The hardest test had this "easy" question with a surprising answer

[@ThicciJo]

I just spent about 25 minutes plugging values into a calculator to prove this wrong...huge waste of 25 minutes. It's somehow exactly right.

[@mike1024.]

Well, the algebra says so. :-) There's something special about percentages that can be written like 3/4, 4/5, etc.

[@SavageSalvageSmithing]

I plugged this into chat gpt. It's answer: "No, if someone's accuracy is below 75% and later improves to above 75%, they have not necessarily shot exactly 75% during any specific instance. Their accuracy rate could have varied between below 75% and above 75% over different attempts or situations."

[@SavageSalvageSmithing]

Chat gpt's proof: Let's consider a scenario where someone's accuracy fluctuates between below 75% and above 75% over multiple attempts. For simplicity, let's say they have attempted 8 shots. Here's an example of how their accuracy could change: 1. First 4 shots: 2 hits out of 4 attempts, accuracy = 50%. 2. Next 2 shots: 2 hits out of 2 attempts, accuracy = 100%. 3. Last 2 shots: 2 hits out of 2 attempts, accuracy = 100%. In this scenario, the person's overall accuracy would be: (2 (hits) + 2 (hits) + 2 (hits)) / (4 (attempts) + 2 (attempts) + 2 (attempts)) = 6 / 8 = 75%. However, it's important to note that during any specific set of attempts, their accuracy was never exactly 75%. It was either below or above 75% in each instance.

[@mike1024.]

​@@SavageSalvageSmithing Well, I guess that goes to show chatGPT cannot be trusted to do math haha.

[@SavageSalvageSmithing]

@mike11986 I wouldn't trust chat gpt with my socks, math is a new level of trust.

[@secondengineer9814]

Here's another way to think about it. You have two piles, one of missed throws, one of made ones. The pile of made throws is less than 3x the size of the missed ones. You can continuously add throws to one or the other pile, with the goal of passing 75%. Because you only add one at a time, you would have to reach 3x the size of the missed throws in order to pass it, and adding to the missed throws only makes the goal higher and further away

[@alemfi]

This is definitely a very intuitive way to present the problem!

[@b3njamMMVII]

this helps to picture the problem a ton!

[@ForcefulDragon]

This is how I pictured it as well. You will eventually hit 75.00% because there will always be a point where the made pile is exactly 3x the miss pile. And you will always hit 80.00% because there will always be a point where the made pile is 4x the miss pile. And you will always hit 95% because there will be a point where the made pile is 19x the miss pile.

[@sundareshvenugopal6575]

Yes, but it is more unlikely than likely that the last free throw he either scores or misses in a match would make his average exactly 75%, at the end of the match, since his scoring rate is fairly consistent at 75% or thereabouts. So the chances of him ending a match with an average of exactly 75% is very unlikely. Again since at the end of each free throw both his throw tally, or his score tally can vary, so his average at the end of any throw being exactly 75%, itself is theoretically and practically very unlikely. Initially, we can assume for every four throws the number of throws he makes would be more likely, (3, 2, 1, 4), which means out of every 4 throws, it is least likely he makes all 4 throws, slightly more likely he makes only 1 throw, and more likely he makes between 2 and 3 throws with 3 throws being the most likely. At the end of the season it is probably (4, 3, 2, 1 ) or (4,3,1, 2), if he has significantly improved his scoring, during off season. No one who does a bit of math will ever be involved in betting or gambling, or playing games of chance, since most games are highly rigged against the player, so the player loses money heavily apart from being addicted. It is not a wise thing to do. The same applies to online trading, as well, probably even worse. The market crashes and the poor investors are left high and dry but the fund managers are plush with funds.

[@NoLongerBreathedIn]

@@sundareshvenugopal6575Except poker - the house just takes a cut there.

[@JalebJay]

This was one of the problems our professor liked to give us when practicing for the Putnam every year in our problem solving group.

[@commanderkuplar3790]

Jaleb okay buddy

[@yourlordandsaviouryeesusbe2998]

@@commanderkuplar3790 what is that supposed to mean? Are you insinuating something?

[@commanderkuplar3790]

My Lord, I completely agree with you Lol this was in no way to provoke anything or anyone.

[@yourlordandsaviouryeesusbe2998]

@@commanderkuplar3790 alright. It sounded a bit like you were doubting the veracity of his comment. But anyways, my bad. You have a good day.

[@tgrizz213t]

My Lord, I completely agree with you, Please answer my prayer

[@FloraOfTheCats]

he just sounds so happy to answer a good math question

[@Braden1005]

Watch this: he hasn’t made any free throws (0/0) and then makes one (1/1). It goes from 0% to 100%

[@AAAAAA-gj2di]

@@Braden1005 🤣🤣🤣

[@AAAAAA-gj2di]

@@Braden1005 0/0 is indeterminate, better you take some mathematics classes

[@The_Untitled_Py]

It wont be 0 over 0 tho, ite the number of successful free throws over the total number of attempts on free throws so it will be 0 over 1 and then next turn it will be 1 over 2 attempts..

[@daniele_93]

Wow, this was really unexpected

[@skig11]

@insert username 606th like

[@mikekaleiwahea7337]

I remember hearing this problem a decade ago and decided to try to work it out while on my daily walk. But the version I heard used 80% as the FT percentage. It took me the whole hour on the walk to figure it out but what I enjoyed about the problem is its counter-intuitiveness. Yes, at first I thought the answer must be no, there must be a way to "hop over" 80% so I went searching for a specific counter-example. Of course, no matter what fraction I started with, increasing both numerator and denominator incrementally by one always ended up with 4/5. Having wasted 15 minutes already, I convinced myself ok, the answer must be yes, the FT percentage must have to equal 80% exactly at some point. Using the same algebra inequalities, I reached the same proof you describe in the video along with the general proof wrt k/k+1. But it also made me wonder what happens if you flip the circumstances and say he had a FT % greater than 75% at the start of the season and less than 75%. Would it then necessarily equal 75% at some point? Applying the same reasoning and algebra you find the answer is now NO and the "must-equal" fractions now are of the form 1/k.

[@Ald3r_]

for the flipped circumstance it's also incredibly easy to provide an example. If you assume he scored his first shot, and then never scored again, he would never hit 75%

[@danquaylesitsspeltpotatoe8307]

NO next easy question!

[@RishabhSharma10225]

That was very interesting. Thanks for sharing it.

[@danquaylesitsspeltpotatoe8307]

MMHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH 73/75 74/76 75/77 Whatever */100 Dont see no 75% there!

[@nivyan]

The question is bad: "Was there **necessarily** a point during the season where his free throw percentage exactly equalled 75%?" Here's a season here he never hits 75%: 0 throws thrown = 0% hits 1 throw, which scores = 100% hits Retires, season over. Never hit 75%

[@SorenHigh]

I thought it was too easy then I realized my logic was wrong. Videos like these always help me think out of the box and learn to breakdown a problem into smaller pieces first.

[@BaneWilliams]

I kind of don't agree with it though. The problem never states that DeAndre ever threw a free throw. S could therefore = 0, and n could therefore also = 0. At that point, if DeAndre was successful every time since, DeAndre would go from 0% success rate (which is below 75%) to 100% success rate (which is above 75%) and never touch 75% itself. There are even times where DeAndre can be unsuccessful, as long as the percentage never dips to 75% or below.

[@FourthDerivative]

Now THAT'S an awesome and counter-intuitive result! Thanks for presenting this problem.

[@omatbaydaui722]

the reason it's counter intuitive is because if it was the opposite, means , that the basket ball player was above 75% at the start of the season and below 75% at the end of the season, then the answer to the question would have been "no", because there could be a point when it will never be equal to 75% in that case. But the opposite is not true, if you start below 75% and grind above 75% then it seems that there will be a moment it will be equal. So in our reasoning we use the first case scenario to intuitively conclude the second case scenario, so it's nor "really" counter intuitive, it's more like we are mislead by intuition because we do not take every case scenario into consideration, so we mistakenly use the conclusion for the first scenario to conclude the later

[@redwings13400]

This actually does work both ways, if a player starts at 80% and goes down to 60%, a player will have to be at 75% at some point. The reason for this is that 75% is at a specific point where the amount of shots made is a multiple of the amount of shots missed. Instead of 75%, let's ask this question about 50%. Can you go from below half of shots made to over half of shots made in a single shot? Nope! This one is more intuitive. If I've taken an even amount of shots, my percentage will be at, below, or above 50%. If I've taken an odd number of shots, my percentage will be above or below 50%, and my next shot will make me take an even number of shots. If I'm 300/601, and I make my next shot, I'm at 50%. If I miss it, I can't get back to 50% without making at least another 2 shots. Another way of saying this is that shots made and shots missed are either equal or unequal, and if you only add one at a time to either category, they will equal each other before changing the lead. For 75%, imagine this made/missed category. I reach 75% when the amount of make shots is triple the amount of missed shots. I've currently made 10 shots and missed 4, putting me below 75%. To reach 75%, I need to increase the amount of made shots to triple the amount of missed shots. To reach above 75%, I need to get more than triple the made shots to missed shots. If I've missed 4 shots, a 75% rate will be making 12 shots. If I miss another shot, I will need to hit 15 shots to hit 75%. To get above 75% without hitting 75% Would imply that I somehow make more than 12 shots without hitting the 12 mark first. This is true in any percentage where the number of shots made is directly proportional to the number of shots missed. Two shots made for one shot missed is 66%, so that's another unstoppable point. 80% is another. 5/6 is another. 6/7 is another. And so on.

[@redwings13400]

Wait, oops, I'm wrong, it actually doesn't work both ways. Weird. Sorry, just looked at it and realised that.

[@omatbaydaui722]

@@redwings13400 yeah, I fell for it the first time too. that's why it looks counter intuitive. Weirdly enough, missing shoots doesn't give the same % as gaining them. This most likely has to do with the fact that when you'r missing a shoot, your success shoot count (let's say it's "m") goes along with the total number of shoots (let's say "n"), so at every success, the % goes up by m+1/n+1, whereas when you miss a shoot, it goes down with an m/n+1.

[@redwings13400]

Yeah, basically, if we look at made shots to missed shots, and we start at, say, 50%, we have 5 made shots and 5 missed shots. If we double the amount of missed shots, we get 5 made shots and 10 missed shots, or 33% made shots. 5 made shots and 15 missed shots, or 25%. If we make another shot, we have to go to 6/18 or 7/21 before we can go down further. So, if you start above a certain amount and go down, you'll experience the same phenomena of always crossing certain points, but at different points. 1/3, 1/4, 1/5, and 1/6, etc. So if we start at 35% free throw percentage, and then go down to 17%%, we will inevitably hit 33%, 25%, and 20% along the way. If we go back up to 50% again, we will not necessarily hit those numbers again.

[@ian5804]

Okay, okay, this is a nice problem, but we all know that DeAndre hits all of those shots, no 75% bs.

[@Father2462]

Did I find you on KZbin as well

[@AaronHe]

If you're curious the original problem had the percentage at 80 and the basketball star is Shanille O'Keal.

[@AndrewBlechinger]

For those wondering, the answer is still yes because 80% = 4/5 and the general form of the proof outlined in the last part of the video applies.

[@AAAAAA-gj2di]

@@AndrewBlechinger we know right

[@agfd5659]

Did you mean O'Teal?

[@MegaKillMeister]

@@matteoooo4791 Nope it says Shanille O'Keal in the problem

[@AaronHe]

@@agfd5659 I did not.

[@musanim]

Here's how I would answer it (without proof by contradiction). A success rate is 75% means there are 3 times as many successful (S) free throws as unsuccessful (U) ones: S=3U If the success rate is less than 75%, then S3U. For each successful free throw, S increases by 1 and U stays the same. S is an integer. 3U is an integer. S must equal 3U at some point (on its path from S3U).

[@DrHades]

I solved the problem, but I was shocked as my intuition was completely off. One of your best videos of all time.

[@cubemaster1987]

I took the Putnam exam and got one problem right for 10 points back around 2009. The problems are very difficult, but always have solutions that are easy to follow. Usually the problems can be solved without the need for high end mathematics

[@SMA2343]

I honestly think it has more to do with little common sense rather than high level math. You over think it, and thus make it so convoluted that you get it wrong

[@jaggeriscoughmedicine]

@@SMA2343 yeah . Probably it needs more of algorithm knowledge

[@harrisonsloan6036]

κων/νοσ φωτ that question you definitely need to know some algorithm and basic number theory

[@gamma-bv6ty]

@@jaggeriscoughmedicine It's less of an algorithm knowledge, than just general problem solving experience. The problems pretty much all have some "trick" to them, that is, they are all easy with the right approach. You just have to experiment and try different things until you find what that approach. Algorithm knowledge won't really help as there's generall no obvious algorithm to use. Though thinking algorithmically is a similar skill to being able to write good proofs which is definitely a requirement.

[@gamma-bv6ty]

​@@SMA2343 Most of the problems do need at least at some higher level math. They set it up so the "easiest" problem can be solved with just highschool knowledge, but some of the harder ones can require things like abstract algebra or real analysis.

[@bradzhang2195]

I think it's easier to make sense of the result if we think about it in terms of shots missed. The only way to decrease the missed free throw percentage is to increase the denominator without increasing the numerator (e.g. 2/7 missed -> 2/8 missed). If the missed free throw percentage goes from above 1/4 to below 1/4, at some point the denominator has to be exactly four times the numerator since the denominator only increases by 1 every free throw.

[@ssrini2002]

@@augustelalande5144 he clearly said shots missed, not shots taken - so it's like 26/100, 26/101, 26/102, 26/103, 26/104, 26/105 and 26/104 is the point where he misses 25% of his shots

[@bradzhang2195]

​@@augustelalande5144 Of course, as you have shown, it is possible to go from above 3/4 free throw accuracy to below 3/4 free throw accuracy without having exactly 75% free throw accuracy. However, we are looking at the complement here: going from *below* 3/4 to *above* 3/4 free throw accuracy (problem text) is the same as going from *above* 1/4 free throw misses to *below* 1/4 free throw misses. I claim it is impossible to go from above 1/4 to below 1/4 free throw misses without having exactly 1/4 free throw misses. 26/100 missed = 0.2600 missed = 0.7400 accurate, 26/101 missed = 0.2574 missed = 0.7426 accurate, 26/102 missed = 0.2549 missed = 0.7451 accurate, 26/103 missed =0.2524 missed = 0.7476 accurate, 26/104 missed = 0.2500 missed = 0.7500 accurate. Why is it impossible to go from above 1/4 to below 1/4 free throw accuracy without passing through 1/4 free throw accuracy, but not from above 3/4 to below 3/4? This stems from the fact that that for any integer numerator _n_ , 1/4 can be expressed as _n/(4n)_ , where the denominator is guaranteed to be an integer. As we increase the denominator, the denominator must take the value of _4n_ in order to decrease the free throw miss percentage below 1/4 (increase free throw accuracy above 3/4). This is not true for 3/4: for an integer numerator _n_ , 3/4 can be expressed as _n/(4n/3)_ . Clearly, if the numerator _n_ is not divisible by 3, the denominator is not an integer and cannot be taken as a value for the total number of free throws, which is demonstrated in the example you have shown.

[@akaRicoSanchez]

Yeah I solved it the same way. Let me try to rephrase it to see if it helps. You can take the original question and reformulate it: DeAndre missed more than 25% of his free throws, he got better and better and at some point he missed less than 25% of his free throws, is it possible for him to never had exactly 25% of missed free throws? Everytime DeAndre will makes a free throw his total attempts increase by one and his total of missed shots stay the same. Obviously you can't attempt half a freethrow nor miss half a free throw: we are dealing in whole numbers here. Even more obviously, all whole numbers can be multiplied by 4 to get another whole number. So there is no scenario where DeAndre shoots a free throw, makes it and his total attempts goes from just under 4 times his missed shots to just over 4 times his missed shots: 4 times his missed shots is a whole number and DeAndre shoots his free throws one after the other so his total attempts go through all the whole numbers without skipping any: he must have had 25% missed shots, so, to answer the original question, he must have had 75% made shots.

[@ecatuccio]

@@akaRicoSanchez ty for explaining this proof clearly.

[@algebranograzie1396]

Thanks. This is much more intuitive.

[@kiarashseiedi7255]

the way you solve it was remarkable. Good job!

[@carlosd.zapata7306]

Nice problem. The part of "at some point during the season" misled me because initially I thought that the averages were calculated post game, case in which the exact 75% is not necessarily achieved. But by calculating them after every free throw you can't avoid the +1.

[@snared_]

can't avoid (k-1)/k, integer k>1, if starting below and eventually surpassing it.

[@saikat93ify]

This is one of those facts that is so unexpected that I don't understand it even after understanding it.

[@elchingon12346]

It's one of those math problems where the intuition for understanding why something is true is straight up more involved than the math you need to prove it. Presh makes a point about consecutive integers that he doesn't actually go into too deeply over, but is a major reason why the proof works the way it does, and why the intuition makes sense. I'll give it a shot if you want a better explanation of it.

[@Midnight_Rider96]

@@elchingon12346 I understand it reasonably well now after writing out the fractions and trying it with different percentages. The key is consecutive integer fractions, since if you try to creep up on the value you will always hit it if you try to go over, since the graduations in percentage shrink as the denominator grows. And because the denominator is consecutive to the numerator it can't be reached until you get to the next factor value of the denominator, which would cause you to hit it. Very counter intuitive but once you see it it makes sense.

[@MrBrain4]

This only took me longer because it says the result is "unexpected". But my initial answer was "yes", and I spent a while trying to figure out how I might be missing something. I simply made a graph y=3x, where y are the makes, and x are the misses. Since there is an integer value y for each x, it's clear that if you are below the line (below 75%), and you start making free throws, you can't get above the line without landing on the integer value y=3x, meaning you have exactly 75%.

[@ryannoonan5518]

That is a nice way to think about it!

[@cooperhichkad444]

@@noomade 0,0 isn't a valid point due to the constraints in the original problem, the shot percentage in terms of the y=3x scenario would be y/(x+y), and a 0,0 you're dividing by 0. It's no different than trying to refute the solution by using negative numbers, it's not within the constraints of the original problem.

[@templarknight7]

@@noomade guess you've just proved why the original comment here would not get full marks on this problem.

[@cadewatkin7086]

Me looking at thumbnail: Lol obviously the answer no, this must be an ez test lol. *7 min 22 sec later* Me: 😮

[@ihatetobethatguybut7175]

I still can't understand it even when he is explaining

[@jkxss]

Bamboozled.

[@Fildoggy]

Cade Watkin wait whaaaat

[@Anime_Edits_TN]

Amazing feeling bro, same here....

[@Daniel-q9f8p]

As Michael used to say, "You miss 100% of the shots you don't take." This means that we can define the free throw accuracy of not taking any shots as 0%. With this definition DeAndre's free throw percentage can bypass 75% by hitting 1/1 shots by the end of the season.

[@jacketofdiamonds]

0/0 = NaN% which is not less than 75%.

[@Daniel-q9f8p]

We define shot accuracy as a piecewise function f(x,y) = z on the non-negative integers where x

[@Batatero1]

Using some logic, you can notice that it's impossible to reach 100% accuracy with a miss, which means that, if you hit 1/1 shots, it wouldn't be possible to be 100% if 0 was considered a miss. This way, 0 shots taken have to result in 0 misses and therefore 100% accuracy.

[@Mythraen]

​​​@@Batatero1As the other person said, 0/0 is not a number. It isn't 100% and it isn't 0%, because those _are_ numbers.

[@bgugi]

Ahh yes, the classic "six rings" proof.

[@Outwardpd]

The last thing I needed was to be reminded of the hell that is proofs

[@sawyerwest3990]

The ceiling of pi is equal to 4 implies pi is about equal to 4. Bam, something worse than proofs.

[@JonSmith-hk1bq]

This is one of those cases where I can easily answer the question in my head, but would struggle to explain it mathematically.

[@catburner1896]

2:18 from what I can tell, it depends whether you consider the function continuous or discrete, because necessarily if the function is discrete, the function has to have equalled exactly 75% due to the intermediate value theorem.

[@RomanBelisarius]

If the function is continuous, then intermediate value theorem is applicable, you mean?

[@Languste]

Damn, this is one of the most surprising results of any riddle videos I watched; such a tough, interesting and easy to state question! Love it :3 By trying out I came to the conclusion that the answer should be "yes", but I definitely would have not been able to proof it.

[@perc-ai]

indeed we need to do these putnam tests!!

[@Languste]

@@perc-ai I am German, so I am unable to participate :b Though I could do it when I am in the US for half a year. But if we are honest... I would probably solve none of those questions and waste 6 hours 😂

[@abdelfattahkrita7874]

OMG, I was thinking the answer is no because the S(N) is not continuous and you can't apply the intermediate value theorem. Thank you

[@sahilbabbar8859]

Yeah I was thinking the same, absolutely stumped by this amazing problem tbh

[@isaacjg8]

I solved this by imagining the difference between the current number of scores and the number of scores needed to make 75%. In formal terms, this would be N*0.75 - S(N). Every time you miss, N goes up by 1, so N*0.75 goes up by 0.75. But S(N) doesn't change, so the difference between the two increases by 0.75. On the other hand, every time you score, N*0.75 still goes up by 0.75, but S(N) increases by one. So the difference between the two decreases by 0.25. Going from below 75% to above 75% would mean, in this context, making the difference go from positive to negative. If you can only add 0.75 and subtract 0.25, there's no way to cross over without hitting 0 at some point. You would need to achieve a difference that isn't a multiple of 0.25, which isn't possible using only these two options. Using this method to generalize the problem tells us that any percentage P will be unavoidable if it is evenly divisible by 1-P. Hope that makes sense. Very fun problem, thank you for sharing it!

[@LJCana]

At first, I also thought that it was just going to jump from below 75% to above it without ever being 75% exactly. But then I decided to give it some thought. I just decided to express 75% accuracy into a ratio of successful to failed attempts, so it would be 3:1. According to the ratio, DeAndre just needed to make 3 throws per 1 fail. For him to go above 75%, he will eventually have to reach a 3:1 ratio and then go make more shots. There's no way 3:1 can be skipped, therefore at some point his free throw percentage was exactly 75%.

[@bscutajar]

I don't really follow your logic. To say there's no way a 3:1 ratio can be skipped is just restating the original problem in a different way

[@olixx1213]

​@@bscutajarit's really obvious to show that this ratio can't be skipped Before you have to skip it , you need a certain amount of failed throws k, and a number of successful throws n, just before skipping it This means that n < 3k

[@Antoine.Comeau.Racing]

That’s a much better solution

[@DimkaTsv]

@@bscutajar As he can only increment number of shots by 1 each time, he won't be able to skip. But if incrementation would've been by other numbers... Like total throws at end of game and not for each through. Then achieving 75% exactly would've been different story as you could've skipped it.

[@jrodartec]

Brilliant proofs and explanations by Presh and also by a lot of his viewers down in the comments. This problem strucked me as truly counter-intuitive, and for me counter-intuitive truths are something not only interesting but also useful. Thank you all (Presh and viewers) for the time invested in making these explanations as crisp and clear as possible.

[@varshamedhekar8012]

Counter-intuitive problem. I thought it would obviously jump from below 75 to above it

[@abdulbuc7209]

At first, I was sure that the answer is NO but then I started thinking and tried to give a counter-example and realised how wrong I was! This is actually a great question that opened my eyes a bit and made me wonder about other percentages. The "wow" effect occured.

[@colonelb]

Interestingly enough, I think the answer is "no" if coming DOWN instead of going up. A player makes the first four shots (100%) then misses the fifth (80%), and then misses the sixth (66%) and skips over 75%. It's interesting how increasing and decreasing have such different calculations

[@stevenfallinge7149]

When counting down, you can't miss the 1/n fractions. Because you need to have n-1 misses (an integer number) for every success.

[@howlinmad03]

I don’t think you can get below 25% without hitting it exactly though…

[@zanti4132]

Yes, the arguments mirror each other. You have to hit exactly the (n-1)/n fractions when going up, but not necessarily when going down. You have to hit exactly the 1/n fractions when going down, but not necessarily when going up. The only exception is 1/2, but that is because it is both a (n-1)/n fraction and a 1/n fraction!

[@jkvoot]

I thought of a different solution. You can only get exactly 75 % every fourth time you have thrown the ball, bacause you can only divide every fourth number in four and get a whole number. This also means that for every four number, the divident will increas by three in order to get 75 %. If your divident is one value under the needed value to get 75%, you have to gain five values till the next fourth time, in order to get over 75%, but the most value you can get is four. That means that if he ended up with over 75 % he had to be at the 75% point at one point.

[@BaneWilliams]

Except, if his prior rate was 0% because he had never thrown a free throw (thereby having 0%), then he would go to 100% if his first one was a hit. At that point it is trivial to keep above 75% and never hit it.

[@jkvoot]

@BaneWilliams Wrong, his precentage is not defined if he havent thrown yet.

[@Tehom1]

Yes, even though the function is not continuous. Let a/b be the first free-throw percentage that equals or exceeds 75%, in lowest integer terms. Since it just increased and can only increase by 1 throw at a time, his previous ratio was (a-1)/(b-1). So: a * 4 >= b * 3 (a-1) * 4 < (b-1) * 3 a*4 - 4 < b*3 - 3 a*4 - 1 < b*3 This narrows the bounds, and since we said a and b were integers, we have the exact solution: a*4 = b*3. Therefore the first free-throw percentage that "equals or exceeds 75%" is exactly 75%. Therefore we always exactly hit 75%. Similarly, we always exactly hit 50%, 66 2/3%, and every other percentage of the form (n-1)/n if we cross it.

[@sagittarius5466]

It’s logical, lets suppose it was 50% instead of 75. In this case it’s logical it was exactly 50% at some point. So then it also has to be true for 75% (and 87.5%, 93.75% etc.)

[@kirilgeorgiev751]

Need to comment... I got caught off guard by that one. Good one! My brain still can't process it.

[@Bokery1000]

Thanks for publishing such an intriguing clip. Your channel is fantastic.

[@alexzhlebinkov3966]

This really reminded me about the "squeeze" theorem that we just went over in Calc 1 class for some reason.

[@theegyptiangod123]

Alex Zh I vaguely remember that. I haven’t watched the video yet, but I would expect him to use IVT

[@Markymark10]

I thought about that too but the IV/"squeeze" theorem requires the function to be smooth and continuous, which as mentioned in this video, free throw percentages are not.

[@TheMrFrontera]

i remember that garbage. i remember my teacher teaching us about the lemma, and i was like no way she would put that on the test. and it was on there

[@dimitristripakis7364]

PROOF BY CONTRADICTION (REDUCTION AD ABSURDUM): Assume the answer is "NO". This means that he has "skipped" the exact 0.75 percentage. Now if his shots are X/Y (X =successful, Y = total), then at some point there was a pair X, Y where both these were true: X/Y < 3/4 (this is the shot before the 0.75) AND 3/4 < (X+1)/(Y+1) (this is the next shot which passed the 0.75) These two yield: 4X < 3Y AND 3Y+3 < 4X+4 Solving these two for 3Y, yields: 4X < 3Y AND 3Y < 4X + 1 Or together: 4X < 3Y < 4X+1 Now if X is an integer and Y is an integer, then 4X is an integer and 3Y is an integer. But it is not possible to have an integer inside non inclusive (4X, 4X+1) (which is two consecutive integer numbers). Since the initial assumption leads to an impossibility, it means the assumption can not hold true. Final answer = We can not avoid exactly 0.75. Amazing! (I am not a mathematician, tell me if this is correct or not. I believe it is. A big thanks to the author, this was fun!)

[@thedank0r162]

Mr. Talwalkar could you please put your videos on dark mode. It makes it easier on the eyes and especially when watching at night. Thank you hopefully!

[@channel_B5]

Nah, math problems are typically presented and solved on white paper with black ink/pencil...

[@supercoolguygamer9239]

TheDank0r its actually helpful it'll prevent you from falling asleep while solving😝😝

[@ObsidianParis]

@@channel_B5 Never used a chalk and a black board ? :-)

[@channel_B5]

@@ObsidianParis true... :)

[@ousimanie1869]

@@channel_B5 Ok, so?

[@mike1024.]

Wow, I truly expected we'd find some sort of counterexample, but the algebra definitely works. If the denominator is one higher than the numerator, you end up with that + 1 after distributing and combining the constants. I've seen this integer style contradiction used in other proofs too. Proving a quantity is strictly between 0 and 1 yet is an integer is used in proving several irrational and transcendental proofs.

[@jkid1134]

I kept trying to think of counter-examples and I couldn’t come up with any. I was thinking of rational numbers just over 75%, so 4/5, 6/7, 7/9, etc. and I realized that, as I was making my denominators larger and larger, the change to the percent per shot became smaller and smaller, making it less and less likely that I would eventually find one. Since there wasn’t one at the smallest numbers of shots, and since I was making it less and less likely that I would find one by increasing, I started to think there wouldn’t be any counter-examples, but it’s still a shock.

[@elchingon12346]

I usually listen to your videos when I'm going from point A to point B Presh, and I was so sure that you could skip the value like you said, but now that I saw the proof I wish I would have worked it out, I think I would have gotten close! 😤

[@NoahWillCrow]

It is important to note that this is only true if n >= 1. For instance, DJ could have 0 free throws made because he has shot 0. But then DJ could shoot 80% the rest of the season without ever going back below 75%. This is technically shown in the video since one of the ratios is S(n)/n, which is undefined if n = 0, but the actual question did not specify this.

[@tarotanaka9984]

0/0 isn't "less than 75%" so no need to specify.

[@MisterAssasine]

@@tarotanaka9984 0/0 is undefined

[@tarotanaka9984]

@@MisterAssasine I don't get what your point is lol. Noah said you need to specify N > 0 for this statement to be true. But "free throw percentage below 75%" already implies N > 0, since like you said 0/0 is undefined and is not "below 75%". In the more specific language, the condition is S(N) < 0.75N so you can clearly see N must be greater than 0 (since S(0) = 0 and 0 < 0.75*0 is false). So a jump from 0/0 to 1/1 does not violate the statement and there is no need to specify N > 0.

[@shrankai7285]

Quickly thought yes. If you have a certain amount of misses, then there would be a time where your amount of misses is one fourth of your attempts. For example if you are 5/12, you have 7 misses. 7 misses is one fourth of 28, and you would have to pass that at some point if you don’t miss. If you miss more, then the shot number that is 4x your misses changes as well

[@cp3190]

If you can't skip over a percentage in the smallest values you cannot skip over it at higher values.

[@kiyokiyoko]

lol thats not true at all

[@cp3190]

@@kiyokiyoko Okay give me an example. Always up to learn something new.

[@peterlopinto]

@@cp3190 I think what you're saying is if you miss the first shot, then never again miss a shot, each percentage thereafter is unskippable, which was proven in the video to be true. In this case, each fraction you're creating has a numerator one less than the denominator, which is the proof essentially. (As in after 0/1: 1/2 50%; 2/3 66.6%; 3/4 75%; 4/5 80%; and so on. Each percentage is unskippable.)

[@zserf]

How I thought of it: Another way to think about this. Let your score be your number of successes minus 3 times your misses. So, if you've made 4 and missed 2, your score would be 4 - 3*2 = -2. If your score is negative, you have a FT% < 75%. If it's positive, your FT% > 75% and at 0 it's exactly 75% (since you have 3 times as many successes as misses). When you miss a shot, your score decreases by 3, but when you make a shot it only increases by 1. So, if you're at -2, a made basket would bring your score to -1, but a miss would bring it to -5. Because making a shot only increases your score by 1, you can't skip 0 when going from a negative to a positive and you have to hit a score of exactly 0, and therefore 75%. This also shows that the problem doesn't work in reverse. If you've made 4 out of 5 and are at 80%, you can miss your next shot and drop to 4 out of 6 and 67% without hitting 75%. Your values in this case go from 4 - 3*1 = 1 to 4 - 3*2 = -2. Because you drop by 3 when you miss, you can skip over 75% when missing shots, but not when making them. The opposite phenomenon applies at 25%, where you can skip it on the way up, but not on the way down.

[@ruinedmirage9722]

There is one scenario, though. If a player never took any free throw shots in their career, their percentage is 0%. Then they get their first free throw attempt and never miss any shots. It jumps to 100% This was the first thought that came to my mind, since the question was vague enough to allow this situation (starting below 75%)

[@aliaksandrprus8473]

0 / 0 == 0 ? really?

[@RexxSchneider]

@@aliaksandrprus8473 Sure. Zero is as good a value for 0/0 as any other. What do you think their percentage success rate was before they took their first attempt?

[@Johngram1000]

By that reason, might as well call it 100% before the first attempt, which neutralises the supposed counterexample. "Undefined" is better.

[@keith6706]

@@RexxSchneider Doesn't work. The percentage is defined as what fraction of the number of free throw attempts were successful. If there were no attempts made, by definition there can be no percentage, not 0%. Those are two different things.

[@RexxSchneider]

@@keith6706 You claim: "If there were no attempts made, by definition there can be no percentage, not 0%" No. By _your definition,_ the fraction is 0/0. That does exist, but has an indeterminate value. That's not the same thing as non-existent.

[@idk_6385]

Thanks for making math fun again Presh

[@tfofurn]

75% = ¾. You can only hit exactly 75% on throws that are a multiple of four. We can consider any set of four shots: n MOD 4 will cycle through 0, 1, 2, and 3. If DeAndre scores on every 1, 2, and 3, but misses on 0, he can avoid 75%. Each set of four has a local percentage of 75. Even if he managed that miss-hit-hit-hit cycle perfectly for a large number of sets, he can't exceed an average of 75% no matter how long he tries! To exceed 75% would require making all four shots in at least one set, and that would mean hitting 75% exactly when n MOD 4 is 0 in the set that crossed the threshold, right?

[@Lillarion]

Much easier solution: Fact: 75 % is 3/4, that means 3 times as many successes as the number of misses in this case. Suppose he have M misses and G successes before the switch from S(n) being below 75 % of n. We thus know that G

[@pigbumpy]

Of course it had to be a proof by contradiction and give me flashbacks to discrete mathematics

[@sagov9]

Another way to think about this maybe makes it easier to understand why this is. The fractions at which this true 2/3, 3/4 etc indicate that the number of successes is an integer times the amount of misses. The number of misses is always an integer. As your number of successes can never increase by more than one at a time, this means you can never skip a multiple of the number of misses.

[@blakdeth]

What if at the beginning he never made any shot and his percentage was 0, then he made only one shot and his percentage was 100?

[@MichaelDarrow-tr1mn]

his percentage wasn't 0, it was indeterminate.

[@blakdeth]

@@MichaelDarrow-tr1mn you miss 100% of the shots you don't take

[@MichaelDarrow-tr1mn]

@@blakdeth that means after his percentage should also be 0 because he only took one of the shots and he still has all the shots he didn't take

[@blakdeth]

@@MichaelDarrow-tr1mn dang it

[@anuragramachandran5169]

That was actually a great problem and I would never think about it like that man. Amazing job Presh.

[@AAAAAA-gj2di]

That's why I had subscribed to this channel. But from the last few months, he was just bwating ariund the bush. This was not too tough but this video is awesome. Now see almost everyone is praising Presh in the comments section. This is what we want.

[@YogeshPersonalChannel]

I loved the subtle QED you place at the end after solving. (The black square)

[@AndrewTyberg]

What is QED? I was wondering what the black square was.

[@Theraot]

@@AndrewTyberg It means that the proof is complete. QED stands for "quod erat demonstrandum" (thanks Wikipedia) meaning roughly "which was demonstrated".

[@JesseLH88]

I solved it a different way. let x/y = S(n) < 3/4 is there a p such that (x+p)/(y+p) = 3/4? solve for p: p = 3y-4x Since x and y are integers, p is an integer. And since x/y < 3/4, p is also positive. The Talwalker solution is more elegant, though.

[@toast_recon]

Sorry, I don't see how this works at all. If I understand correctly, you're saying that he makes p free throws in a row without missing any in order to reach 3/4 free throws. Proving that it's possible to reach 3/4 with some set of free throws is a very far cry from showing that you *must* reach it with every set of free throws, so this is probably a 0 or 1 point answer on the test. I didn't solve it myself so no shame to you if that is the case. Again, if I've understood you correctly. You haven't given much explanation about what any of your letters mean.

[@JesseLH88]

@@toast_recon ty for the response (i am just glad someone actually read my comment). Yes, it would need some more formalism to be accepted, but I do think it works. Yes, this assumes he gets p throws without ever missing. But if he misses, he is just at x/(y+1). But since x and y are arbitrary, we can let y' = y+1 and follow the same logic. There are some additional proof-by-induction steps that I did not include because this is a YT comment and not an actual exam. Informally, for every shot he misses, p will increase by 3. Every shot he makes decreases p by 1. If you transform the problem into p-space, it becomes: given positive integer p, is it possible to make negative using only these two operations: p' = p-1 p' = p+3

[@toast_recon]

@@JesseLH88 ok, it still took me a couple minutes to make the final connections which I didn't get from your text. Since you showed that for every current state, there is a positive integer p number of free throws you can make to get to 3/4, and that remains true every time he succeeds or fails, eventually you will reach 3/4 if you're trending up (which we know we are because we're going to pass through 3/4) I'd agree with you now that you solved it (assuming I haven't just made errors in reasoning), but I think it's a bit of a bitch writing it out in a convincing way in the alotted 30 minutes. Please correct me if I mischaracterized your argument or made errors of my own.

[@mikemondano3624]

So, people who hate basketball get it wrong. It doesn't mean they don't know the math - it means they don't know basketball, like probably the majority of intelligent people.

[@Felixkeeg]

I realized the answer was "yes" when I threw in random numbers, but I suck at proving things in a mathematical precise way

[@MacstersUndead]

plugging in numbers can remain a good way to start a problem to see if there's a pattern.

[@fk319fk]

I took this test in the '80s. I do not recall what I got, but I think it was in the 10's or 20's. What I remember is a guy that did not sign up so his score was not official, but he got all but one, he said so. We were talking about the problems and as each of us had worked on it we had familiarity with the problem. So what did I remember? he was able to explain each solution for each problem in a couple of minutes.

[@zelda12346]

You can simplify the question as follows: Does there exist natural numbers a, b such that a/b < 3/4 and (a+1)/(b+1) > 3/4? No matter what free-throw pattern you have, you have to cross the threshold. Hence, you can think of the before and after points as the arbitrary start and end points of the sequence length 2. Proof by contradiction - Assume the theorem is true. Then we can rewrite the inequalities as 4a < 3b & 4a+4 > 3b+3. Bring the 3 to the other side, and we get a three part inequality: 4a < 3b < 4a+1 Note that 4a and 4a+1 are consecutive integers. b is an integer, which implies 3b is an integer. However, there cannot be an integer between two consecutive integers. Contradiction. Hence, (a+1)/(b+1)

[@michaelempeigne3519]

Z-Statistic wouldn't applying the mvt for derivatives be faster method to solve this problem?

[@isaacaguilar5642]

Michael Empeigne i was thinking the same thing but i dont believe that function would be continuous so it wouldnt apply

[@AnonymousAnonymous-ht4cm]

Another approach is to consider the misses. Before we have m misses and less than 4m total shots. Since any miss will not bring us over 75%, we can ignore those. Any succes will increase the total shots by 1. However since we always have an integer number of total shots, we cannot avoid having 4m shots at some point.

[@perc-ai]

@@AnonymousAnonymous-ht4cm Natural number theory and proof by contradiction

[@Antox68]

Now that's an interesting problem! I hope you make more like these

[@lppunto]

As a Putnam participant the past three years, I wish the problems were usually this easy!

[@JohnRandomness105]

This was easy. If one is below 75% at one point, after the next through, he is either below, at, or above 75%. If he does cross 75% upward, then when he crosses it, it must before be below, and after be at or above. So let's consider below then above: k/n < 3/4 < (k+1)/(n+1) -- that last throw made it through the basket. Cross multiply: 4k < 3n, and 3(n+1) < 4(k+1). Subtract three from both sides: 3n < 4k + 1. Combine the inequalities. 4k < 3n < 4k +1. But no integer is between two adjacent integers -- 3n can't exist between 4k and 4k+1. So this case is impossible. Consequently, while crossing 75%, the fraction must be exactly 75% at one step.

[@matteo-ciaramitaro]

Intuition told me the answer would be no, but math says otherwise. if you have x misses and hit all remaining shots, eventually your ratio will reach 3*x / 4*x, at which point you will have 75% accuracy let n be the number of shots, h be the number of successful x = n - h our current accuracy is h / n n and h are natural numbers, h < n. Let K be a natural number Solve for K (h + K) / (n+K) = 3/4 h + K = 3/4 n + 3/4K K = 4(3/4 n - h) = 3n - 4h Since n and h are natural numbers, our solution for K is an integer. Since h/n < 3/4, 3n - 4h > 0 Then K is a natural number. Since K is a natural number, then from any starting percentage below 75%, if we hit 3n - 4h more shots successfully, our percentage will be exactly 75%

[@SnowFaceChamcham]

I couldn't even begin to solve it but I found it funny how the question posed in the thumbnail feels like it should be answered "obviously no" and then the layman version in the video feels like it should be answered "obviously yes, that is possible."

[@fencejumpa2365]

God . .all your problems make me wanna pull my eyes out before you explain them XD

[@SteveNeubauer]

And what about after?

[@fencejumpa2365]

@@SteveNeubauer I pretend I understand it.

[@timothymclean]

It seemed pretty obvious to me, so I wrote my solution/proof below. Assume DeAndre starts the season by missing four free throws in a row, followed by succeeding at his next sixteen free throws. S(12) = 75%. This scenario meets the criteria outlined in the problem; S(5) = 20% < 75%, and S(20) = 80% > 75%. Therefore, it is possible for S(N) to be exactly 75% at some point in the season. This is where my proof broke down. I tried to construct scenarios that passed 75% at any N that was not a multiple of four, assuming that it would be trivial to do so. It was not. I broke out a _spreadsheet_ so I could instantly calculate S(N) at any stage of any arbitrary sequence of free throws, which would be cheating if I was taking the Putnam Exam, but I'm not so _hush._ I kept getting 75.0% in one cell or another, so I started to suspect that it _was,_ in fact, impossible to _not_ hit 75% exactly. But I didn't understand _why_ that was happening, so I couldn't prove that hunch. I see why the Putnam Exam is so hard.

[@nekogod]

So if I've understood correctly, it's impossible to skip 1/2, 2/3, 3/4, 4/5, 5/6 etc when going from below to above.

[@jasonterry1959]

.

[@Momogamer9]

i actually dont understand why it is possible to skip 1/2 2/3 etc when going from above to below

[@jasonterry1959]

@@Momogamer9 I find it better to approach it from a win : loss ratio, as in gaming / sports / etc. If you have 30 wins : 0 losses, and you *eventually* somehow manage to go 30 wins 40 losses, i can promise you that at one point you went 30-30 (50%). 30 wins 9 losses. One more loss is 75%. 31 wins 9 losses. Odd number of wins, you cannot have 10.33 losses, so you will never get 75% if you keep adding to the "loss" column. I believe the same idea works from below 1/2 if you keep adding to the "win" column. 0 wins 31 losses, no matter how many wins you get in a row, you will never reach 1/5, 1/4, 1/3, but you will reach 1/2. (you can't skip 1/2 ever, just adding 1 game at a time. Above 50% you can skip by losing, Below 50% you can skip by winning.)

[@yoseftreitman7226]

Exactly. You understand perfectly.

[@yoseftreitman7226]

@@jasonterry1959 Yeah, I was thinking the same thing.

[@pdpgrgn]

If the question was asked without any context, my immediate response would be no, because percentages jump so 75% can be skipped. But as the video started with explaining how difficult the test where the question was asked is, I thought the answer had to be Yes. My reasoning was that if the answer was no, providing one example where the percentage skipped 75 would be sufficient, but proving that the percentage can't skip 75 would be difficult.

[@arikwolf3777]

The N where the percentage equal 75% is the number of misses X 4.

[@axelgilbert7220]

It's always good to point out that n is positive before cross-multiplying. Fun problem anyways.

[@wolfie6175]

Yeah usually we never cross multiply instead we just shift the entire thing to the right (or left) . Unless we know all terms >0

[@crowbar_the_rogue]

I mean ... We're dealing with natural numbers here.

[@Unknownpony]

To be honest. If question with obviouse answer is on such test. It's almost guaranteed to be opposite to what you initially thought. It's kind of funny that unexpected thing is expected in certian circumstances

[@timopeukert1301]

Well, the hard thing about this was less figuring out which of the answers is correct, but how to prove that it is correct. Ot was obvious from the beginning that it had to land on 75% because otherwise the problem wouldn't be hard enough for Putnam, but that still doesn't mean you'd be ablr to prove why.

[@ObsidianParis]

To my eyes, there's a even simplier way to explain the problem : - To reach 75%, we need to complete 3/4 successful throws, or in others words, « three times the number of failed attemps » : success = 3×failed. Since both « 3 » and the number of failed attempts are integers, the product will be an integer too. And since the number of successful throws is lineary browsing each value of ℕ (« 1 successul throw, 2 successful throws, 3 successful throws… »), we'll forcibly encounter a number that matches a 3/4 ratio. And this is actually true for every integer ratios (e.g : 2/3 = 66,666% = twice the number of failed attempts).

[@kurozawa6]

Wow. To think of it as the ratio of # of shots made to # of shots missed (instead of involving the total # of shots) is quite out of the box. This solution that I believe to be more 'elegant' deserves another video. Well thought. Here's a rephrasing that I tried to make as clear and 'complete' as possible: 75% means 75% of the shots are successful while the other 25% are not. 75% is exactly thrice of 25%. Thus, to make 75% (from less than 75%), he needs a total of 3 successful shots for every 1 missed shot. All he needs is to keep making successful shots. If he misses another shot, he'll just need to compensate with 3 extra made shots. Let N be the number of shots to make to reach a 3:1 ratio (of made shots to missed shots). N becoming zero means 75% has been reached, while N becoming negative means he has surpassed 75%. Making a shot decreases it by 1, while missing a shot increases it by 3. 3 is a multiple of 1, so missing any number of shots won't make the goal mathematically impossible as N would always be an integer (more of this later). Since he can only decrease it by 1 one at a time, it would be impossible for him to make N negative without first making it reach 0. Therefore, he can only surpass 75% (negative N) by first reaching exactly 75% (N is 0). Reaching exactly 75% is necessary before surpassing it.

[@kurozawa6]

Now, about N being always an integer, this is where it gets interesting. Let's say the free throw percentage started with below 25% instead and we're interested to know whether reaching exactly 25% is necessary before surpassing it. 25% implies that the # of shots made is one-third of the # of shots missed. So there's 1 shot made for every 3 missed shots. If he makes a shot, N decreases by 1, and when he misses THREE shots, N increases by 1. What happens if he misses one shot? Well, he'll have an extra 'point deficit' of 1/3. N increases by 1/3. Let's say N is now 1. His 'point deficit' to reach 25% is 1. Then, his next shot misses. So N is now 1 and 1/3. Then, his next shot hits, so now N is 1/3. Take note, 25% is still not reached. From this point, he can surpass 25% by hitting his next shot. N will be -2/3, and it would never have become 0. Therefore, in this case, reaching exactly 25% is not necessary. This is why whether the final shots made to shots missed ratio is an integer or not matters to whether the target % can be skipped. We can generalize that as long as this ratio is an integer, the target percentage cannot be skipped before surpassing it.

[@TomRiddle-yo3pt]

*_I see pictures that look like the S.H.I.E.L.D logo, I click._*

[@shaungordon5104]

I solved this in a completely different way. I constructed a plane of y = number of successful throws and x = number of successful throws. X and y are always integers. I then drew a line (A) of y = 3x showing the points where his success rate is 75%. Movements alone the plane can only happen (0,+1) or (+1,0) (successful throw or unsuccessful throw respectively). I then start at some random point of integers below A. In order to get above A, you have to have successful throws otherwise you get further away from A For all values of x (x = n), moving upwards along that line will always meet a point along A as y = 3n and if n is an integer, so will y be. In short, if you plot this and use it like a road map, you'll see that if you're Below the 75% line, the only way to travel above it is to go through an intersection representing the 75% mark

[@bennettsavard6744]

Exact same thing I did! Took me barely anytime at all

[@fingamersroup]

I think you could make a case that if andrew hasn't thrown any free shots his procentage would be 0. Tho mathematically it would result in dividing by zero in the proof so its not very strong case.

[@TheSpyroplayer]

Yeah, diving by 0 is undefined, yet the problem defined the previous percentage as "below 75%."

[@vulblhotdiessfi]

0/0 does not equal 0

[@unclecreepy4185]

(Before watching the video). It seems the question is asking is there a scenario where the situation is true. If at some point he made 3 out of 4 FT, then his average is exactly 75%. If he missed his previous FT shot, he would have been 2 for 3, which is below 75%. He then made another FT and ended the season making 4 out of 5, which is more than 75%. So let’s watch the video to see what I’m missing.

[@unclecreepy4185]

I guess I did misunderstand the question. The guy seems to suggest the question was more about is there a way to skip over 75% or will it be exactly 75% at some point in any scenario. You could make a case for any number bigger than 0 where the % is below, exactly, and above that number, but could most likely be skipped over. Whereas certain numbers can’t be skipped over.

[@hypnovia]

I would graph throws vs successes on an integer lattice and show that every diagonal has a point on the line y=3x/4

[@nikogruben9573]

How do you graph throws vs successes, there’s infinitely many possibilities

[@BreatheManually]

We start below 75% by missing first and then add successful shots, until it goes above 75%. We can vary the number of missed free throws to see if we can skip 75%. At 75%, the first fraction is 3/4, you can find the denominator that the free throw will hit 75% by counting the misses (here is 1) 1/0.25 = 4. Since 0.25 is the proportion of missed shots at 75%. So if you only miss one free throw, you will hit 3/4 and yes you do hit 75%. Since we want to miss first, 0/1> 1/2>2/3>3/4(75%)>4/5. Another example is 2 misses, 2/0.25 is 8. 6/8 also hits 75% What we want to find if any integer is divisble by 0.25 to not result in a positive integer. This will be the number of misses in which that is the case. Since all positive integers can be divided by 0.25 to give another positive integer, there is necessarily a point in which 75% is hit. Assuming it was less than 75% before and more than 75% after.

[@sleepingsaucer3801]

If only there was an integer between 3 and 4 called bleem...

[@Rekko82]

It is 3.333....

[@radreonx5386]

Haha, yes! 'The Secret Number' :)

[@Rekko82]

@@radreonx5386 The number Illuminati is hiding from us!

[@djdjsjdhbd8092]

Ferdinand Rafanan there is pi

[@Rekko82]

@@djdjsjdhbd8092 Well, they wanna eat the pi.

[@AidenHardy596]

You can miss throws to try and change up the fraction, but because the percentage of shots made is found by (shots made/total shots taken) instead of (shots made/shots missed), this means whenever the numerator increases by one, the denominator does too. So if you had a number that you would think can't reach 75%, such as 7/11, then you make a shot, so your made shots becomes 8, and your total shots becomes 12, giving you 8/12. Then 9/13, then 10/14, then 11/15, then 12/16. 12/16 is the same as 75%. You can try this with any fraction and the same thing will happen

[@ponyboycurtis8490]

Questions like this is why I skipped math and went to the gym to shoot free throws. 🏀🏀🏀

[@crowbar_the_rogue]

Yes, but now you know that if you were any good at basketball, there was definitely a point in your career where you hit exactly 75% of your free throws.

[@rmsgrey]

I proved it by a different route: After some number of throws, n>0, there have been m missed shots, with m/n>1/4 (that is, at some point the miss percentage is greater than 25%). Rearranging gives n

[@simon_patterson]

DeAndre Jordan's free throw percentage increased because he practised smarter, not harder. He worked out a pre-free-throw routine that helped him psychologically, and followed it religiously. That was the difference, not him working any harder than before.

[@niketsrivastava2423]

This is a straightforward application of discrete continuity... An easier example is like say you have an infinite staircase of natural numbers where each step represents each natural number... you were at 3 initially and then were at 5 finally... now assuming you jump each step one at a time... you will have to cross step no."4" at least once before coming to 5

[@ssrini2002]

So it can never skip values like 1/2, 3/4, 5/6, 6/7 etc- is that right?

[@LeoAr37]

Correct, that's what he did when he replaced 3/4 by (k-1)/k and reached the more general result. Basically 50% 66,6% 75% 80% 83,3% etc...

[@ssrini2002]

@@LeoAr37 so also interestingly, it can never happen for any values below 50% whatsoever

[@mikekimura2532]

@@LeoAr37 ...and I believe the only Integer Percentages are: 50%(1/2), 75%(3/4), 80%(4/5), 90%(9/10), 95%(19/20), 96%(24/25), 98%(49/50), 99%(99/100)

[@grinreaperoftrolls7528]

I did this a little differently, but I dunno if it’s very rigorous. I started with S(N)/N (/=) 3/4. And N>S(N), all variables are integers. Then I added D to the top and bottom representing successful shots. Then I did algebra until it became D(/=)3N-4D. There is a value of D that will make the two sides equal for any value of N and S(N) within the initial criteria (n>s(n) and all variables are integers

[@slicewinds5583]

I thought this was simply an Intermediate Value Theorem answer

[@TayaTerumi]

I thought this also, but quickly remembered that this distribution is discrete. We have no continuous function to apply the Theorem to.

[@Hon2fun]

@Partha Vemulapalli Is this not tchebysheff's theorem ??? It seemed fairly obvious to me that you would solve it exactly as such since this exact problem is asked in a statistic book.

[@KazmirRunik]

Used a simple, intuitive solution to solve this one in seconds. I imagined him starting with 2 bad shots out of 5, or 60%. To get above 75%, he has to have good shots until he gets to over 8 total shots. He hits 75% at 8. Then, I imagined 3 bad shots. Now, he has to go past 12 total shots, making him get to 75% at 12. This established the pattern that you will hit exactly 75% when the total shots are 4 times the missed shots (with the trivial case being 1 bad shot out of 4). The missed shots are only ever an integer value, so 4 times that number is also only an integer value. You can't skip that number which is 4 times the number of missed shots by counting one at a time, so you do necessarily have to hit 75%. Took longer to write than it took to come up with the solution.

[@michaelweiske702]

I saw the thumbnail and thought about the problem a bit. I accidentally rebooted the page so the video disappeared, leaving me without a solution (the horror). I then spent about 10 minutes on my chalkboard thinking about this problem, and here was my solution: N and M are integers. M is the # of attempts and N the number of successful attempts. Overall, N/M

[@YTSparty]

Great proof. My simple logic was that all integer multiples like 1/2, 1/3, 1/4, 3/4 would eventually have to happen on the F(x) = x/y where x, y are integers. Since you only make one free throw at a time, x/y will eventually have to pass 0.5, 0.3333, or .75 etc. But your proof is great to show it mathematically.

[@zym6687]

You can only make one free throw at a time but you can have more than one attempt removed from your record at a time so it is not *necessary* to have passed through 75%.

[@YTSparty]

@@zym6687 Not sure what you're saying. His proof is more elegant than my logic. There is no way to step over a fraction multiple of 3/4. You could prolly do a series for how you progress like Xn/Yn etc.

[@firefly7076]

You forgot to state that N and S(N) are positive integers

[@iainrobinson4280]

You've always made an integer number of misses A (for airball) and an integer number of buckets (B). So if you're improving your average, B is going up one at a time, and you always reach a position where your B is 3A because both are integers. It's also the case for improving to 50% (B=A), 80% (B=4A) and 90% (B=9A). If your average is descending, though, your airball could take you from 10/13 (76.9%) to 10/14 (71.4%), so you'd miss the 75% figure. But you'd always hit 25%, 20% and 10% on the way down because they are the inverse of the sums above. If you can express the target as a fraction with a 1 on top and an integer denominator you'll hit it exactly. You always hit 50% exactly, whether you're getting better or worse.

[@bpark10001]

This problem assumes the free-throw percentage is calculated after every throw. This would not be true if the percentage were to be calculated after every game, week, or season. Are free-throw percentages really calculated after EVERY SINGLE THROW? The only way to establish whether the problem is "correct" is to contact sports rule-makers, and ask how free-throw percentages are OFFICIALLY calculated. The other way to separate the math from the sports rules is to specify how the percentages are to be calculated. That is not stated!

[@diprajbhattacharjee4462]

I had the Same point in my mind

[@rafi7397]

It says at some point, there is no reason for that "some point" to not be in the middle of a game

[@charleswajda-gotwals3690]

Think about two groups, one of missed and one of made shots. Each miss is paired up with three made shots (3/4 shots in that group are made, thus 75%), but because the accuracy is

[@minizilla100]

Well but if it was his first free throw? Bevor the shot it was zero... an after the freethrow it could be 100

[@mikezheng33]

0/0 isn't 0% it's indeterminate

[@kobybattema3144]

@Brian I'm curious as to what your explanation is for why 0/0 isn't indeterminate?

[@kobybattema3144]

@Brian I understand the logic, as a/a = 1. However, a=0 is the exception I believe. According to the mathematical definition, 0*(any number)=0, 0/0 cannot equal any single number, therefore 0/0 is indeterminate. I realize this explanation is rather difficult to read but mathematically, it's true. There's some people on the intermwebs who explain it better than I do. Cheers!

[@mikezheng33]

@Brian it's math not religion, what you believe doesn't make it right or wrong

[@mikezheng33]

@Brian k

[@crowbar_the_rogue]

I solved the actual problem fairly quickly and then spent much longer trying to prove that this only works for fractions that can be written as (k-1)/k. It's actually fairly simple, if we write the percentage as m/M, we get 0 < m•N - M•S(N) < M - m. We're looking for cases where M - m > 1, so m•N - M•S(N) = 1 avoids any contradictions. But I have no idea how to prove that such N and S(N) exist. Edit: I figured it out. If we make sure that m/M is in its simplified form, then n(M-m) will not be divisible by M as long as n is smaller than M. It shouldn't be too hard to prove that the remainder of n(M-m) // M will be different for each n

[@pepehimovic3135]

Hi. Is there an online course where I can learn this?

[@gourmet_salad]

Solved it, the answer is yes. Assume for the sake of contradiction that at no point the ratio is exactly 3/4. Since earlier the ration was less than 3/4, and later it was more than 3/4, then there exists a free throw which before he took his ration was less than 3/4, and after he took it, his ratio was more than 3/4. Assuming his ratio before this free throw was a/b, this means a/b 3/4 Rearranging both of these, we obtain 3b>4a And 4a+1>3b Since 4a and 3b are integers we may change the “greater than” to “greater than or equal” and add one to the right hand side. So 3b>=4a+1 And 4a+1>=3b+1 Adding these two together we get 1>=2. Contradiction. So indeed, at some point his ratio was exactly 3/4.

[@johnchristian5027]

nice