Can you solve the basketball robot riddle - Dan Katz
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Date: 2024-06-13
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Comments and reviews: 20
immortel_0
so heres how i did it, cd be wrong havent seen the answer yet. so, the human can win the match in these ways, going sequentially
human dunks.
or
human misses, robot misses, human dunks,
or
human misses, robot misses, human misses, robot misses, human dunks
and so on.
probability of human dunking is p, so probablity of human missing be 1-p, lets denote it by p’
similarly, probability of robot missing is 1-q, denoting it by q’
so the probability of the first case is p
second case is p’q’ p
third case is p’q’p’q’ p and so on
we get an infinite GP whose first term is p and common ratio p’q’
so equating the GP to 0. 5 we get
q = p/(1-p)
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so heres how i did it, cd be wrong havent seen the answer yet. so, the human can win the match in these ways, going sequentially
human dunks.
or
human misses, robot misses, human dunks,
or
human misses, robot misses, human misses, robot misses, human dunks
and so on.
probability of human dunking is p, so probablity of human missing be 1-p, lets denote it by p’
similarly, probability of robot missing is 1-q, denoting it by q’
so the probability of the first case is p
second case is p’q’ p
third case is p’q’p’q’ p and so on
we get an infinite GP whose first term is p and common ratio p’q’
so equating the GP to 0. 5 we get
q = p/(1-p)
reply
MineCraeper
I solved it with the second method.
See the game as a series of two shots, the first one taken by a human and the second one by the robot, there are four possibilities:
The human and the robot both miss [no one wins]: (1-p(1-q)
Only the robot scores [robot wins]: (1-p)q
Only the human scores [human wins]: p(1-q)
Both score [human wins]: pq
Since the robot should be winning 50% of the time, and the human should be winning 50% of the time, p(human wins) = p(robot wins)
Thus
p(1-q)pq = (1-p)q
p-pqpq = q-pq
p = q-pq
p = q(1-p)
q = p/(1-p)
Thank you for these amazing riddles TED-Ed!
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I solved it with the second method.
See the game as a series of two shots, the first one taken by a human and the second one by the robot, there are four possibilities:
The human and the robot both miss [no one wins]: (1-p(1-q)
Only the robot scores [robot wins]: (1-p)q
Only the human scores [human wins]: p(1-q)
Both score [human wins]: pq
Since the robot should be winning 50% of the time, and the human should be winning 50% of the time, p(human wins) = p(robot wins)
Thus
p(1-q)pq = (1-p)q
p-pqpq = q-pq
p = q-pq
p = q(1-p)
q = p/(1-p)
Thank you for these amazing riddles TED-Ed!
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Skully935
If there’s one video I NEVER skip from Ted-Ed, it’s the riddle videos, they are always just so interesting and I always have the riddles playlist on in the background when I’m doing other things.
Even tho I could never solve the riddles, I’m always so fascinated by the solution.
Ted-Ed riddle videos are at the very top of S-Tier videos to watch for me, I do wish they’d upload riddle videos more frequently tho, but as long as Ted-Ed keeps posting riddle videos I can handle long gaps between those videos
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If there’s one video I NEVER skip from Ted-Ed, it’s the riddle videos, they are always just so interesting and I always have the riddles playlist on in the background when I’m doing other things.
Even tho I could never solve the riddles, I’m always so fascinated by the solution.
Ted-Ed riddle videos are at the very top of S-Tier videos to watch for me, I do wish they’d upload riddle videos more frequently tho, but as long as Ted-Ed keeps posting riddle videos I can handle long gaps between those videos
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terencemah8521
This riddle can be solved without all the complicated algebra. Think about it this way: if the human has probability _p_ of making a shot, then out of every 100 games, they will win on the first shot _p_ times. To have the same chance of winning, the robot must make _their_ first shot _p_ times out of the remaining ( 100 - _p_ ) events, giving the required robot probability of p/(1-p. In the event that both miss their first shot, the problem now simply collapses back into the initial problem.
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This riddle can be solved without all the complicated algebra. Think about it this way: if the human has probability _p_ of making a shot, then out of every 100 games, they will win on the first shot _p_ times. To have the same chance of winning, the robot must make _their_ first shot _p_ times out of the remaining ( 100 - _p_ ) events, giving the required robot probability of p/(1-p. In the event that both miss their first shot, the problem now simply collapses back into the initial problem.
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daveshockey86
I normally like these riddles, but I didn’t think this made much sense. Most people shoot over 50%, so fair games are impossible, and is the goal to have the collective result be 50% wins, or 50% of any individual (which we already know is impossible) If it’s the former, you need to preemptively know every individual’s %s. All in all, not loving this riddle. Hopefully next one will be more clear.
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I normally like these riddles, but I didn’t think this made much sense. Most people shoot over 50%, so fair games are impossible, and is the goal to have the collective result be 50% wins, or 50% of any individual (which we already know is impossible) If it’s the former, you need to preemptively know every individual’s %s. All in all, not loving this riddle. Hopefully next one will be more clear.
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bigF94
I dont think this contest is good for perceived fairness because you can only win or lose. Doesnt matter the probability, a game of luck will never be fair. With so few attempts its impossible, its like a penalty shootout, I don't think anyone thinks about it being fair. If they were playing a whole game of basketball robots vs humans and some kind of skill values assigned, maybe I could see it.
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I dont think this contest is good for perceived fairness because you can only win or lose. Doesnt matter the probability, a game of luck will never be fair. With so few attempts its impossible, its like a penalty shootout, I don't think anyone thinks about it being fair. If they were playing a whole game of basketball robots vs humans and some kind of skill values assigned, maybe I could see it.
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rokaq5163
Q can't be equal to P because if the human always goes first and has a 100% probability they would always win.
They are going to always win, regardless. If the human has a 100% I could make the robot fail every single time and it wouldn't make a difference.
I feel like there's a better way to exemplify geometric series.
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Q can't be equal to P because if the human always goes first and has a 100% probability they would always win.
They are going to always win, regardless. If the human has a 100% I could make the robot fail every single time and it wouldn't make a difference.
I feel like there's a better way to exemplify geometric series.
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risks9024
I was close on this with the formula, but forgot to also consider the losing probably of current person in previous round
p p(i-q) p(1-q)2 p(1-q)3. = q(1-p) q(1-p)2 q(1-p)3. => q=p/sqrt(1-p)
But I am happy I am getting closer to enlightenment
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I was close on this with the formula, but forgot to also consider the losing probably of current person in previous round
p p(i-q) p(1-q)2 p(1-q)3. = q(1-p) q(1-p)2 q(1-p)3. => q=p/sqrt(1-p)
But I am happy I am getting closer to enlightenment
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zmaj12321
I did the first approach, but when I got the suspiciously clean answer of success/failure, I figured there was a more elegant reason.
I like how mundane the end of the story of this riddle is compared to the usual dragons and aliens and whatnot.
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I did the first approach, but when I got the suspiciously clean answer of success/failure, I figured there was a more elegant reason.
I like how mundane the end of the story of this riddle is compared to the usual dragons and aliens and whatnot.
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onion1940
I feel this one was worded poorly. I was immediately confused by the fact that it's impossible for this to succeed if p > 50%, thus making it impossible to ensure that each human wins 50% of their games. Oh well still cool ig
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I feel this one was worded poorly. I was immediately confused by the fact that it's impossible for this to succeed if p > 50%, thus making it impossible to ensure that each human wins 50% of their games. Oh well still cool ig
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SeltzerFountain
Nice to see we actually get a better job in this riddle after the fact
I feel a certain game-tallying, vote-counting, dragon-land-dividing and maze-game-creating someone feels pretty jealous over that, though
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Nice to see we actually get a better job in this riddle after the fact
I feel a certain game-tallying, vote-counting, dragon-land-dividing and maze-game-creating someone feels pretty jealous over that, though
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patrickbutsmart
This is cool and all but given how you dismisses q = p due to if p= 100%, the person have a 100% chance of winning. This problem still arises from the solution gave so shouldn't that solution also be dismissed
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This is cool and all but given how you dismisses q = p due to if p= 100%, the person have a 100% chance of winning. This problem still arises from the solution gave so shouldn't that solution also be dismissed
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LameytheClown
I brute forced some values in excel and then figured out what formula would match the curve. I guess I will find out if there is a logical non mathy solution once I watch the video.
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I brute forced some values in excel and then figured out what formula would match the curve. I guess I will find out if there is a logical non mathy solution once I watch the video.
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user-jc2lz6jb2e
My internal clock is so tuned, I was thinking it's been long enough since the last TED-Ed riddle video (it's ben 3 months, and so I checked and this was psoted 2 hours ago.
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My internal clock is so tuned, I was thinking it's been long enough since the last TED-Ed riddle video (it's ben 3 months, and so I checked and this was psoted 2 hours ago.
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Zackforsure
I appreciate that pretty much every Ted Ed riddle these days is just a convoluted math lesson disguised as entertainment. Kind of a good way to trick people into math I guess.
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I appreciate that pretty much every Ted Ed riddle these days is just a convoluted math lesson disguised as entertainment. Kind of a good way to trick people into math I guess.
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SanketAlekar
For probability p > 0. 5, you can change the game and have the robot go first, with probability q where q/1-q = p. So q = p/1p which will have a solution for all p.
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For probability p > 0. 5, you can change the game and have the robot go first, with probability q where q/1-q = p. So q = p/1p which will have a solution for all p.
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JH-cp8wf
Look, I'm gonna be honest, I didn't even really understand the question here, never mind the answer. But good job I'm sure this would be a fun video for a smart person!
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Look, I'm gonna be honest, I didn't even really understand the question here, never mind the answer. But good job I'm sure this would be a fun video for a smart person!
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Sovreign071
Finally!
I wonder: could you do a riddle on how to solve the classic color password game
Edit: Game is Mastermind. Sorry, was too focused on the video!
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Finally!
I wonder: could you do a riddle on how to solve the classic color password game
Edit: Game is Mastermind. Sorry, was too focused on the video!
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jameswang489
Ben Simmons participated in the game, your robot won every single round, your CEO made billions and wrote a book about how he was self made to success.
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Ben Simmons participated in the game, your robot won every single round, your CEO made billions and wrote a book about how he was self made to success.
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QuickScienceYT
Ok, but how did my team find out the exact people who would come, and where did they find their chance of shooting a basket This cannot be ethical.
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Ok, but how did my team find out the exact people who would come, and where did they find their chance of shooting a basket This cannot be ethical.
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