Simple, yet counterintuitive mathematics | Why numbers don't always mean what you think 

Ironically, we intuitively guessed that the guy on the right saved more money on gas, because you wouldn't have asked otherwise...

Damn, I can't believe you actually wrapped a cable around the globe just to make this video!

Oh man, the extension cable example made going up a belt size seem so much worse :(

The first problem shows why measuring fuel efficiency in L/100km is much more useful. Since the units are volume per distance rather than the reciprocal as with mpg, you MULTIPLY your distance by fuel efficiency to compute the volume of fuel consumed. So, if you rephrased this question for someone not using wonky hillbilly units, it would read "One person trades in their 9.4 L/100km car for a 6.7, while the other trades in their 29.4 for a 23.5. Who buys less fuel compared to before?" The answer is then intuitive and obvious.

Hey, you should watch your youtube statistics to see if your views go up over the next few weeks. The main thing stopping me from watching your videos is other time commitments. But now that schools/colleges are being canceled, I think a lot of people will have enough free time to watch more of your videos. I know I sure will. Thanks for making great content. Your job is my dream job.

The metric measurement for gas consumption is L/100km, so the green car (lower left) gets 9.44 L/100km, and the yellow (top left) gets 6.71 L/100km, thus saving 2.72 L/100km. The grey car (bottom right) gets 29.4 L/100km and the blue car (top right) gets 23.5 L/100km, thus saving 5.8 L/100km if you switch. Measuring gas consumption in a unit that actually measures CONSUMPTION of gas per amount driven is much more meaningful for these types of comparisons. MPG (or km/L) is not actually a "consumption" but more like a "range" measurement.

Just discovered this channel and I'm hooked!

It's amazing how in the third example the original radius doesn't matter at all.

There was a sports show called “Numbers Never Lie” where they used a lot of stats to talk about games and make predictions, etc. But they were wrong so many times because of the simple fact that while numbers never lie, people do. They would constantly misinterpreted or misunderstand the stats and numbers. This is especially true when dealing with a small sample size where adding 1 or 2 can double the percentage.

A mother is 21 years older than her Child. In 6 years, the Child will be 5 times younger than the mother. Where is the father?

I’ve told people the circumference puzzle before, and once you do the math, you figure out that of course it is the non-intuitive answer. But I also am looking for ways to visualize it that make it more intuitive. One way that I think makes it easier to picture is if you imagine the earth being a cube rather than a sphere. It is then much easier to picture what adding 6 feet does to the height of the cable, since it gets 1.5 feet on each of its four sides.

Simple intuition for the last ace thing: The probability for a position to contain the last ace is the probability that it contains an ace, times the probability there is no ace after it. The former a constant 1/13 (4 aces / 52 cards), the latter is 1 only if there are no cards after this one.

I appreciate the growth of your channel along with the content over time. Keep it up 👍🏻

the first example is so confusing because the guy on the left IS saving more on gas in terms of percent. He's saving 40% instead of the other guy saving only 25%. but ofc the guy on the left's gas bill was already so low saving 40% on it doesn't matter that much. very interesting though, fooled me! o:

Hey man, it always great to get an enlightening on what it means to have numbers in interpretation! I was wondering if you could do a video on Number theory actually

mah boi sneakin'

The metric way of expressing fuel economy is in liter/km not km/liter. Converting into those units makes the first exercice not counter-intuitive at all! :)

You should do a "How easy it is to lie with statistics Part 2" I woiuld love that, and im sure many others would! You could use real world political examples too.

the strawberry example is the same as stock prices. if a stock drops 99% vs 98% the difference in price between the two is 100%. you'd have twice as much $ if it only dropped 98% instead of 99%

The card game actually illustrates really well the Monty Hall riddle, especially when downed to 3 cards only

The way we solved the cable around the Earth one when I was at school was "We're not told the radius of the Earth, so presumably it doesn't matter. That means we can set it to zero, and the answer is simply the radius of the circle whose circumference is the length of the cable."

Zach I'm going to be honest with you: this is way more interesting than most math in school. You're an excellent explainer and I love ur vids

The reason why it doesn't work is because the cost depends on the amount of gasoline burned, which is part of the denominator, so you first have to switch from kilometers per liter to liters per kilometer so that the amount of gasoline burned is in the numerator, then you can do a simple calculation to get the right answer.

3:05 I got this one right by intuition without doing any maths. my thought process was like "if the total mass were 98g it wouldn't make sense because 1g is a small value and there's too much water and too little "other stuff" so that will not even scratch. the subtraction of mass must be something huge to make a difference so maybe half of the initial mass"

I never never comment, but Zach your videos are so crisp and clear, with beautiful animations and intuitive explanations, I had to let you know. I'm about to finish undergrad and am so glad I found these, they'll keep me learning beyond school. Keep it up!

The main thing about the "cable around the earth" part is indeed psychology - or at least our frame of reference. We are used to driving 100 miles (or kilometers) to our hometown, and it really isn't that far away if compared to, say, international travel. Unless you live in Vatican City, but then you'd probably not know how to use the internet and wouldn't be here. However in height, we probably don't go above 10km for the level at which a plane flies (or, talking orders of magnitude, the Mount Everest). We just don't have anything to do with the 6000+ kilometers of Earth before it reaches what we call the floor. If the question was "we have a circle with a radius of 6000km and make it longer by 1m, it will raise a couple of cm" ("longer by a yard it will raise a couple of inches") nobody would be surprised at all. The cable is longer by a negligible amount in context, so the radius also increases by a negligible amount.

I thought about the ace question like this: For every position, there is a 1/13 chance that it is an ace, so this is the unimportant part. So the important question is, if it is an ace, will it be the last one. This chance increases the further you go in the deck, so 52 (where the second chance is 1) has the highest odds of winning

La probabilidad de que aparezca un as en la posición nro 52 del mazo es exactamente 1/13, como corresponde a cualquier otro número de las cartas desde el as hasta el rey. En cambio, para que aparezca el ultimo as (o cualquier otro número) en la posición nro 51, la probabilidad ya es algo menor, ya que al 1/13 hay que restarle la probabilidad de que la última carta tambien sea un as. Y así sucesivamente retrocediendo en el mazo, las probabilidades serán decrecientes. Por lo tanto, sí, la posición nro 52 es la más probable para un nùmero de carta determinado, sea este un as, un 2, un 3, un 4, etc etc. Translated for english speakers: The probability of an ace appearing in the 52nd position of the deck is exactly 1/13, as corresponds to any other number of cards from ace to king. However, for the last ace (or any other number) to appear in position 51st, the probability is already somewhat lower, since at 1/13 you have to subtract the probability that the last card is also an ace. And so on by backing into the deck, the odds will be decreasing. Therefore, yes, the position 52nd is most likely for a given card number, be it an ace, a 2, a 3, a 4, etc.

Thank god he said he was kidding at 7:47 or else I would've believed him.

About the aces problem, think of it like this: The chance that any given position is an ace, is equal for any position. However, if the last card IS an ace, it's guaranteed to be the LAST ace. For any other position, there is a chance that it's the second to last, etc. ace. So your best pick is the last card.

Your videos are so interesting and insightful, please keep it up🙏💯

Great stuff Zach! Thanks for that video.

I kind of guessed the MPG problem intuitively. The guy on the left is already spending very little on fuel, so even though the percentage decrease is bigger, it is negligible. Going from $3 to $2 gives a better net saving than 0.5$ to 0.25$.

For th car question I got completely confused, we always measure consumption with L/km here, not km/L.

Here's some even more counterintuitive maths: Since he apparently got some 40,000 km of cable lying around (and he wouldn't have lied to us about the actual lenght of the cable shown in the video??) i did a rough calculation: Here in Europe, for 100 m of double stranded copper cable, the cheapest variety, the cost would be about 25 € (or 30 Dollars). So it's about 300 Dollars for 1 km of cable. Multiplied by 40,000 km, this is about 12 million Dollars worth of cable (!) All spent for a single RU-vid video. This man is a true hero of the internet! ^^

I love your maths videos man, Keep it up🧜🏼‍♂️💛

It’s cool you also put metric 😃

The rope around the earth was a pleasant surprise for me, that soon became very obvious. As for the last Ace problem, I needed to find the probability distribution for each position. Turns out, it's not that hard. Let: N = total number of cards (N = 52, here) a = number of aces (a = 4, here) X = position of the last Ace Now, we can readily see this constraint: a

I've recently had this guy in my feed. Good stuff. Can't believe the cable around the earth thing.

Dang you really had me on the radius one. I literally just finished coding some stuff working with rings and circles and thought for sure it would be very small. Great example.

Oh right, that cable-around-the-earth-problem I remember that being a question on a test in school. When my classmates and I exchanged results, the good ones (minority) were scratching their heads because of the counter-intuitive answer and the fact that the majority got a different answer. And the students that were not good at maths were terrified considering the fact that everyone who is good at maths got the same answer, but they didn't. I'm sure the teacher liked hear the chaos unfold, he kind of disliked our class because many did not pay attention. x²-64=0 came up on three consecutive tests with 50% or more not being able to solve it.

Amazing as always.

Learning how to use deceptive numbers in video games sounds like something I should pay really close attention to. I could have a lot of fun tricking players with rules that play off of counterintuitive examples. Some like turn based RPGs already do by lying about statistics for most difficulties, but giving the real stated odds on the most hardcore difficulties and suddenly people feel like they're missing a lot more often.

Better way to thing about the “cable around” the world question INTUITIVELY, is the following: Imagine Earth is a square (or cube). Shift the cable by 1 feet on 1 side of the square by adding 1-1 feet to the 2 sides. So 1/4 of the cable will be 1ft above Earth. Repeat it on all 4 sides. So we added 8 feet, and lifted the cable by 1 feet all around. This is a good approximation for the case of circle/sphere too.

I remember seeing one. It was a heart disease meditation advertising that it was 67% more effective than placebos at reducing the risk of a second heart attack. In actuality, the original risk was, let's say, 90%. The placebos made the risk go down to 87%, and the medicine reduced it to 85%. 5% is 1.67 times 0.03, or 3%. Note: I don't remember the exact numbers but you get the idea.

3:04 this can be misleading; the way you phrased it implies that you are removing a lot of watter, but it could be phrased in a way such as this is not obvious, for instance, if you say you're removing 1% of the watter contribution to mass (or, as you said it, "removin some such as now it accounts for 98% of the mass"), it means you're removing a lot of watter, if you say you're removing 1% of watter, it means you're removing 1% of watter, or 0.99g of watter, thus implying that it did not lose 50% of it's mass, only 1%. I've seen this problem before, and I'm glad you knew how to phrase it correctly so your calculations actually mean what you said.

With the card game if you choose a spot (1->52) before this is correct. But if you can make your decision while the dealer flips, you discard all the cards including the third ace. Then you have equal permutations for the following ace location.

Thought of a way the problem at 7:10 relates to entropy. Or to find the answer using entropy. Entropy is a function that is maximised when a system is in its most probable state, that also increases linearly with the size of the system. The equation for entropy is S = sum( - P_a * log(P_a) ) where S is the entropy and P_a is the probability that a function is in state “a”. These probabilities are free to change except that they must sum to 1. To find the most probable state we then maximise the entropy, to some constraints, which represents things we can measure on the system. If we constrain the system to states with a defined maximum ace position and to that the number of aces in each states is 4 (so we discard states that have more than 4 aces). For each of the microstates (which include states that give equivalent measurements.) we will model them as being 4 aces in 52 random positions but any 2 aces cannot be in the same positions [they are fermions I guess]. Now to find the most probable maximum ace position we can calculate the entropy for a certain position of the maximum ace. As the entropy is higher the more probable a macro state is, we can find the most probable maximum ace position by finding the maximum ace position with the highest corresponding entropy. The number of microstates that have the given maximum ace position is (A-1)! . We can then say that the probability of each microstate is then 1/(A-1)! if we constrain the system to have a given A. Calculating the entropy for a given A we find that the entropy always increases with a given maximum ace position, therefore the most probable maximum ace position is the one that is the highest possible i.e 52

"And let's say its total mass is a 100g" WOW THAT STRAWBERRY IS HUUUUGE

A different way of looking at the problem with aces is this. To win you have to pick a a location with an ace, and there must be no aces after your spot. By symmetry all spots have 1/13 chance of being an ace. It is also clear that the more cards there are after your pick the larger the chance it contains one of the remaining 3 aces.

The thing about the cable is that a 1 foot increase in radius is very small compared to the original radius of the earth.

For the final ace problem, I think of it as the following: If the an ace is at the last position, the other cards can take any order. Since the position I choose needs to be the final ace, any card that comes after the position cannot be an ace, which decreases the number of suitable permutations. Therefore, there are the most number of permutations the cards take with the final ace at the last position.

Question of curiousity, How would one go about optimizing their guess for finding the third ace?

As a Brit, I heartily approve of you slipping in and out of unit. "If the circumference of the earth is ....km then we can lift it .... feet". Yup, that's the UK alright. Don't bat an eyelid at someone approximating a length as "about 1 metre six inches".

The 4th ace is the same as the 1st ace if you look at the deck from the other end. So you immediately see that the odds of winning is 1/13. The odds of the second card being the first ace (the second to last being the last ace) is 12/13 (first card not an ace) times 4/51. 4/51 is only slight more than 1/13, but you multiply it by 12/13 which lowers your odds much more than what 4/51 vs 4/52 increased it.

Could you please do a video on the history of trigonometry. Sure i could read about it, but your voice just butters my bread.

If it's too easy you are doing it wrong -Mathematics-

“it’s not always the blatant lies we need to look out for. it’s often the misleading truths.” put that on a bumper sticker

Excellent! Monty Hall problem is another counterintuitive example.

I was expecting the probability of the card guessing game to increase near the last 1/4 of the deck (cuz assuming an even distribution it's most likely there) but I didn't expect the last one to be the best

Got to learn new thing

The chance of the 4th ace coming at the last card is simply the chance that the last card is an ace, at all, or 1/13, or 7.69%. Meanwhile the chance that it comes as the 4th card, is 4/52x3/51x2/50x1/49=0.000369%

In the Czech Republic (not sure if it applies across Europe too), we don't use miles per gallon (or kilometers per liter) but liters per 100 kilometers. That representation turns out to be much more convenient when solving the first problem - because it tells you directly how much fuel each to car needs to travel the same distance. Hearing about miles per gallon is always quite confusing to me :D

If you use metric conversion you can see the difference immediately in benefit to the vehicle user on the right. I live in a metric country and to me, the first 'problem' was very obvious to a point where I was looking for a catch as to why I could be wrong.

Here's an interactive graph of the "find the ace" problem: www.desmos.com/calculator/3t8ftm2tmg

is it good or bad that a lot of these arent unintuitive to me?

Really love your content. We need more critical thinkers like you in the world.

Thank you enjoyed your tutorial and explainations

I think that there is no permutation between the aces because their "identity"is irrelevant, they should be considered as "just an ace", hope I'm clear LOL 😆 .

A lot of these are unintuitive but, to even an average high school student, are solvable. The strawberry question (or as I've seen it before, the watermelon question), is one that I've asked students to work through before because of how simple the answer is once its worked out but it will wring your brain out to get there if you don't understand it immediately. Understanding unintuitive data is a brilliant exercise and your examples at the end summarise it beautifully.

The percentage thing reminds me of an online argument i have seen recently concerning the pandemic: Person A : it is pretty bad though, 1 in 50 people die from it. Person B : You are lying, they say there is a 98% survival rate. which shows again, that humans really really really suck at interpreting probabilities other than 50/50 on an intuitive level.

Great vid as always zachary

The odds of getting the final ace in the 52nd spot is one in 13 as each card type is equally likely to appear.

10:54 win% on choosing 1st: 0/6 2nd: 2/6 3rd: 4/6 so by picking 4th u would win 6/6 times, but thats cheating

6.46 just wanted to add that the end result is correcct but the math behind it isent.... C - Cirumferance (after added cord) Ce - Cerumference eart Re - Radius eart Ra - added radius Xc - Cable extension K - 2 * pi [a constant] gives Ce + Xc = K * ( Re + Ra ) [Ce depends on Re, Xc depends on Ra] wich also gives gives Ra= Xc/K (growt of circumferance) substituting in Ra gives C = K * ( Re + ( Xc / K ) ) [as a general formula to solve any similar problem]

Great job!

yep, mathematics are great

Keep your units consistent. Do not give a length increase in feet and compare it to an overall length in kilometers. That is just sloppy.

Wonderful video! Thank you!

For the final question, isn't it just an n choose k problem? Ie., how many combinations are there for 4 aces in different segments of the deck? Let A be the range of aces (from the beginning of the deck to the end) and C be the number of cards. Here, the amount of permutations (not counting repeats of different aces or other cards) can be expressed as C!/(A!(C-A)!), this number will obviously shrink as A becomes larger. I think this makes it somewhat apparent? (Lmk if I messed up anywhere here).

My favorite misleading statistics is: say you have a 2% risk of getting hit by a car. Say texting while walking increases that risk by 50%. you now have a 52% chance of getting hit right? nope, it's 3%

Why is Mathematics always so interesting? Is it because of Zach Star?

In my country we describe fuel efficiency using liters per 100km, so taking the inverse was the first thing that came into my mind. I didn't expect the choice of units to make the reasoning that much easier and more intuitive.

Great content. You earned a subscriber!

Stating fuel"economy" is just making it needlessly unintuitive and errorprone. There is a reason why the vast majority uses the actual useful metric of fuelconsumption per distance traveled. Cause in general you have fixed distances and want to compare how much fuel you need for that distance. And given in this normal way it becomes quit obvious: Car A goes from 9.4L/100km to 6.7L/100km, and car B from 29.4 to 23.5 - also all but the upgraded car A are horrible.

first at 3:00 AM lol

4:41 We had that at school. And intuitively I would still answer incorrectly although I now know that it doesn't matter what radius the initial circle has. The difference to the new radius is the radius that results from the length of the extension as the circumference of the circle. The obvious reason for me to reject common sense as a justification.

For the cards, every spot in the deck has the same probability of being any of the values ace->king, I.e. 1/13. For the last card, all of the 1/13 for ace corresponds to it being the last ace, while for every other position in the deck, the 1/13 is shared between the first, second, third, last in some way.

The fact that you're doing physics with imperial units saddens me.

obviously, the last one's always a better pick, because you've seen all the previous cards, so it becomes a 100% chance if you go down to that end without seeing the 4th ace before. if the 3rd one is the one before the last card, there's also a 100% chance that the last one is an ace, so your odds only go up when you don't see an ace, until the 4-1 last cards depending of how many ace you've already seen. 😎 Assuming that there are 4 ace in the deck and the cards are revealed 1 by 1 instead of all at once.

This is awesome

I have an interesting example similar to the power cable thing: In Australia, it's popular for retired people to buy a camper or caravan (US: mobile home or trailer) and drive around the entire continent, usually in an anticlockwise direction (US: counter-clockwise). These people are nicknamed "grey (US: gray) nomads" and many will tell you that you should swap your tyres (US: tires) right with left and left with right because over such a long distance one side (the one farther from the centre (US: center) of the circuit will do many more kilometres (US: miles). I once told one of these grey nomads that it only travels about 9 1/2 metres further but no amount of demonstrating the maths (US: math) or logical argument could convince him.

Regarding the casino riddle - if you ask the question what number has any ace, any number you give has the same odds. By picking the last card you maximize the odds that actually matter

Another good one (though covered many times on YT, and probably most viewers already know) is the three doors or Monty Hall problem.

The last problem (the one with the cards) has a really neat prove (at least I hope so, but if I'm wrong, the comments will surly tell me): To prove that guessing that the forth ace is in the last spot is the best strategy, I will prove, that the number of possible permutations with the last ace being in the n-th spot strictly increases, when n increases (obviously only for n > 3, but since up to n = 3 there are no permutations, these can be discarded anyways). This means that the last probability will be the best (which than even could be generalized for any number of aces and cards). Lets call the number of permutations in the n-th spot Pn. P(n+1) than must be greater than Pn, since for the last n cards we have at least Pn possible permutations (all of them having the ace in last place, because we can just take the permutations with the ace in the n-th spot and add another card in the front). But then we can additionally shift the card from the front through the stack up to the second to last card (the last one always has to be the ace), creating more permutations. qed.

Great videos man, tbh when you were in the sand my furst thought was you were going to try a new beach style platform to present video lol

I want you to know I took your idea of the string around the earth to prepare a math lesson I will give to highschool students. Very fun problem for students to mess around with and try to figure out the answer both intuitively and mathematically

Two cars are making a trip of the same distance. One part of the trip includes going up and down a hill. One uses cruise control and drives the same speed the whole trip. The other car drives the same speed as the first car, except on the hill. It goes 10mph slower uphill and 10 mph faster downhill. Assuming the length of the uphill portion is the same as the length of the downhill portion, do the two cars arrive at the same time?

4:18 A more detailed explanation: Take x grams of water out, and you have 99-x grams of water, and 100-x grams total. To figure out the new percentage of water for any x, divide (99-x) by (100-x) For the purpose of this problem, we set (99-x)/(100-x) to 98/100. 99-x = 98 * (100 - x) / 100 = 98 - 98x/100 -x = (98 - 99) - 98x/100 98x/100 - x = -1 x - 98x/100 = 1 (1 - 98/100)x = 1 2x/100 = 1 2x=100 x=50 grams New mass = 100 - 50 = 50 grams.

One of my favorite counterintuitive math examples is from the 2015-16 NBA season, where for most of the season, the Warriors had the best three-point shooting percentage AND the best two-point shooting percentage, but it was the Spurs who had the best overall shooting percentage