Can you solve the basketball robot riddle? - Dan Katz
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- Опубліковано 12 чер 2024
- Practice more problem-solving at brilliant.org/TedEd
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You’ve spent months creating a basketball-playing robot, the Dunk-O-Matic, and you’re excited to demonstrate its capabilities. Until you read an advertisement: “See the Dunk-O-Matic face human players and automatically adjust its skill to create a fair game for every opponent!” That’s not what you were told to create. Can you recalibrate your robot to make it a fair match? Dan Katz shows how.
Lesson by Dan Katz, directed by Igor Ćorić, Artrake Studio.
This video made possible in collaboration with Brilliant
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“Skimmed an article about AI and overpromised”
- every tech company in the 2020s
Agreed
Hey boss check out this documentary on AI!
Oh how wonderful
*_looks at first five minutes_*
alright you’re being replaced now
I'd stop whatever I'm doing for TED-Ed riddles.
fr tho
That video about Ted Ed be like:
RELATABLE🗣️🗣️🗣️🗣️🔥🔥🔥🔥🔥‼️‼️‼️‼️‼️
Me too 😂
I’m stopping whatever I’m doing for a TED-ed riddle.
You should set it to 100% and if anyone complains, explain they don't understand probability
Dam, that's exactly what I'd do too
Honestly finding easy solutions like that is more fun than solving it as intended
@@metal_pipe9764object show fan?
@@SixtyStone yeah? Not sure how that's relevant
@@metal_pipe9764 I just like pointing out when I see sumone is a fan of sumthing I'm a fan of lol
I like how this ends with the character getting a better job
Agreed
And a regular spot in the basketball team.
Better fate than that of the guy from Dormor 😆
I was like "this is the most unrealistic part of the whole video"
"oh wow a riddles i love ridd"
ALGEBRA
NO it's not only Algebra but also probability which makes the sequence more complicated
@@arkmoncushman1824 You can do it without probability, with some basic logic. Like if opponent needs 10 throws to win, then robot should also get win within those 10 throws i.e. 9 throws, so you can deny always go first advantage by finishing 1 turn faster (you can see it as increasing robot chance to win by "human go first advantage percent"). Formula would be something like q = 1/((1/p) - 1) and if you simplify it would be the same q = p/(1-p)
Riddles so good it will put the riddler to shame
I Àgreed.
Batman already did that with the ultimate Riddle. (No not "Who is The Batman", but rather how he survived that explosion. It drove the Riddler to Arkum faster than the police could.)
Cool robot and puzzle but I’m more interested in the puzzle of how she managed to find a better work place and navigate all that legal trouble.
What legal trouble?
@@randompastahandle Since she made the robot while working on the company she was leaving, probably on her contract was stated that everything she made was company's property, so she needed to somehow get the company to not get the rights for her basketball robot so she could use her robot at the better work place
That's what I THINK it meant, i don't have any expertise in this area so i could be totally wrong, and my english is not very good so if there's some phrase that sounds wrong, sorry for that
@@randompastahandle companies typically don't let intellectual property go easily, and depending on the contracts that were signed, keeping your creations when leaving a company you worked for can be straight up impossible.
That's why you always read the terms and conditions before you agree to them.
She told her boss she had green eyes and then asked to leave.
This isn't a riddle, this is a math problem
Same difference
They always are
Only a few of them are math based, most of them are logic.
@@ExzaktVid math is logic with symbols
@@slept5971 Ok but you know what I meant though
If I would get a dollar for each time I see brilliant sponsoring a video, I would afford that basketball robot
You could afford 10 if they were pennies
@@davidbailis8415?
I would be able to afford a house probably, and honestly, I'm getting quite tired of it. I'm so glad I have something that auto skips all adds and sponsor segments
You'd be able to afford making what the boss promised
ANOTHER RIDDLE!!!!! I LOVE THESE!
Me Too! 😊
Same
They are backkk
IKR
Me too, it always awesome every time he makes a new one
You know the riddle was tough when you can't even understand the answer after it has been explained.
Here's a summary of the simplified version near the end:
Let's say the human has a 30% chance of making a basket, so in 3 out of 10 tries, they win on the first try. If you adjust the robot to win in 3 out of the remaining 7 tries (3/7 = about 43%), then things are even:
* Again, out of 10 tries, 3 times the human wins on their first try.
* 3 times, the human misses and then the robot wins on its first try.
* The other 4 times, they both miss, and basically the whole thing starts over. And because it was even up to this point, it'll stay even, no matter how long it takes for someone to win.
phew i thought that it was just me.
Since the human goes first, your robot should be set slightly higher than the human for a 50% chance to win, since the human can win without the robot ever getting a turn. How much higher? That's what the video is about.
I like that the robots are called Dunk-o-Matic's
Me too
I was literally doing my math homework, finished and wanted to take a break with a riddle. I see this and think “I JUST DID THIS!”
This felt more like a complicated math problem than a riddle...
The geometric series solution is, but the second approach, starting at 4:43, is more about finding a framing that makes the solution simple.
Almost all of Ted ed riddles are
While mathematically correct, I feel like your explanation went purely for the mathematical sentencing and didn't quite reach the bottom line where real-world understanding lies.
I thought it out like this in my head and solved it within a minute or two:
1. Whenever the human scores a basket, the robot doesn't get a chance to throw.
2. If the human has a 1/2 scoring ratio, then statistically for every two rounds the robot may only play once and would thus have to have a 100% scoring ratio (i.e. 1/1) to even the score.
3. Also with every other human scoring ratio, the robot will play one turn fewer for every set of rounds where the human scores once, so if for example a human with a 1/4 scoring ratio scores 1 in a 4 round game, the robot will need to score 1 in its available 3 attempts; if a human with a 1/10 scoring ratio scores 1 in a 10 round game, the robot will need to score 1 in its available 9 attempts, etc.
4. Thus, if p = 1/x, then q = 1/(x-1), and x >= 2.
Woah that was really good. I like this one better ngl
@@LightYagamiK You're welcome. Not sure why they didn't provide any less-convoluted explanation that's easier for the average viewer to mentally decode.
They way I solved it was the way they mentioned that bypassed the series; that the probability the first shot occurs on either of the first two turns must be equal, so p = (1-p)q.
The series thing is cool but it's like using a sledgehammer to crack open nuts.
Can you solve...
Me :*clicks immediately*
So true
Step 1: Confirm that you have green eyes
Step 2: Ask the basketball playing robot to leave
Took a while before one of these showed up.
“Hey [Basketball playing robot], can I leave?”
“…Ozo”
@@MothmanOfficialWvatoo bad we still don’t know what that means
Then we bet 24 gems on the silver hexagon and figure out if the wind crystal is lying
Uh... wrong puzzle, buddy.
I literally just completed an algebra II course with the unit of probability in it and then I watch this video on the same day and nailed it in the head. They were right when they said we'd use those skills in real life.
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 😎
I Àgreed.
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/1+p which will have a solution for all p.
But you arent allowed to do that. So just set it to 100%
Boss be like:
Eh , she can probably handle it.
She: spends about five minutes WITH the help of ted ed
Waiting for this day.... We need more such riddles
I Àgreed.
Taught some students probability this morning and here I am learning something new about it.
I wish they'd call these problems something other than riddle because they're really not
Finally another riddle!
I Àgreed.
I'm subscribed EXCLUSIVELY for the riddles
Same here
real
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
Get online boys. New TedEd riddle just dropped.
I Àgreed.
The riddle gods have smiled on us!
I Àgreed.
0:02 Probably not relationship advice... maybe.
Wise words
Keep in mind that droids don't rip people's arms out of their sockets when they lose....
I say let the Wookie win.
Agreed also I think you mean rookie
@@yellowstarproductions6743 Star Wars reference
The riddle can be more easily solved using the expectancy of a geometric distribution.(1/p) Then subtracting 1 since the robot goes second and calculating q from known expectancy and distribution.
Ted-Ed: "Can you solve the basketball robot riddle?"
Me: Yes. Through the power of perseverance I will (hopefully) succeed.
That's the spirit!
@@TEDEd 😮
Back with the riddles I love them even though I can't solve it but i am proud to watch them and i can proudly say I solved the cursed temple riddle also back with the old style I love it they should continue the riddle & style
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.
We were trying to keep this on the down low
Wow hi 🤩!
@@TEDEdwhat was the answer? I was enjoying the animation!!!!!!
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-pq+pq = q-pq
p = q-pq
p = q(1-p)
q = p/(1-p)
Thank you for these amazing riddles TED-Ed!
Yes! More riddles, please!
Yeah, I figured this out with the second method of making the robot's first-shot chance equal to the human's, since it resets after each set.
Also, if the robot shoots first, then the same logic gives q = (1-q)p = p - pq, or q + pq = (1+p)q = p, or q = p/(p+1). Unlike the human going first, this works for any shot probability.
Finally , another riddle , love'em , waiting for next one Ted-ed
2:26 this should be |r| < 1 or -1 < r < 1 not r < 1.
i always take copious notes whenever TedEd drops a new riddle just in case I ever find myself in the same situation
A simpler solution would be to consider the first turn only.
The human has probability p of winning on his first throw. The probability for the robot to win on his first throw is the probability that the human missed his throw, times the probability of the robot scoring on his, that is, (1 - p) * q. The human and robot should be equally likely to win on their first turn, which gives the equation p = (1 - p) * q which when solved for q gives q = p / (1 - p).
Ignore it and screw this riddle, literally do consider how much tries needed for a human win, add that tries + 1.
If you want a good decent robot , after x tries score 100% and that's it.
It is still fair and decent as long as the bot doesn't score in the first few tries.
this animation is so cool! and the jokes + puns are so actually funny
So....... I still don‘t understand the riddle
It's hard to even think about it brother
Try looking at the edge cases. If the robot is playing against a human who has a fifty percent chance of sinking the basket and the human goes first, and the robot only gets to shoot if the human misses, then the robot has to be 100% accurate to have a 50% chance of winning.
If the human's chances of making a basket on a single shot is greater then 50%, then the rules don't allow a way to make it fair for the robot.
If the human is zero percent accurate, then if the robot has a non zero percent chance of scoring, then given a potentially infinite number of tries the robot wins. No way for the robot to give the human a fifty percent chance of winning.
Therefore, for any chance of P of the human making the shot, where P is greater than zero and less than or equal to half, Q (the robot's chances to make a basket on a single shot) must be greater than zero but less than or equal to one.
It is a math problem not a riddle, ignore and move on with your life :)
I dedicate two whole hours to understand this riddle just because I love TED ED riddles too much
the marketing guy scewing the programmer is the closest any riddle has ever come to reality
There is a certain aesthetic to these riddles, that’s why I always watch these but never be able to solve this 😁
I realized the answer was going to involve some series shenanigans and didn't want to bother with that, so i looked at it differently:
I made it so that the robot has a 50% chance of scoring on their first shot, and you set q to be 0 after. This garuntees that the player will score eventually, and so the total proability of winning is just based on the first throw of the robot and player.
The probability of the robot winning on their first shot is (1-p) * q, and so if we set that probability to 0.5, we get q = 0.5 / (1 - p). And so you set the initial q to be this for the robot, and if it misses, set q to 0, garunteeing the player wins 50% of the time.
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.
What happens when the Robot is paired with Steven Curry?
That match will never end.
I find your riddles very interesting ! I love them ! Can you do another one ?
When we needed them most the riddles returned
So well created depicting current scenarios
Always waiting for Ted ed riddles ❤
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
Honey, a new TedED riddle just dropped!
I solved it instantly using a much faster (and easier) way:
Note that in the first two shots, one by the human and one by the robot, each player should have an equal chance of scoring. This is because after the first two rounds, the situation is identical to the initial situation, so each player should also have 50% chance of winning overall.
So we know that p=(1-p)q, so q=p/(1-p).
Also, for q
It's great that they're continuing this legendary series 😎
I've been waiting a while for a new riddle :D
Going to be honest, I don’t even try to solve these riddles. I just like the senecios and hearing the explanation at the end.
Surely their must be a way without infinite sums. Here's my thinking: Suppose the opponent has a 40% chance of winning (and that's constant, so even if you go a million rounds before either of you wins, the next shot the opponent takes has a 40% chance of success). 40% of the time you lose on the first shot, ergo 40% of the time you don't even get to make a shot. So, 60% of the time, you DO get to make a shot. If, when you get to take a shot, you succeed 66 2/3% of the time, then, as of the time before any shots were made, you have a 40% chance of winning on the game's second shot (your first shot), because you only get the opportunity 6/10ths of the time, and OF those 6/10ths times, you win 2/3rds of the time, so that's 4/10ths or 40%. And that means that your odds of success, immediately prior to any two shots, are the same as your opponents. Now, 60% of the time your opponent shoots and fails to win the game. Of the remaining 60% of cases, you fail 1/3 of the time. So, right before anyone takes any shots, there is a 20% chance (6/10ths times 1/3rd) that the game will proceed to a second pair of shots. But then since it's stipulated that your opponent has the same 40% chance of a win on THAT shot, you need to have a 66 2/3% chance of success on THAT shot too. No matter how many iterations we go through, your odds of success, to give you an overall 50/50 chance of winning the game, need to remain at a constant 66 2/3%. This ends up in the same place as the video ends up (i.e. it's true that 66 2/3% is equal to 40% divided by (100% minus 40%), but without infinite sums. Suppose that we say that as the arena will close soon we will allow only two rounds today, and if nobody wins, we come back tomorrow. What are the odds then? Well, we have
40% of the time, opponent wins on first shot;
60% of the time, robot gets to make a first shot;
2/3 of 60% of the time (40%), robot wins on first shot;
1/3 of 60% of the time (20%) , robot misses first shot and opponent will be allowed a second opportunity;
(20% times 40%) of the time, or 8% of the time, game ends with opponent winning on their 2nd shot)
(20% times 60%) of the time, or 12% of the time, opponent misses and so the robot gets a 2nd shot
(2/3 of that 12%) of the time, or 8% of the time, robot sinks 2nd shot and wins
(1/3 of that 12%) of the time, i.e. 4% of the time, robot misses and opponent will be allowed a 3rd shot but not until tomorrow.
What are the odds each has of winning on that first day. For both the opponent and the robot, the sum of the chances of winning on the 1st attempt plus the 2nd attempt are 48%, meeting the specification that the odds should be equal. SOMEONE will win, therefore, 48% plus 48% of the time, or 96% of the time, meaning that 100%-96% of the time (4% of the time, agreeing with the result obtained above) the game will have to be continued tomorrow.
But you can get to the same position of equal odds for both players without doing the sums over two shots. The truth is that after any number of complete rounds (i.e. at any time immediately after the robot shoots and misses to continue the game), the computation can be the same as when it was before any shots were taken.
I solved it but not in such a big brain way. I just found some solutions for some easy values for p which made me figure out the correlation between q and p allowing me to make the formula and simplify it to q = p/(1-p).
Ted ed riddles are really challenging for me but it always feels satisfying when I manage to solve one
PLEASE DO RIDDLES MORE OFTEN 😢❤
omg I did the geometric series solution and I’m so mad I didn’t realize the elegant solution. That’s nice.
I remember I watched this channel 4 years ago. So much nostalgia.
oh boy a new riddle
I like when the narrator said "It's geometric series time" and started talking about geometri serieses
I did one of these right first try! Finally! This is a high no mortal being should be allowed to experience!
I just enjoy the realistic and oddly specific premise that someone added a random new constraint to an engineering project at the last second, and also you happen to be able to kinda meet that constraint through some janky nonsense because of course you can, you're an engineer
FINALLY ANOTHER ONE!!!!!! LETS GOOOOOOO!
all you have to do is give the robot green eyes, then it will leave and no one will complain
You've hit the nail on the head here
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!
I've done that before. The trick is to assign each color a digit then each round find the lowest number that doesn't contradict any previous clues.
What's more interesting is that if the robot goes first, we can always ensure a fair play irrespective of human players, with robot's winning probability of p/(p+1) ranging from 0 to 0.5. She got lucky, it would be wiser to persuade the organisers to let the robot go first.
robots are developing so fast that it's scary
if you're worried about the 1st-player-advantage, you add a policy of randomising (to 50%) who starts - you don't try & solve the problem with the original single parameter.
Finally the first riddle I actually solved by myself
OMG A NEW RIDDLE IM SO HAPPY I LOVED TED EDS RIDDLES
Who needs Brilliant when you have TedEd riddles
Congrats on 20 million subs!!!
I honestly love this channel's riddles, they arent like "sally dad has 3 kids January, February then what is the name of the 3rd child?" They actually put thought in their riddles and make us use our brain.
This riddle can be solved without all the complicated algebra. Think about it this way: if the human has _p_ % chance 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). This makes it such that in the event both miss their first shot, the problem simply collapses back into the initial problem.
Needless to say, the highest value of _p_ for which this is possible is 50%, which requires _q_ to be 100%.
Ted Ed’s riddles are my childhood 😂
NEW TED ED RIDDLE DROPPED
The way I thought about it: Just make the probability of the robot's winning on its first try be the same as the human winning on their first try. And compensate for the fact that the robot's probaiblity is multiplied by (1-p).
YES FINNALY ANOTHER RIDDLE IVE LITERALLY WATCHED ALL OF THEM ON REPEAT
More riddles please!
You mean more math problems or real riddles that uses logic and not much complex math involve?
The happy ending was a nice touch. :)
My whole childhood was watching Ted ed
Time flies so fast
Finally another riddle where no one is at risk of dying
It is a beautiful day, ted-ed uploaded a riddle
That's the first riddle I could solve on my own with no help. I thought, this day may never come! 😂
Just tell the people that at least one of them has green eyes, so they get distracted trying to figure out who and end up missing half the shots
This was a great riddle and I was happy to have been able to solve it, but I felt that a simpler solution could have been used, although I may have cut some corners.
First off, I thought that there really is no need to calculate the probability of a tie as all that was important was that the probability that the human winning versus the robot winning just had to be equivalent, along with the fact that only the first win mattered, so it was probably unimportant on which round either one won in.
In order to compute the probabilities of each one winning a round, the human's chance of winning was simply p
P = human chance of winning
However, the robot's chance of winning was dependent on the human losing, so it came out to
R = (1-P) * Q
Solving for Q when these two probabilities are equivalent gives us the following result
P = (1-P) * Q
P / (1-P) = Q
Q = P / (1-P)
So, I was just wondering if I still solved the riddle correctly, or if I got the right answer the wrong way.
New riddle lets go
If your brain hurts from this, kingdom hearts would give you an aneurysm
babe wake up! ted-ed posted a new riddle!!
He is back to making ridles yay
I like the riddles
NEW TED ED RIDDLE JUST DROPPED
And I was hoping to solve this
Does anyone know a universal surface formula for any n-sided polygon, even with non equal sides?