When a genius misses an easy question

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  • Опубліковано 9 гру 2024

КОМЕНТАРІ • 431

  • @gordonweir5474
    @gordonweir5474 5 днів тому +101

    The video and what I see in the comments make it seem as though Einstein was fooled by the question. And yet at 11:31 he correctly calculated that there was no time available for the trip down. I would say that he got it!

    • @juncheok8579
      @juncheok8579 5 днів тому +11

      "Not until calculating did i notice..." implies he got tricked, even if it was for a few moments

    • @ThorfinnBus
      @ThorfinnBus 4 дні тому

      It means he was fully going calculation mode and wanted to subtract t trip-tascent. He was never tricked. He had the right method throughout​@@juncheok8579

    • @sonicwaveinfinitymiddwelle8555
      @sonicwaveinfinitymiddwelle8555 4 дні тому +3

      @@juncheok8579 yeah exactly, that's the human error everyone has when encountering a problem like 7 * 8 you have to calculate it in your head for a few moments

  • @jcortese3300
    @jcortese3300 5 днів тому +40

    Translation: "If I have to be at work by 8am, and I'm only halfway there when the clock strikes 8am, how fast do I have to drive to get to work on time?"
    YOU CAN'T. You will be late to work no matter how fast you go.

    • @Bigchickenburger
      @Bigchickenburger 4 дні тому

      If u go at infinite speed u can reach or just light speed

    • @Hunni125
      @Hunni125 4 дні тому

      or it could still be 8am when you arrive.

    • @Yehan-xt7cw
      @Yehan-xt7cw 4 дні тому +1

      What if the work place is in a timezone west of you? (could be as simple as a short drive crossing a border)

    • @tychozzyx9439
      @tychozzyx9439 4 дні тому +2

      You are not thinking with portals

    • @mrosskne
      @mrosskne 3 дні тому

      @@Hunni125 No, it could not.

  • @dinoeebastian
    @dinoeebastian 4 дні тому +11

    I kept trying to do Einstein's question, getting to the divide by 0 part, and assuming I did something wrong and starting over for like an hour

  • @TheRealMirCat
    @TheRealMirCat 5 днів тому +38

    Therefore, Officer, it was impossible for me to be speeding.

    • @scarletevans4474
      @scarletevans4474 4 дні тому +3

      I like how asking someone very bad at math could actually have them solve it immediately!
      Me: "I want you to solve this problem: [...]"
      Friend (bad at math): "Impossible!" (as impossible for him)
      Me: "That's the correct answer!"

    • @dylanwolf
      @dylanwolf 2 дні тому

      I drank six glasses of 5% ABV beer, you honour. But I only drank 95% of each glass.

  • @ovalteen4404
    @ovalteen4404 4 дні тому +11

    The sum of dice property is used to great effect in table-top games such as D&D. For instance, when rolling a character's stats, you roll 3 6-sided dice and sum them. The probability creates a bell curve that makes the midpoint between 1 and 18 the most common, which works out well at making 10 and 11 the most common stat, with either extreme (3 or 18) the least likely. So just as the "geniuses" are least common in reality, the rolling of a genius stat is also the least likely.

    • @asdfqwerty14587
      @asdfqwerty14587 4 дні тому

      Just a correction there - it is not a bell curve. With 2 dice it's actually just an upside down V with no curves at all, just 2 straight lines. With 3 dice it's a bit more complicated, but still isn't a normal distribution.
      Also, the mid point with 3 dice is half way between 3 and 18, not 1 and 18.

    • @mrosskne
      @mrosskne 3 дні тому +1

      @@asdfqwerty14587 He was referring to 3 dice, not 2, and he didn't say it was a normal distribution. He said it was a bell curve, which it is. This is an impressive amount of being wrong and simultaneously not responding to what the comment you're replying to actually said.

    • @asdfqwerty14587
      @asdfqwerty14587 3 дні тому

      @@mrosskne A bell curve is the normal distribution. That is the dictionary definition of it. Google the definition of it and link on the front page will say that it refers to a normal distribution, including the one coming from a dictionary.

    • @mrosskne
      @mrosskne 3 дні тому +2

      No, it's any curve shaped like a bell, sorry.

    • @Stubbari
      @Stubbari 3 дні тому

      @@mrosskne Astonishing how bad your reading comprehension is and how wrong you are.
      Bell curve means normal distribution, not a "curve shaped like a bell".
      And the comment replies to what OP said. Corrects the mistake of it not being a bell curve and corrects the range to be from 3 to 18 not 1 to 18.

  • @timsmith8489
    @timsmith8489 5 днів тому +72

    There's a story about someone giving John von Neumann this puzzle: Two trains 180 miles apart are on the same track going opposite directions toward each other, one traveling 10 miles/hour and the other 20 miles/hour. A fly starts on the front of one train and flies toward the other at 60 miles/hour. As soon as it reaches the other it turns around and flies back at that same speed to the first train. When it reaches that it reverses, and so on, until the fly finally is crushed when the two train collide. Question: how far does the fly fly?
    von Neumann instantly gives the correct answer. The person then said "Oh, you saw the trick! Most people just try to sum an infinite series!".
    von Neumann than said "I did sum the infinite series! There's a trick?"

    • @michaelz6555
      @michaelz6555 5 днів тому +10

      Just so we're all on the same page, it's 360 miles, right?
      Of course I solved the answer in my head and I also summed the series. But the series was the sum of the two speeds giving me the relative speed between the trains (30mph), giving me a time to collision (6hr), and the total distance traveled by the fly (60mph x 6hr = 360miles).

    • @akuunreach
      @akuunreach 5 днів тому +6

      I did it by combining the speeds, 30mph, 180miles, 6hrs, then how far the fly can get at 60mph.
      After solving, I realized, I could have solve as follows.
      Combined the speed, observe that the fly is traveling at double the speed, and note that it would cover double the distance, and arrived at 360mi

    • @markbothum4338
      @markbothum4338 4 дні тому +3

      Well, 180 miles covered at a combined speed of 30 mph takes 6 hours. Fly always travels at 60 mph during that time = 360 miles. Easy peasy.

    • @Nobody-tu5wt
      @Nobody-tu5wt 4 дні тому +2

      Harder question would be, how many times does the fly turns?

    • @akuunreach
      @akuunreach 4 дні тому +7

      @@Nobody-tu5wt As the 2 trains get closer and closer, the number of turns approach infinity if I'm not mistaken.

  • @nicholasharvey1232
    @nicholasharvey1232 5 днів тому +81

    6:57 I've seen this one before. In order to average 30 mph for a 2 mile trip, the car has to make that trip in 4 minutes. But the car already took 4 minutes to travel the first mile, therefore the car would have to teleport the rest of the way!
    If you average a speed of x for the first half of any trip, it is impossible to average a speed of 2x for the full trip.

    • @Sonny_McMacsson
      @Sonny_McMacsson 5 днів тому +3

      Einstein was figuring out HOW to do it, considering he's gonna need a wormhole.

    • @AlbertTheGamer-gk7sn
      @AlbertTheGamer-gk7sn 4 дні тому

      @@Sonny_McMacsson And then, there has been rumors that Einstein's spirit got cloned by magical equation sheets, so when he was proclaimed dead, his clone spirit quantum materialized a clone body and teleported underground into a base, where he would then change his name to "Doc Brown", and then he would then suffer a blow to the head which will cause him to use his theory of relativity to figure out the way to create a flux capacitor to make wormholes, as well as a hyperinfinipotent liquid known as "Time Juice" to power it, and then created a DeLorean Time Machine to the future, where he then drank a potion of immortality.

    • @logx-ow1us
      @logx-ow1us 4 дні тому +1

      Oh that’s why I was so confused. I kept using d=rt while looking at he thumbnail

    • @stevehorne5536
      @stevehorne5536 4 дні тому +5

      I've seen this before too, and my answer is that "average" is ambiguous for this - I'll stick with the arithmetic mean anyway, but if you average WRT distance instead of time you get that for the second mile the speed must have been 45mph. Remember - you can average anything WRT any variable you want. For what averaging speed WRT distance is (in the general case) that you integrate the (varying) speed WRT distance, then divide by total distance. When you're given averages for intervals that sum to the overall interval, that simplifies to a weighted average of those discrete intervals. When those smaller intervals are all equal, that simplifies to the sum-all-the-values-and-divide-by-the-number-of-values average - in this case (15+k)/2 = 30 which solves to k=45.
      Just because averaging WRT time is what you'd normally expect for problems involving speeds, doesn't mean it's the only possible option. People will say that speed is by definition an average given by distance per unit time, but they're wrong - speed is by definition the (calculus) derivative of distance WRT time, which can vary continuously, and while it's natural to average WRT time when using a derivative WRT time (giving you a convenient integral-is-the-inverse-of-derivative simplification), you can average WRT any variable you like.
      ["speed is by definition the (calculus) derivative of distance WRT time, which can vary continuously" is a slip-up, and unsurprisingly confusing. As mentioned in a much later comment, think in terms of a graph of speed WRT whatever - here, either time or distance, and averaging the height of the curve without caring what units are along the x-axis. The average is a speed irrespective - the units of the y-axis. The average WRT whatever based on integration is the integral of speed WRT whatever, divided by the total interval of whatever. The integral of speed WRT distance is in different units (distance squared per time) vs. the integral of speed WRT time (distance), but then you divide by the interval width along the x-axis in whatever units apply to that particular graph, and get back to speed either way. So WRT "but it's correct" below - well, no it wasn't (sigh) but that's an error in writing it out while half-asleep, not in the underlying principle.]
      It's not a popular answer, based on the last time I gave it, but it's correct - averages aren't even always arithmetic means (median, mode, geometric mean, harminic mean, many other kinds of mean) but, assuming the arithmetic mean, you can average over intervals of any kind you want, or you can even have averages of a list of numbers with no intervals specified (so assuming equal weight to each number) - again (15+45)/2 = 30. Personally, I'd consider my answer wrong normally (we generally know what is meant by "average speed" - ie. WRT time), but if someone asks an absurd question, I say exploit any ambiguity because ignoring conventional interpretation is the lesser absurdity.
      Averaging per distance travelled makes a lot more sense in reality when looking at fuel use - many road conditions vary with the interval of distance you're travelling on (others varying with time due to weather and traffic, others varying depending on the mood of the driver). Gallons per mile times miles = total gallons.

    • @nicholasharvey1232
      @nicholasharvey1232 4 дні тому +2

      @@stevehorne5536 He explained why 45 is incorrect, you must go by total distance divided by total time to get average speed, even if the car drove 200 mph down the hill, the 4 minutes required for a 30 mph average speed are already "used up" so as additional seconds tick by, so the maximum possible average speed of the trip ticks down from that 30 mph target. After 8 minutes, it similarly becomes impossible to average 15 mph for the trip, because the required time threshold has already been passed. And as additional minutes pass from there, the maximum possible average speed gets lower and lower, no matter how fast the trip is completed.

  • @plentyofpaper
    @plentyofpaper 5 днів тому +18

    The hill problem is something that I've often thought about as both a bicyclist and a runner.
    The problem is intuitive if you're talking about differences in speed over fixed amounts of time, but not distance.
    This is basically why dealing with hills, even on a round trip is more of a pain than just a level path.

    • @roberteltze4850
      @roberteltze4850 4 дні тому

      When I was young I used to run such calculations while staring at my head unit on my bike. These days I just ride without caring much about the numbers.

    • @mrosskne
      @mrosskne 3 дні тому

      The trip problem has nothing at all to do with hills, that's basically a red herring. The result is the same if you're on flat ground.

    • @plentyofpaper
      @plentyofpaper 3 дні тому +1

      @@mrosskne That would imply that the speed increase of going downhill is greater than the speed decrease of going uphill.
      Do you assert that to be true? I can't say it's wrong for sure, bit it strikes me as counterintuitive.

    • @mrosskne
      @mrosskne 3 дні тому

      @@plentyofpaper It doesn't matter. The point of the problem is that you travel one speed for the first leg and a different speed for the second. It could be talking about a submarine or a spaceship. The problem's flavor is irrelevant.

    • @plentyofpaper
      @plentyofpaper 3 дні тому

      @@mrosskne The point of the problem is that if an equal distance is traveled at 2 different speeds, the slower speed will be closer to the average than the faster speed. Because the slower trip takes longer.
      I honestly have no idea what you're trying to get at.

  • @markofdistinction6094
    @markofdistinction6094 4 дні тому +17

    I once thought that I made a mistake ... but I was wrong.

    • @c.l.4895
      @c.l.4895 4 дні тому +1

      As a middle aged white man on the internet, I’ve never been wrong.

    • @flanjunk
      @flanjunk 3 дні тому +1

      The correct expression is:
      "I'm always right. I thought I was wrong once, but I was mistaken"

  • @Serg_144
    @Serg_144 День тому +2

    The Feynman question has a problem in it. a, b, c are not independent of each other. That's why you get such different answers.
    Construct only a segment of length = a. From there if you draw a line straight up to the intersection with the circle, and from there strictly left, you can see that b and c are uniquely determined because of just fixing the a value.

  • @CatholicSatan
    @CatholicSatan 5 днів тому +19

    Hah! That Einstein problem... I was getting annoyed with myself because I was getting an infinite speed/zero time for the descent and thought I was calculating it wrong. And then I thought, nah, it's a trick question 🙂

    • @hrayz
      @hrayz 4 дні тому

      Like reverse psychology, it's a reverse trick question. Ie: Not the trick you're thinking of!! 🤣

    • @Goabnb94
      @Goabnb94 День тому

      I was sure there was an answer, and so after confusing myself with the fact both require the same time, I watched the video, to find I was not actually mistaken.

  • @andrewewart7166
    @andrewewart7166 5 днів тому +5

    For the first puzzle, given the 2 results you can get a third result for r in terms of a and c only which comes to r = (a+c) + sqrt(2ac) or r = (a+c) - sqrt(2ac) (both values work as a,c > 0, so a+c > sqrt(2ac))

    • @Cassu1987
      @Cassu1987 4 дні тому

      However, for the second solution the side lengths of the rectangle are a - sqrt(2ac) and c - sqrt(2ac) and at least one of them is guaranteed to be negative, which we probably don't want to accept.

  • @reinisliepins6177
    @reinisliepins6177 5 днів тому +30

    I feel smart when i instantly got the dice problem.

  • @acem82
    @acem82 4 дні тому +3

    6:46 ...I "invented" this problem a few years ago, trying to figure out why if I could ride my bike at 20 mph on the flat, why it was so hard to do so on a hilly course, if you get back all the energy you gained by going up the hill. I realized, that if the first half of the trip was up a hill you could only go up at 10 mph, you'd need to teleport down the hill to average 20 mph, and that allowed me to realize why the 20 mph assumption was wrong!

  • @ashtonaughtband
    @ashtonaughtband 4 дні тому +5

    Technically with time dilation it is possible and you wouldn't need to hit the speed of light because there is also very very very slight time dilation on the ascent as well. So it would be like (4 -(10^-32)) mins for the accent and 1O^-32 mins for the descent.

    • @ericpaul4575
      @ericpaul4575 4 дні тому

      So what speed is that on decent?

    • @tezzerii
      @tezzerii 4 дні тому

      @ashtonaughtband Except that you're in an old car that can only manage 15mph up the hill. =o)

    • @ashtonaughtband
      @ashtonaughtband 4 дні тому

      @@ericpaul4575 so I think you save 1.44 e-13 seconds on the ascent so I think you'd have to travel the equivalent of 6.944e12 miles per sec (without time dilation) I'm not sure how to convert that back to relative light speed (with time dilation) or I guess unwilling to figure it out.

    • @OhhCrapGuy
      @OhhCrapGuy 4 дні тому

      ​@@ashtonaughtbandyou just need to play with the Lorentz factor. I'll calculate it when I get up from my nap, if I remember

    • @mrosskne
      @mrosskne 3 дні тому

      Yes, that's why the simple arithmetic problem didn't mention time dilation. Because you were supposed to consider time dilation. Christ.

  • @planethedgehog2427
    @planethedgehog2427 3 дні тому +9

    5:15 "Imagine we roll one *dice."* Nope, I can't. But I can imagine rolling one *die.* 🎲

    • @ollyrukes
      @ollyrukes День тому

      “Dice” is now totally accepted as a singular term.

  • @dhpbear2
    @dhpbear2 5 днів тому +5

    "I was going 70 miles an hour and got stopped by a cop who said, "Do you know the speed limit is 55 miles per hour?" "Yes, officer, I know but I wasn't going to be out that long..."
    -- Stephen Wright

    • @lpr5269
      @lpr5269 День тому

      I think that would work in court. 😂😂

  • @cret859
    @cret859 День тому

    3:53 Your equation is ambiguous, you may have write it as r =0*a + b + 0*c or r = (0)(a)+ b + (0)(c) nut not r = a(0) + b + c(0) since a(0) and c(0) may be confuse with "value of function a and c" at abscisse 0 exactly as we often write f(x).

  • @mjorge0alves
    @mjorge0alves 4 дні тому

    I love your videos. They are beautifully designed and presented in an absolutely clear way. I'd just point out two mistakes:
    1. The symbol for hour is just (upright or roman) h
    2. Since time depends on the observer's perspective, going at light speed does solve the problem, assuming the one person interested in completing the trip on such average speed is the driver. From his perspective, going at light speed does take him to the 2-mile mark instantly, whereas an observer (e.g., someone who had an appointment with him) would be pissed due to him arrving a fraction if a second late.

    • @tezzerii
      @tezzerii 4 дні тому

      Here's the thing though - it's an old car and can't manage more than 15mph up the hill ! So I reckon light speed is literally out of the question. =o)

    • @mjorge0alves
      @mjorge0alves 4 дні тому

      @tezzerii that seems likely. Also, technically, a wheeled vehicle can only reach half light speed. This is because the highest speeds on a vehicle are those of the upper part of the wheels. If those where at light speed, their center would be at light speed divided by two.

    • @mrosskne
      @mrosskne 3 дні тому

      It's an arithmetic problem, not a relativity problem. Get a clue.

    • @mjorge0alves
      @mjorge0alves 3 дні тому

      It is a physics (kinematics) problem, @mrosskne. Make sure you are at least right before correcting others.

  • @dache85
    @dache85 5 днів тому +19

    7:59 i remember one video where you argued "half of two plus two" is 1/2 * 2 + 2 and not 1/2 * (2 + 2), because 1/2 (2 + 2) would be said as "half of *the sum* of two and two" or "half of *the quantity* two plus two" but look where we are now

    • @mrosskne
      @mrosskne 3 дні тому

      Words aren't precise, that's why we don't use them for math.
      You can avoid ambiguity by saying "one half, multiply, open bracket, two, add, two, close bracket". But then you might as well just use the symbols.

    • @dache85
      @dache85 3 дні тому

      @@mrosskne i just remember watching the "half of two plus two" video and they sounded very serious about it not being 1/2 * (2 + 2) and spent like half the video arguing so

    • @mrosskne
      @mrosskne 3 дні тому

      ok. do you have an argument to contribute?

    • @dache85
      @dache85 3 дні тому

      @@mrosskne search "half of two plus two" on their channel

    • @Goabnb94
      @Goabnb94 День тому

      @@mrosskne I believe the video being questioned was the use of the divide sign, AND juxtaposition of a(b+c), and whether people treat that as a*(b+c) verses (a*(b+c)), and was a lot of controversy because people learn to treat those differently, thus pointing out the problem with using such symbols. So to the point OP is making, half of 15 plus x, hasn't been explicitly stated as the sum of 15 and x, and should, by strict adherence to BEDMAS (or whatever you learned), be treated at half of 15, plus x.

  • @solandri69
    @solandri69 5 днів тому +20

    Averaging speeds over the same distance requires taking a harmonic mean.
    2/v_avg = 1/v1 + 1/v2
    The reason is that you've defined the problem as saying the two legs of the trip cover the same _distance._ So to calculate it, the _distance_ needs to be in the denominator (since they're the same distance), just like when you add two fractions they need to have the same denominator. Since miles/hr has distance in the numerator, you need to flip it to put distance in the denominator before calculating the average.
    If the problem had been the car spends 1 hour driving 15 mph, and 1 hour driving v2 mph, then the time is equal for both legs, so time needs to go in the denominator. Which it already is for mph, so you can solve _that_ as v_avg = (v1 + v2)/2. And v2 = 45 mph would be correct.
    Same issue crops up with fuel efficiency in the U.S. (which uses miles per gallon). If you've got a car which gets 20 mpg and another car which gets 40 mpg and you drive both cars the same distance over the year, then the average mpg is the harmonic mean. Resulting in an average mileage of 26.7 mpg.
    OTOH, if you put 15 gallons in the 20 mpg car, and 15 gallons in the 40 mpg car (same gallons, instead of same distance), and drive both until their tank is empty, then the average mileage will be 30 mpg.
    Since the vast majority of driving you do is based on distance, not how much fuel you have, the rest of the world uses "liters per 100 km" to avoid this inversion issue (puts distance in the denominator).

    • @Llortnerof
      @Llortnerof 4 дні тому

      ...and because they use metric instead of (Hamburger*Eagle)/Shotgun^Footballfields.

    • @mrosskne
      @mrosskne 3 дні тому

      None of this is necessary.
      If you want an average speed of 30 mph (~45 fps) over 2 miles (~10600 ft), the trip must take ~236 seconds.
      You traveled 5300 ft at 22.5 fps, so it took you 236 seconds. You have zero seconds to travel the remaining distance. If you want to reach an average speed of 30 mph, it will take you more than one mile of additional travel to reach it, no matter how fast you accelerate.

    • @johnbaldwin2948
      @johnbaldwin2948 3 дні тому +1

      I agree with you...time has nothing to do with this problem. It's all about speed and distance. You go 15 for half and you have to go 45 the second half to average the SPEED. Who cares that one half takes 1/3 the time to complete. Speed is all that was asked and all that matters. Stick to the same unit of measure...don't bring in something that doesn't matter and use it to "prove" the correct solution wrong.
      If you are driving down the highway at 50 for a mile...and then for another mile you travel at 60...your average speed is 55. How long it took isn't the question.

  • @Vex-MTG
    @Vex-MTG 4 дні тому +2

    The first question felt so obvious that I was SURE I was doing it wrong, especially when you went into your explanation. Nope, turns out it was just obvious. ;)

    • @bobross7473
      @bobross7473 4 дні тому

      It’s a trick question. The diagram is not an accurate representation of a quarter circle so there can be no radius

    • @Vex-MTG
      @Vex-MTG 4 дні тому

      @bobross7473 diagrams in math questions are rarely to scale. That's not a trick, that's a convention

  • @powerofk
    @powerofk 2 години тому

    I actually came a variation of the Einstein problem decades ago (as a kid), so I already knew the answer to it. But it’s also a good teaching tool-the “average” isn’t always the arithmetic mean. Especially when dealing with complex units (a complex unit is one that combines 2 or more basic units).

  • @danielbranscombe6662
    @danielbranscombe6662 5 днів тому +7

    So Feynman's solution actually presents a restriction on viable values for a,b,c. If you instead use b for r in his equation and then solve for b you get that
    b=a+c-sqrt(2ac)
    so for any other value of b given a,c then the problem is unsolvable because the combination is invalid.
    For example if a=2,c=9 then we get b=17. If instead we are given instead b=19 then there is no possible rectangle with diagonal 19 such that to given quarter circle has the extra lengths a,c as given.

  • @RobbieHatley
    @RobbieHatley 2 дні тому

    Regarding Feynman's blunder, an interesting result is found when starting with the assumption that b=r, then expressing r in terms of a and c, by using the Pythagorean formula, then completing the square instead of using the quadratic formula:
    (r-c)**2 + (r-a)**2 = r**2
    r**2 - 2cr +c**2 + r**2 -2ar +a**2 = r**2
    r**2 - 2cr +c**2 -2ar +a**2 = 0
    r**2 -2(a+c)r +a**2 +c**2 = 0
    r**2 -2(a+c)r +a**2 +c**2 + 2ac = 2ac
    r**2 -2(a+c)r +(a+c)**2 = 2ac
    (r-(a+c))**2=2ac
    r=a+c+sqrt(2ac)
    Seeing as how a=r*versine(θ) and c=r*coversin(θ) (where "versine" is 1-cos and "coversine" is 1-sin), the above result would seem to imply the trig identity (1-ver-cvs)**2 = 2ver*cvs. But is that actually true? Let's check:
    (1-ver-cvs)**2
    = (1 -(1-cos) -(1-sin))**2
    = (-1+(cos+sin))**2
    = 1 -2(cos+sin) +(cos+sin)**2
    = 1 -2cos -2sin +cos**2 +2cos*sin +sin**2
    = 2 -2cos -2sin +2cos*sin
    = 2(1-cos-sin+cos*sin)
    = 2(1-cos)(1-sin)
    = 2ver*cvs
    QED. New trig identity learned: (1-ver-cvs)**2 = 2ver*cvs

  • @reddblackjack
    @reddblackjack 4 дні тому

    I must be better at math than I thought. I had the b = r almost immediately. I've played craps in Vegas after reading a book on how to do it and already know that 11 is twice as likely as 12. I thought that the last one was a trick right away, too. It just didn't register exactly how it was tricky until the math was done. Another great video Talwalker!

  • @satrajitghosh8162
    @satrajitghosh8162 14 годин тому

    Trivially r = b
    ( b - a)^2 + ( b - c)^2 = b^2
    b ^ 2 - 2 ( a + c) b + a^2 + c ^2 = 0
    [ b - ( a + c) ] ^2 = 2 a c
    b = a + c + √ ( 2 a c) ,
    a + c - √ ( 2 a c) ,
    Hereby radius is
    r = b = a + c + √ ( 2 a c) ,
    or r = b = a + c - √ ( 2 a c) ,
    Second problem
    Uphill distance = downhill distance
    = D, say
    2 D / 30 = D / 15 + D / v
    so it would require v = INFINITE

  • @kane_lives
    @kane_lives 4 дні тому +3

    "Never try! Trying is the 1st step to failure!" -- Grandmaster Ben Finegold

  • @johng.1703
    @johng.1703 5 днів тому +4

    looking at the car problem, to get a 30mph average, and you can only do 15mph for the 1st half of the trip you would have 0 time to do the 2nd mile, it takes the same amount of time to travel 2*D at 2*T as it does to travel 1*D at 1*T.
    this problem is deceptive in that it is really a time problem, not a speed problem.

  • @AudioMusicElectronics
    @AudioMusicElectronics 13 годин тому

    For the ones who have been averaging the speeds: The cause of confusion here is that there are two possible physical quantities that this special case question can talk about.1. Equal distance segments: You cannot simply average the speeds. 2. Equal time segments: say, 15 miles per hour for 1 hour and 45 miles per hour for 1 hour (as an example), here you can simply average the two speeds.

  • @kenmore01
    @kenmore01 5 днів тому +5

    Einstein got it right. Thats what the video said. Also, he had time left to make bagels!

  • @RavenMobile
    @RavenMobile 3 дні тому

    I came up with 45 mph for the second half, just like you mentioned. Then when you said that was the wrong answer most people get, I whipped out my calculator and started poking at it to see what I could come up with. I worked out that the second half of the trip would need to be infinitely fast.

  • @SiqueScarface
    @SiqueScarface 5 днів тому

    When looking at the radius problem, I didn't notice the r=b shortcut at first. But when I looked at the quadratic solution, I immediately saw the 2ac - a² - c², and in my head converted it to -(a-c)², and now I had 2b²-(a-c)² and thus b² + b²-(a-c)² = b² + (a+b-c)(a+b+c), and slowly, it dawned me, that I missed something.

  • @SmileyEmoji42
    @SmileyEmoji42 2 дні тому

    IMHO it is best to consider units. In the wrong solution the 2 must be miles (or else you woud clearly be assuming that the different distances were irrelevant, which is obviously wrong) and mph/m = 1/h which is not a speed.

  • @ryotoiii
    @ryotoiii 16 годин тому

    A way that I think of the dice probability problem is through rolling one dice at a time
    Going through cases, if you roll a 5 first, then you only have a chance to roll an 11. 12 is now impossible
    If you roll a 6 first, then you have a chance to roll a 12, BUT you still have a chance of rolling an 11.
    If this is still hard to visualize, you can also imagine the only numbers on the dice are 5 and 6 and from there do the classic exhaustion of every possibility (don’t know if exhaustion is the right word, but it seemed to fit)

    • @Stubbari
      @Stubbari 15 годин тому

      You could also make it a coinflip. Which is more likely: to flip one heads and one tails or to flip 2 heads.

  • @SJrad
    @SJrad 4 дні тому

    well because the other diagonal would also be b, it would be a straight line going from the center to the edge of the quarter circle, which means b is the radius of the circle. as for a and c, idk lol maybe somehow would be able to compute it in terms of r and b using trig functions.

  • @silver6054
    @silver6054 2 дні тому

    So the first question, the answer is independent of a and c, which is not immediately obvious in the solution of the quadratic. I guess there is some interplay between a, b and c that would make the a and c terms cancel

  • @obiwanpez
    @obiwanpez 2 дні тому

    I opted to change the relative speeds to 30 mph for the first half, and 60 mph average. Makes it easier to think in terms of miles/minute.

  • @Billyblue98
    @Billyblue98 2 дні тому

    The thumbnail question caught my attention, so I thought I'd do it out in my head before watching the video, but . . . . . This is infinity?
    Okay, so. . . 15 mph for 1 mile. Trip lasts another mile, 2 miles total. Average speed for total trip is 30 mph
    In my head I went "Okay, so 2 miles would be 1/15 of 30 miles. 1/15 of an hour is 4 minutes. So the total trip has to be 4 minutes"
    Then I looked at the first mile "Okay so that's also a rate of 1/15, so the first mile has to take 4 minutes. . . Wait a minute"
    Lemme do the problem another way just to be sure I'm doing it right (though, from the title I think I'm doing it correctly)
    We know the speed and distance for both the first mile and the total trip
    D/T=V. 1 mile / T{1} = 15 mph. 1 mile = 15 mph x T{1}. 1 mile / 15 mph = T{1}. 1/15 hour = T{1}
    2 miles / T{2} = 30 mph. 2 miles = 30 mph x T{2}. 2 miles / 30 mph = T{2}. 2/30 = T{2}. 1/15 hour = T{2}
    The first mile takes the same amount of time as the total trip of two miles. The second mile must be completed instantaneously. The second mile has a velocity of infinity
    1 mile / 0 hours = V{3}

  • @killerjg
    @killerjg 3 дні тому

    Last question depends on if you are the observer or not, if you are not the observer you just need to instantly travel to the bottom. You can go at the speed of light if you are the one in the car and get to the bottom instantly. (assuming the car is massless.)
    You don't need to go faster than the speed of light you just need to travel at the speed of light if you are not the Observer.

  • @antonvanopstal
    @antonvanopstal 4 дні тому

    I've seen this riddle before and had hoped for a relativistic approach to the solution

  • @Kualinar
    @Kualinar 4 дні тому +1

    Question in the thumbnail... I went directly to : How much time it take for the first part (4 minutes) and how much time do I have for the full trip (also 4 minutes) ?

  • @jaykoni
    @jaykoni Годину тому

    You forgot about time dilation. If you travel at c on the way down, no time will pass for the car, so total trip time is still 4 minutes.

  • @Hypericus2
    @Hypericus2 3 дні тому

    Strictly the Leibniz problem is ambiguous: if you don't assume commutivity then you can interpret "a sum of" to be "a specific sum of" so that 5+6 is not the same sum as 6+5.

    • @Stubbari
      @Stubbari 2 дні тому

      How is it ambigous? Rolling 11 is more likely and that can easily be tested.

  • @WhoStoleMyAlias
    @WhoStoleMyAlias 4 дні тому

    Haha! This was a dead giveaway to me but out of interest I subjected my family to this riddle and for sure my wife responded 45mph. Next my kid who is in second grade middle school gave the exact same answer, so I asked him if he could calculate what the resulting times for both distances would be for the given average speeds. For sure he did calculate it was 4 minutes for both but due to the nature of the question he then concluded that the additional mile had to be run in less than a minute and ended up answering 60mph. So next I asked him if he could use seconds instead of minutes and the big question mark started rising above his head - ehm 3600mph? Granted I don't really know if he already learned the concept of infinity slash division by zero, I think back then I already had it in first grade, but I believe he really did get it when I gave him the answer.

  • @flamesintheattic
    @flamesintheattic 4 дні тому

    The speed for the second mile increases exponentially with every second spent driving under 30mph and hits infinity at 4min. If the car spent even a tiny fraction of a second driving faster, then a correct speed could be calculated to solve the puzzle. For example 15.0627mph for the first mile would leave 1 second at a speed of 3600mph to average out to 30mph for the whole trip.

    • @tezzerii
      @tezzerii 4 дні тому

      In an old car that can only manage 15mph up the hill . . .

    • @flamesintheattic
      @flamesintheattic День тому

      @@tezzerii Well the whole question doesn't make any sense since a car has to accelerate/decelerate in the first place.

  • @angrytedtalks
    @angrytedtalks 5 днів тому

    I didn't overthink the first one, it seemed obvious using trig that r=b.
    The second one is deeply obvious unless you think 6+5 and 5+6 are different.
    The last one takes a moment to realise the average speed is already sucked up by the ascent time.
    Sometimes overthinking or expectation stops the otherwise obvious from being apparent.

  • @kilroy1964
    @kilroy1964 2 дні тому

    I came across the radius problem on an IQ test, many years ago. I solved it instantly. Take that Feynman!

  • @tomekiriazi8736
    @tomekiriazi8736 5 днів тому

    The "famous right triangle theorem" made me LOL 😂😂

  • @BsktImp
    @BsktImp 2 дні тому

    11:04 There's a finite probability that the car was always on the other side of the hill, so to an observer it looked like it made the journey in 4 mins. Tunnelling. Ba-dum-tss!

  • @robertjarman3703
    @robertjarman3703 4 дні тому

    You know, I actually thought about it using the second method you were talking about, with the idea of 4 minutes. I am not the best at math in my head for equations like this, but I did notice that it seemed like there wasn't enough time remaining to go that fast although I wasn't sure exactly why.

  • @panyachunnanonda6274
    @panyachunnanonda6274 4 дні тому

    Wow , so amazing technique of puzzle #1.

  • @dxjxc91
    @dxjxc91 22 години тому

    My "haven't watched the video yet" answer:
    Isn't the answer undefined (Basically infinity)? Speed is distance over time, to double the speed you need to double the total distance or halve the total time. We covered half the distance already, so the total distance is double, but we also used 100% of the time. We need to travel 1 mile in 0 time to double the average speed. 1÷0 mph. Undefined.
    The instinctual answer is probably 45, but that averages 22.5mph because time is the denominator and the time is not constant between the 2 legs.

  • @DrAndyShick
    @DrAndyShick 3 дні тому

    ultimately, the equation becomes 1/30 = 1/30 + 1/(2v). In this, 1/(2v) must equal zero, which is not possible

  • @Omnifarious0
    @Omnifarious0 4 дні тому +1

    Thumbnail problem: Infinity. The car would have to instantly travel the remaining mile in order for it to average 30mph for the whole trip.
    Circle radius problem. The answer is just r = b. The two diagonals of a rectangle are of equal length. One diagonal is the radius. The other you've labeled b.

    • @bobross7473
      @bobross7473 4 дні тому

      Circle radius is a trick question, there is no quarter circle because the two radii are unequal

    • @Omnifarious0
      @Omnifarious0 4 дні тому

      @@bobross7473 - Perhaps. I've encountered other trick problems that are effectively "I purposefully drew this inaccurately in a misleading way.".

  • @malcolmtaylor1224
    @malcolmtaylor1224 4 дні тому +36

    It isn't necessary to travel at an infinite speed for the trip down. Just travel at the speed of light and time stands still.

    • @Dysan72
      @Dysan72 4 дні тому +10

      At that point you need to ask what clock you are going by. Because by the clocks of the rest of the universe it still takes you some time to travel that last bit, even at the speed of light.

    • @rogerkearns8094
      @rogerkearns8094 4 дні тому +1

      Love pedantry. ;)
      Still, either way, the given problem has no practical solution..

    • @simpleminded1uk
      @simpleminded1uk 4 дні тому +3

      The answer is C. Even though it isn't a multiple choice question.

    • @tezzerii
      @tezzerii 4 дні тому

      In an old car which can only manage 15mph up the hill.

    • @asdfqwerty14587
      @asdfqwerty14587 4 дні тому +2

      There are a few problems with this.
      1) If you're moving at/very close to the speed of light and you're talking about the frame of reference of the car, then the distance will not be 1 mile anymore. Relativity causes the lengths to change too, not just time. Even if you were talking about a wacky frame of reference the math would still work out the same and it would still be impossible. The only way you could come up with different answers is if you're talking about things like trying to calculate the speed of an object by taking the distance from 1 frame of reference and the time from a different frame of reference, which is a nonsense measurement which doesn't actually calculate anything useful (by that logic you could come up with literally any number, you could even say the car is travelling backwards if you wanted to).
      2) If you're talking about the frame of reference of the object itself, then "the car isn't moving at all" - there's no such thing as measuring the speed of an object from its own frame of reference, because obviously the car is always at the same position relative to itself.

  • @marcelluswallace6240
    @marcelluswallace6240 4 дні тому

    There are several lines of thought to immediately get the answer to the Einstein puzzle. One of them is: If you you have a certain speed for a given distance and you want double the speed on double the distance then of course the time has to be exactly the same. This is true for any non-relativistic speed and any non-relativistic distance.
    If it's easier for you to calculate with certain numbers. Just imagine the first part of the hill to be exactly 15 retard units. Then the ascent takes 1 hour. The total distance would be 30 retard units and hence you also have 1 hour for the full distance to get the desired average speed. Which is of course impossible unless you are a massless particle...

  • @paulnieuwkamp8067
    @paulnieuwkamp8067 4 дні тому

    Just look at the hill from a physics perspective instead of a math perspective. Due to significant figures you might make it if you go REALLY fast (for a car) on the descent: if the distances are 0.96 and 1.96 miles and the averages are 15.4 and 29.6 mph, but all written down with two significant figures, you would make it if you drove 2.6E+2 mph average, or accelerate with 1.6G if we don't account for stopping at the end (if I did all that correctly. Knowing my prowess with math, I probably made an error or two :P ).
    On the other hand, you could also be out of time before you even get to the top. Your mileage may vary...

  • @didgeoridoo
    @didgeoridoo День тому

    There isn’t enough information to answer the hill question, because it doesn’t specify a reference frame.
    From the perspective of an outside observer, the fastest the car can cover the distance is limited by the speed of light, so it is indeed impossible to cover the distance in zero seconds.
    However, from the perspective of the car itself, there is a solution. If the car instantly accelerated to the speed of light (ignoring the infinite energy required to achieve this), it would subjectively experience no passage of time, and the distance would contract to 0. This would allow the remaining distance to be traversed instantaneously.
    Surprised Einstein didn’t trollishly pull this one out of his hat as soon as he found himself dividing by zero.

    • @Stubbari
      @Stubbari День тому

      Never heard an OLD car being able to go the speed of light

  • @WillRennar
    @WillRennar 4 дні тому

    I refer to it as "a 5th-gear brain working on a 2nd-gear problem." If you think of the brain as a car engine, you can think of working on higher-end problems as being in a higher gear. Solving a simple problem with your brain in low gear is like trying to drive at a slow speed in a low gear; the engine handles it just fine. Trying to drive slow in a high gear, however, causes so few RPMs that the engine stalls and dies out, so it can't do it. ...A bit of a weird analogy, I admit, but it seems to get the point across.

  • @TheRealMrBlackCat
    @TheRealMrBlackCat 6 годин тому

    "Tricked"... in other words deceived intentionally, or unintentionally, so the failure belongs to the person stating the question.

  • @victorfinberg8595
    @victorfinberg8595 4 дні тому

    6:35
    mark antony disagrees:
    "the evil men do lives after them; the good is oft interred with their bones"

  • @robertdubard7959
    @robertdubard7959 5 днів тому +1

    Every one of these is trivially easy, IF you look at them a certain way.

  • @srf2112
    @srf2112 3 дні тому

    My guess is 45 mph. I did it in my head reading the thumbnail. I have not watched the video. Am I right? Time to see.
    Update: Dohhhh!🤦‍♂

  • @andyespinozam
    @andyespinozam 5 днів тому

    Damn; you had me with the last one.

  • @mrosskne
    @mrosskne 3 дні тому

    If you want an average speed of 30 mph (~45 fps) over 2 miles (~10600 ft), the trip must take ~236 seconds.
    You traveled 5300 ft at 22.5 fps, so it took you 236 seconds. You have zero seconds to travel the remaining distance. If you want to reach an average speed of 30 mph, it will take you more than one mile of additional travel to reach it, no matter how fast you accelerate.

  • @thomashoglund5671
    @thomashoglund5671 2 дні тому

    "If you make a mistake, it's totally fine. Try, try again." - Presh Talwalkar
    "You tried and you failed miserably. The lesson is: Never try!" - Homer Simpson

  • @d007ization
    @d007ization 4 дні тому

    I vaguely knew about the Einstein riddle but I never figured out why it truly was the case.
    Of course, students might have an easier time understanding it by considering that a single B on their record means they'll never have true straight As.

  • @Raven-Creations
    @Raven-Creations 4 дні тому

    For the thumbnail question, no calculation is needed. If you cover double the distance in the same time, you've doubled the average speed. Therefore, you have to cover the second half in zero time, which means you need an infinite speed. This is conceivably possible if there's a wormhole between the top of the hill and the bottom, but it's otherwise impossible.

  • @hookahkid
    @hookahkid 3 дні тому +1

    If he travels at the speed of light, then no time passes for him, so the answer is approx 670,600,000 mph.

  • @kirkjohnson6638
    @kirkjohnson6638 22 години тому

    The speed would have to be infinite since a 30mph avg over 2 miles takes 4 minutes and so does a 15mph average over one mile. Since the allotted time was already all used up, the second mile would have to cover 1 mile in zero time.

  • @birchy8305
    @birchy8305 4 дні тому +1

    On the 1st one, where does a^2 + c^2 - b^2 = 2 come from?

  • @98.11Deet
    @98.11Deet 3 дні тому

    Just to reword the question before watching:
    If a car travels one mile in four minutes, what speed does it need to go to cover a second mile in zero additional minutes?
    In Star Trek, they call it warp ten.

  • @amethystjean1744
    @amethystjean1744 3 дні тому

    15 mph is 4 min mile. An average of 30 mph is 2 min mile or 4 min for 2 miles. So there is no answer that can work as we have already used 4 min in the first half of the trip.

  • @marvhollingworth663
    @marvhollingworth663 5 днів тому

    I got the dice thing & I said the car would have to teleport instantaneously to the bottom. (I've heard a version of the last puzzle before, but I got the answer then too.) My answer to the 1st puzzle was a bit more complicated, I should have noticed r = b. I said r = sqrt (b^2- (r-a) ^2) +c & r = sqrt (b^2- (r-c) ^2) +a. This seems a valid answer to me, but what do you think?

  • @SarahAbramova
    @SarahAbramova 5 днів тому

    Thank you for this video!

  • @quigonkenny
    @quigonkenny 5 днів тому

    But from whose point of reference are we observing the car going up and down the hill? If that observer is moving fast enough, its quite easy to only take 4 min for the entire trip...

  • @jerimiahbrown9629
    @jerimiahbrown9629 10 годин тому

    Before watching, just looking at the thumb, I’m saying that this is impossible because to average 30mph the 2 mile trip must be done in 4 minutes, however it has already taken 4 minutes for the first mile at 15mph

  • @gonzalotapia1250
    @gonzalotapia1250 4 дні тому

    The first one... if you draw the diagonal in the other direction then you get the radius is....B.
    The second one: Nope. Theres 1 way to get 12 and 2 ways of getting 11. (5-6) and (6,5)

    • @Stubbari
      @Stubbari 4 дні тому

      Wrong.
      There are 2 ways of getting 11: 5+6 and 6+5

    • @Raven-Creations
      @Raven-Creations 4 дні тому

      @@Stubbari He said there were two ways. There was just a typo putting a hyphen instead of a comma. It was perfectly obvious what he meant.

    • @Stubbari
      @Stubbari 4 дні тому

      @@Raven-Creations Yep, I see he edited the mistake.

  • @j4s0n39
    @j4s0n39 2 дні тому

    Dice is the plural word. The singular is die.

  • @rgmolpus
    @rgmolpus 4 дні тому

    If the back side of the hill is a cliff, then the car can make the trip in the required time - 1 mile down won't take that long.

  • @UncleKennysPlace
    @UncleKennysPlace 5 днів тому +2

    My father showed us the first one, a half-century back.

  • @samuelgionet529
    @samuelgionet529 4 дні тому

    My intuition said 60 mph, because if we go half the speed in the ascent, then we must double the speed in the descent. Alas, calculating it gave an overall speed of 24 mph. Very interesting problem!

  • @mashmachine4087
    @mashmachine4087 22 години тому

    I got the b=r one instantly but that's probably because Feynman was primed for it to be a maths puzzle and tunnel visioned himself (I assume, I don't know anything about him outside that)

  • @cjburian1
    @cjburian1 2 дні тому

    Einstein wasn't tricked by the trick question, but neither did he intuitively recognize the problem and jump to the conclusion. Suppose you give yourself enough time to drive to an appointment while going the speed limit, but encounter traffic dragging along at half the speed limit for half the total distance. Then the road clears and you can get away with driving as fast as you please for the second portion of the journey--will you punch it to the floor and actually hold out (futile) hope that you might still arrive on time? Or intuitively realize without doing any calculation that your car can't go infinitely fast?

  • @jdlessl
    @jdlessl 4 дні тому

    Oh thank goodness. I clued in on r=b immediately, but thought I must have missed something.

  • @banibratamanna5446
    @banibratamanna5446 4 дні тому

    for the last riddle actually we have to solve for x in 30=2/(1/15+1/x) ......it has the solution x=∞ which is not practical....

  • @diniaadil6154
    @diniaadil6154 3 дні тому

    Averaging 2v on the whole trip means going twice the distance in the same time, which is impossible

  • @user-co9ge6vc3c
    @user-co9ge6vc3c 4 дні тому

    It's highly doubtful whether these geniuses were really involved in these, or whether these were just apocryphal anecdotes that get passed down because they make for good stories!

  • @daveincognito
    @daveincognito 4 дні тому

    I remember the car riddle from an old Encyclopedia Brown book.

  • @J7Handle
    @J7Handle 3 дні тому

    Correction: the notion that the speed of light is the same in all directions is unprovable in principle (at least, not without FTL capability). The speed of light is considered to be the same in all directions only by convention. Therefore, it is entirely possible for the speed of light to be infinite in one direction and 1/2c in the other. Since the car only needs to travel at infinite speed in one direction, it can, provided that it becomes massless on the descent.
    What’s funny is that Einstein should have noticed such a loophole given that he noted in his own paper that the one-way speed of light is unknowable and he set it equal to c by convention.

  • @Pinkyyybrainn
    @Pinkyyybrainn День тому

    Fgs question 1 is looking for a pythag expression for b, so its SQRT/((r-a)^2)+((r-c)^2) = b = r

  • @kevinmcknight5000
    @kevinmcknight5000 4 дні тому

    Due to time dialiation, I think the car only needs to go the speed of light (with reference to whoever observes it go an average of 15 mph for the first half) to make it down the hill with an average of 30 mph.

    • @tezzerii
      @tezzerii 4 дні тому

      Light speed in an old car that can only manage 15mph up the hill ?

  • @TheADDTinkerer
    @TheADDTinkerer 4 дні тому

    The first one doesn't seem right.
    C and A are supposedly different values which would indicate that it is not a quarter circle. Because of that the diagonal B can never be R.
    Unless C and A are equal of course, then B would be R.

    • @Stubbari
      @Stubbari 4 дні тому

      Yes, a≠c. b is always r.

    • @tezzerii
      @tezzerii 4 дні тому

      @TheADDTinkerer It is a quarter circle. It is possible to place a rectangle of any proportions in a quarter circle. And so long as the one corner is at the circle's centre, and the opposite corner meets the arc, then the diagonal = the radius.

    • @TheADDTinkerer
      @TheADDTinkerer 4 дні тому

      @@tezzerii how can it be a quarter circle if side C and A are different lengths?

    • @Stubbari
      @Stubbari 4 дні тому

      @@TheADDTinkerer Why couldn't it be?

    • @TheADDTinkerer
      @TheADDTinkerer 4 дні тому +1

      ​@@Stubbarisorry sorry. I mistakenly changed rectangle in to a square. Which had me confused.

  • @datapro007
    @datapro007 4 дні тому

    First, I don't believe that any of the geniuses mentioned were fooled by these trivial problems. It's a fable. Second, Presh always finds a way to complicate the solution. For example let's look at the last riddle. My solution: 60 mph = 1 mile per minute. 15 mph for a mile therefore takes 4 minutes. 30 mph for 2 miles take 4 minutes. Hence you can't average 30 mph. Duhhhhh.

  • @dinoeebastian
    @dinoeebastian 4 дні тому

    I got Leibniz's question right, and then I changed my answer to them being equal, and now I'm mad

  • @empmachine
    @empmachine 4 дні тому +2

    so: (a+c+b^2)/2 == b ??? I solved the 1st by saying a+(b^2)+c covers 2r, and then just divided.. am I wrong? (besides being overly complicated?)

  • @johnyoung9649
    @johnyoung9649 4 дні тому

    If I do the second half of the trip at the speed of light, then time would stop for me.

  • @GamingDreamer
    @GamingDreamer 4 дні тому

    6:32 thanks, i needvto hear that

  • @ochayethegnu2915
    @ochayethegnu2915 3 години тому

    I guess this particular genius doesn’t know that you can’t have one dice. Dice is plural.