nice work by the crew to get to that 2/3, especially all the long math guys. new puzzle required viewing the image so probably should just look at the link.Ignoratio Elenchi said:
Am I missing something? Isn't it 50%*50%*50%???Ignoratio Elenchi said: