Can you solve the basketball riddle? - Dan Katz
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- Опубліковано 11 вер 2024
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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.
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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
Not only tech companies, unfortunately.
"Guys I know that blockchain stuff was all baloney but this AI stuff is definitely legit and not at all blown out of proportion"
@@PandaCake978That’s not entirely fair. Comparing blockchain to AI undersells the fact that the entire cryptocurrency industry was a scam from top to bottom. Meanwhile, behind the desperate corporate scramble for money that will supposedly materialize out of thin air is some researchers just doing their work. Before Sam Altman got his greedy CEO hands on it, OpenAI was strictly a non-profit that told investors they were unlikely to ever see returns.
The board failed to kick Altman out this last winter, so there’s little hope left that OpenAI will achieve anything meaningful, but the scientists are still around. Ilya Sutskever left to start over again, and again he’s built a company entirely uninterested in selling products or substance-less dreams of productivity.
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'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.
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"
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.
She built a lawyer robot and set it's odds of winning the litigation to 100%.
"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)
@@FreezyFrogYou literally just described probability and algebra
I was also expecting a "insight answer" instead of a wall of numbers lol
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
Ok I'd say ur right they advertise brilliant in EVERY VIDEO
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!”
for real, I have my mocks in a few weeks and finals in a few months. This is actually a good poractice question haha
I like that the robots are called Dunk-o-Matic's
Me too
despite not dunking once
ANOTHER RIDDLE!!!!! I LOVE THESE!
Me Too! 😊
Same
They are backkk
IKR
Me too, it always awesome every time he makes a new one
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.)
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.
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
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%
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.
@@Dexaan Indeed, since after every basket made the human is next to go, thus every time the human makes a shot they get to take another turn right away. The robot, meanwhile, only gets one shot at a time because basket or not the human is next to shoot.
It just depends on whether you understand the math behind it. For anyone that’s taken a calculus class (or any class where series are taught) it shouldn’t be that hard, but if you haven’t taken that yet then it’s gonna be hard
0:02 Probably not relationship advice... maybe.
Wise words
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.
Went with this same, much more intuitive thought as well and came up with 1/((100/p)-1) which is Mathmatically aquivalent to the Ted ed solution. Nice to have it confirmed :)
Taught some students probability this morning and here I am learning something new about it.
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.
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.
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.
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
Agreed. In its defense, the competition overall seems to focus on difficult shots to begin with, otherwise every time the human makes a basket they get to take another shot (without giving the robot a turn at all).
Can you solve...
Me :*clicks immediately*
So true
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
@@EmperorZ19 The 2nd approach is still difficult and still uses math.
@@InsightSplash1the didn’t use to, at least not to this scale
Boss be like:
Eh , she can probably handle it.
She: spends about five minutes WITH the help of ted ed
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 always take copious notes whenever TedEd drops a new riddle just in case I ever find myself in the same situation
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!
Waiting for this day.... We need more such riddles
I Àgreed.
2:09 guys is that freddy fazbear? ar ar ar ar ar-
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 😮
I don’t understand any of this but I just want to hear the solution
Finally another riddle!
I Àgreed.
I wish they'd call these problems something other than riddles, because they're really not
Had exactly this thought, riddles probably don't involve calculating probability and stats and such
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 gods have smiled on us!
I Àgreed.
I'm subscribed EXCLUSIVELY for the riddles
Same here
real
Get online boys. New TedEd riddle just dropped.
I Àgreed.
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.
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.
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.
all you have to do is give the robot green eyes, then it will leave and no one will complain
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.
I found the solution in a different way. The way I looked at it, if the human has a 1/4 chance of scoring, there are 4 alternate universes, 1 where he scores, 3 where he doesn't. Then the robot would need to have a 1/3 chance of scoring, since in that scenario, in 1 universe he scored and in the other 2 it reset, both have scored in the same amount of universes and so the cycle repeats. You can use this logic with every probability, you just need to find the average tries to score, which can be achieved with 1/p, then substract 1 from the result and divide 1/ the result. In other words :q= 1/(1/p -1). I spent a long time on this riddle
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
Technically speaking, if the player wins on the first shot, it’s not as if you’re showing off the robot. So can those instances really count against you? It’s not as if the robot had a chance to show off its “adjusted skill”.
Not to mention this isn’t a very good way to demonstrate the robot. If we assume each person has over 50% odds of winning first shot, the ability to demonstrate for a skilled player is lost. It’d make more sense to go for “best of”, such as best of 3 or 5. While it’d take longer, it’d more effectively show it could match skill with a skilled player, and lower its skill for less skilled players.
I'm sorry, but not only did I not understand a word of the explanation, it also doesn't make any sense. If our goal is to get a 50% win rate between both the Human and Robot, but the Human ALWAYS goes first, then a Human with a 100% probability to hit the shot WILL always win, since the game ends on the first successful shot. And any probability higher than 50% will also still win more than 50% of their shots. So the entire riddle has a massive fundamental flaw.
Yeah.
Glad I'm not the only one to immediately realise this massive flaw in the setup.
If you watched the video they addressed that
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
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%.
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?
A divide-by-zero error occurred and the machine killed the contestant.
oops. This is why you should not import program changes and release them without unit testing.
@@user-vt9bp2ei1w interesting proposal, I think this could work better but we need a way for the contestant to still win 50% of the time even when dead.
If p>0.5, there isn't any solution. If p
@@NaHBrO733 what if p = 0
@@patrickbutsmart0
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.
I did one of these right first try! Finally! This is a high no mortal being should be allowed to experience!
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
So....... I still don‘t understand the riddle
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 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.
Two problems: One: The riddle states you can set q between players not between shots, meaning you can't change q after the first shot. The riddle also never said you could make a complexe q just a number. Problem two: It would still be very sus is the robot only scores first or never. The actual solution will result in the robot sometimes scores his 5th attempt sometimes the 9th and so on this will seem way fairer. If you only want to force the robot winning 50% of the time that would be simple: Simply alternate q from 0% to 100% based on how many wins he has. But it will be very obvious.
I got the solution because of the following example:
Let’s say the probablity of the human making the basketball shot is 1/3. In this hypothetical world, the human is bound to make their shot after they shoot three times. Since they made their shot in 3 shots, the robot shot two times.
Therefore, in the interest of fairness, we have to make the robot have a probability that makes him make a shot in 2 shots, so a 1/2 probability. Meaning, any fractional probability that the human has, (let’s say 1/4) you subtract one from the denominator to get the robot probability.
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).
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
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.
I dedicate two whole hours to understand this riddle just because I love TED ED riddles too much
Video solved the riddle math-ways first, but don’t be turned off; you can definitely solve it logically with a couple intuitive leaps just like their other riddles. Here’s the line of reasoning I used, hope it explains things well for some people.
1. If the human makes it 50%, we need to be at 100%. In other words, they have taken up 50% of the pie, the “win-pie” and that’s all they’re supposed to get, so we need to immediately take that other remaining 50%. We must take ONE HUNDRED PERCENT of the remaining 50% right away. Obviously anything above 50% for the human means we can’t take an equal amount of the win-pie.
2. As we go deeper and miss more shots, the slices we take of the remaining win-pie get smaller and smaller, of course. It thins out and we eventually take infinitesimally small pieces. We must both “arrive” at 50% of the total by the end. You may be tempted to take bigger slices of the win-pie than the human to make up for an apparent advantage they have by going first and maybe ending the game. RESIST that urge. At the end of our turn each round, we must have the same slice as them. If they took 40% of the win pie, we must end our turn having taken 40% of that entire win-pie out of that smaller portion the left for us. Trust that the infinite sequence will mean we will always get to take our slice; the goal is to end with 50% at the END.
3. Now me must make our equation. If they took 40%, we must GET a full 40% out of the 60% they left us. We must take 66.66% of it and get our total 40% of the win-pie. Our basket rate, q, should be 66%, then we will always be getting the same slices as the human, making it 50/50. We must receive their p percentage, 40%, out of the 1-p, 60%, that they left for us. So, our percentage of 1-p must be equal to their p. In other algebraic terms, q(1-p)=p. Our portion of the remaining pie must be equal to their slice. Divide both sides by (1-p) to isolate our target variable. q=p/(1-p) excellent!
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.
Hey, do you think an average person could solve it? What does it take to solve this? Do you need to learn SAT-difficult questions?
There is a certain aesthetic to these riddles, that’s why I always watch these but never be able to solve this 😁
2:26 this should be |r| < 1 or -1 < r < 1 not r < 1.
Oh man I liked working on this one! I recognized pretty quickly that series could come into play-which was troubling because I was terrible at them! I noticed that the second round would be identical to the first and I came up with the correct answer. Imagine my panic when the first method was shown!
I mean, one could simply set p = q and call it a day. It's a semi-formal demo with human participants, limited throws, and a high degree of randomness. How are they gonna tell that your robot is slightly worse than it theoretically 'should' be?
(Heck, you might even be able to get away with just making the robot mimic the result of the human player's throw and then either throw the game or always throw correctly at some specified game length. That might actually appear _more_ fair to viewers of the demo/participants.)
Slight detail, if p=q=1 (or 100%), then the human doesn't always win. He just wins with probability 1. Or in mathematical terms, he wins almost surely.
Finally the first riddle I actually solved by myself
5:00 that is exactly why I did because each tries are restarted if both fails.
When we needed them most the riddles returned
the marketing guy scewing the programmer is the closest any riddle has ever come to reality
What happens when the Robot is paired with Steven Curry?
That match will never end.
this animation is so cool! and the jokes + puns are so actually funny
I like when the narrator said "It's geometric series time" and started talking about geometri serieses
I just figured that if p= 1/2 then q should be 1 more, so q= 1/3. Similarly, for example, if p= 1/37 then q should be q= 1/38 🤷♂️ did this in my head
The term (p / (1-p)) is undefined when p is 100% or 1. How does this avoid the problem we initially had with the human winning 100% of the times because they go first. Can someone explain this?
When p>0.5, there isn't any solution, it is impossible. If 0
Does anyone know a universal surface formula for any n-sided polygon, even with non equal sides?
If you were allowed to watch and adjust remotely I would simply change the probability between 0.0 and 1.0 depending on who won last.
It's great that they're continuing this legendary series 😎
My first guess was to just set the robot's probability the same as my opponent's so they both have an equal chance, but I figured the solution was more complicated than that.
I actually solved this on my own using the non geometric series method.
My reasoning was as follows
p=1/x
q=1/y
1/x is the probability of winning instantly, so 1-1/x games will enter the robots turn.
If the robot has a higher or lower likelyhood of winning in that turn, it would take a larger or smaller cut of the number of theoretical games, therefore it’s cut of the remaining games should be the same. Meaning 1/y * 1-(1/x) should cancel out and equal 1/x, giving both players an equal probability. Therefore the equation we get is
(1-1/x)/y=1/x
Multiply both sides by y
(1-1/x)=y/x
Multiply both sides by x
x-1=y
Sub in p and q (see equations above)
1/p-1=1/q
Rewrite 1 as p/p
1/p-p/p=1/q
Combine like denominators
1-p/p=1/q
Inverse the equation
p/1-p=q
If your brain hurts from this, kingdom hearts would give you an aneurysm
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.
Here's another way of thinking about it, which I believe is more intuitive:
First, imagine that the human has a 50% chance of making the basket on the first try. Obviously, the robot must have a 100% chance of making the basket, to maintain the 50% overall winrate.
Now, let's think about it with regards to the expected number of tries it takes to make a basket.
50% is 1 in 2, so on average it takes 2 tries to make a basket. The robot needs an equal number of tries, but keep in mind that the human has a head start, so we want to subtract the number of tries expected to make the basket by 1. Thus, we get 1 try, which means 100%. Similarly, with a 1 in 3 human, you want a 1 in 2 robot.
The expected number of tries for the human is 1/p, so you want (1/p - 1) for the expected number of tries for the robot, which means q = 1/(1/p - 1). Simple algebraic manipulation (multiply by p/p) shows q = p/(1-p).
To all the folks who allege that the Ted Ed riddles are not riddles, A riddle can be a conundrum where the answer is a pun or some other lateral thinking challenge, or it can simply be any puzzling challenge. Both are very common and accepted definitions of riddles. There's no authority that says a riddle can't involve mathematical or logical thinking.
For my money, I think the Ted Ed riddles are much better and more interesting than conundrums, because all of the information you need to solve them is contained within the video. I've been watching them for years, and I've seen many people in the comments talk about how they were able to get to the solution using a different path of reasoning than the one the video explains, which is evidence that most people are able to get to solutions without much more than logic and basic arithmetic. Conundrums, on the other hand, typically require you to just know a certain specific piece of information or connection between two ideas. And if you don't already know it, then there's basically no way of actually solving it.
But hey, different strokes for different folks.
You set q = 100% after a player scores a point and back to 0% after the robot scores a point. Problem solved.
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
6:11 if a human always goes first and if he is 100% accurate technically there is no way for a 50-50 chance since the human will always win
Yay more riddles! Please make more of these, I just love listening to these when I’m bored or curious about how these work.
Ted-ed riddles need to be more frequent 😭😭
Finally , another riddle , love'em , waiting for next one Ted-ed
Another way to look at this is that if a human, for example, has a 25% chance of scoring, that is 1 in 4. For the game to be fair the robot will need a chance to score of 1 in 3, since the human goes first so the robot always has 1 less chance to win.
Now, 1/p is the average chances needed to score for the human (4 in the previous example), to which we can substract 1 to get the average chances needed to score for the robot: (1/p)-1. Then get the inverse to get the robots probability to score 1/[(1/p)-1], which would've been 33% in the previous example.
Multiply both parts of the fraction by p to simplify and q=p/(1-p).
Understanding geometric series, while fun and interesting, is not necessary for this problem. Simply understand that each competitor should have an equal chance of winning on each try.
For instance, if p=0.25, then out of every 100 games, the human will make the first shot 25 of the games. Of the 75 remaining games, the robot should win 25 on the first try. So set q=0.33.
This works for every p value at or below 50% and also results in the equation p=q(1-p).
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.
My head is spinning from all this math.😵
Glad I'm not only the one who gave up
This feels like a math problem more than a riddle
Hi Ted team. I have a request to have you guy making video about "invisible hand" and "visible hand" in economy. I haven't seen one that explain them well enough and I'd love to see one. Thank a lot!
This solution is wayyyy overcomplicated! Here's how I solved it, more like a riddle and less like a math problem. With pizza! 🍕
Say the human makes 50% of baskets. This means the human will win on the first try, 50% of the time. The robot can't let the human take any more shots, or the human would win >50% of the time. So, the robot must make 100% of shots.
I then realized that, mathematically, this game is exactly like taking turns eating pizza. The whole pizza represents 100% of all outcomes. The human making 50% of shots is like slicing off 1/2 of the pizza-on the first slice, the human will win 50% of the time/claim 50% of the pizza. The robot can't let the human have anymore pizza, so it takes 100% of what's left.
What if the human makes 1/3 of shots? That's like cutting the pizza into three slices and taking one (1/3). The robot needs to take the same amount of pizza as the human, so it should take one of the two remaining slices-1/2. This process is repeated forever - the human takes the leftover third, cuts THAT into thirds, and takes one piece (a.k.a. makes 1/3 of baskets); the robot also takes one piece of the two remaining (a.k.a. makes 1/2 of baskets). The robot copies the human this way to infinity, hence the phrase "infinite series, finite sum"-the slicing goes on forever, but the pizza has a set area.
The same works with 1/4. The human cuts the pizza into 1/4s and takes one, so the robot needs to take 1 of the 3 remaining slices, or 1/3.
The same works with 1/33. The human cuts the pizza into 1/33s and takes one, so the robot needs to take 1 of the 32 remaining slices, or 1/32.
So, to get the robot's shooting percentage (q), you just have to take the human's shooting percentage (p) and subtract one from the denominator:
If p = 1/2, q = 1/1
If p = 1/4, q = 1/3
If p = 1/33, q = 1/32
It's that easy!
Surprisingly, this is kinda hard to express with algebra. What formula can you use to turn an input (p) of 1/4 into an output (q) of 1/3?
First you have to flip the fraction: 1/4 -> 4
Then subtract 1: 4 - 1 = 3
Then flip the fraction back: 3 -> 1/3
So, q = ___1___
(1/p) - 1
Take p, flip it upside down, subtract 1, and flip it again, and you get q.
To simplify, multiply that mess by multiplying by __p__ and you get q = __ p__ , the solution in the video. There are probably even easier ways, but this one makes sense to me visually!
p 1 - p
oh boy a new riddle
Yes! More riddles, please!