Funnily enough, the tenth guest eats roughly 6.7 duodecillion times more pie than the last one. Assuming that the pie is made entirely of elemental carbon, if the last guest gets to eat only a single atom, the tenth guest eats 1.3 trillion tons of pie.
Based on Mr Sanjay's information (and what I found on the interwebs), this pie weighs almost as much as Lake Baikal (imagine Lake Baikal's rift filled with a pie!) or about 128 times more than Mt. Everest. That's 2^7 Mt. Everests made of pie and maybe even more significant than the pi within the pie!
@@leif1075 It can't really suggest that. #1 gets 1% and the next few get more than 1%. Yet the amounts have to AVERAGE 1%. So clearly they have to start falling.
There is no point in you working it out. It has no practical application. You will never see it again, whether you drop out of high school, or whether you become the next Einstein. It is utterly menaingless. You will not encounter it if you go to one of the top 5 universities for maths in the world. So, don't worry.
@@tryphonunzouave8384 Most (predominantly) English speaking countries use the decimal point as opposed to the decimal comma, but it's more regional driven than it is language. Some English speaking countries do use the decimal comma and Britain itself was almost one of them.
I have no idea if they thought that because it had to do with a mathematical principle that somehow affects the answer here but I'm not surprised that even a cursory knowledge of the history of math could be helpful in some situations since there are so many mathematical patterns everywhere. Or its just coincidence
Devan Burke> I have a feeling that finding pi here is more significant π has a habit of showing up in the strangest places. 3b1b has some videos of it popping up in odd places. pavel goor> It's logical as it's a pie problem :P Ugh, hardly. 😒 It could just as easily have been about splitting a sandwich or a pile of money. 🙄
@@vasilischatzipanagiotou9051 Someone pointed out on other part of the comment but when you try this same problem with 10^4, 10^6, 10^8, 10^12, etc. those numbers will approximate the digits of 2*pi more and more closely. Not coincidence; the reasoning is rather complicated but I think 3Blue1Brown kinda makes an explanation to a not so similar but probably very parallel problem with the bouncing squares.
It’s a very easy problem to solve with an Excel spreadsheet. So I used that, then clicked the video and unsurprisingly see the “no computers allowed.” Oh well
I tried to solve this problem on paper and found it too difficult for me, but on Excel this is child's play as you say… …funny how the first 20 or so guests eat almost all the cake and for the last of the guests there's hardly atoms left...
Haha, did the same thing, using Excel I found out the 10th guy gets the biggest slice in like 1 minute, and it's enough for me. Saved 4 minutes of my time. :D Don't give a damn about the equation, lol.
Actually, not really. That is because there are approximately 10^27 atoms in a pie. Don't ask me why that information is available on the InterWeb, but it is: ( answers.yahoo.com/question/index?qid=20111004110759AAolUCP ) So it turns out that the 90th person gets a single atom of pie, and the rest cannot actually receive even one atom.
Yes. Because if you serve a cake whose volume falls below that of Lake Baikal, then you would have to split atoms before the last guest receives his share. And doing so usually results in some serious explosion.
There's a much quicker way of solving this than the method in the video. First realise that the answer, n, is greater than 1. (easy to verify) Second realise that the answer is n (x - x·n/N)·(n+1)/N Simplifying gives n² + n - N > 0 and the solution is the first integer greater than [√(4N + 1) - 1 ]/2 For N = 100 this gives n =10 No need for factorials and general terms for the nth portion.
I realized the first part just by reading the description without even watching the video xD I didn't bothered tho to procede further (and to watch the video xDD)
@@purplewine7362 I'd say Presh's method took considerably more work, because he needed the full formula for the total amount of pie available to the n'th guest. Whereas this solution observes that it doesn't really matter how much you get in total because you know the fraction of the remainder you and the next person are taking so you can calculate which of you is getting more and that's enough. Ultimately Presh's solution simplifies a lot (hence all the cancellations) but you didn't really even need to include those factors if you had set up your inequality better.
Hey don't you think the 50th person gets to eat the most pie.... Because he is getting 50% of the remaining 50% pie so that is 25% of the original pie.. Please Explain where do I got wrong...
@@harshnair7126 You misread what Ged Langosz said, he doesn't mean that the 50th gets the most he means that it is easy to exclude anyone after as candidates.
@@randomdude9135 Do you even know what poetry is? Think carefully, as examples get very obscure... So what do you do with logical ideas to get maths? One may not be good with poetry of letters and words, but may be great with poetry of logical ideas.
A simpler way is to find the exact tipping point, where slices do not become any bigger anymore: x = (1 - x) * (x + 0.01) with x between 0 and 1. Solving the quadratic equation yields a result obviously above 9% but below 10%. So as long as x is below 10%, the slices will become bigger. Thus the tenth guest will get the biggest slice.
@@jessstuart7495 It is obvious that the pieces are getting bigger in the beginning while they are shrinking towards the end. So there must be a percentage in between (not necessarily an integer) where the current piece has the exact same size as the next one. The x on the left is the size of the current piece. (1 - x) is the part that remains for the next piece and (x + 0.01) is the part of the rest that the next piece would be. Thus the equation defines the exact percentage a piece needs to be in order for the next piece to be of the same size. It is important to notice that is does not matter how much of the cake is actually left when comparing two adjacent pieces.
@@jessstuart7495 The only assumption is that it's not a trick question; i.e. there is a local maximum which is also a global maximum, as asked. The starting percentage and increment (indirectly) are given, corresponding to a convenient %_n=n. How close the starting amount of cake to the amount of cake at the nth guest in question is irrelevant as each step only concerns the amount of cake left after the previous. Algebraically, the fraction of the original cake at any step is a common factor to the steps after it, so it doesn't change the size comparison of two subsequent pieces.
Actually, the occurrence of 2pi is a misleading coincidence. One would hope that with increasing number of guests, the largest share would approximate 2pi better and better, i.e. that the largest piece approximately equals to 2pi/m -- where m is the number of guests. This is not so. But other interesting mathematical constant appears! Generally, for m guests, the formula for the n-th guest piece is: x_n= n(m-1)!/m^n/(m-n)! -- the rules are the same, the first guest takes 1/m of the pie, the second 2/m of the remaining and so on. If the number of guests is a square number, i.e. m=k^2, the largest piece goes to the k-th guest. His share is thus x_k= k(k^2-1)!/k^(2k)/(k^2-k)! which -- for large k -- approximately behaves like 1/k e^(-1/2). Thus the inverse of the square root of e is the correct constant!
@Girisha Polzin It does. If there are m guests, first takes 1/m pice, the second 2/m pice of the remaining, and so on until the m-th guest takes m/m= 100 % of the remaining pie. There would be nothing left.
Moreover if you express in terms of m, person k gets ~.6/sqrt(m) of the pie - so adding more people means they quickly get more than a fair portion of 1/m. The last person gets factorially little of the pie
True. You can see it converge: k=10 → 0.6281... k=100 → 0.6085... k=1000 → 0.60673... k=10000 → 0.606550... e^(-1/2) = 0.606530... Now you can even do the Taylor expansion in 1/k: k → e^(-1/2) * (1 + 1/3 1/k + 2/9 1/k² + ...) Already the first three terms of this Taylor expansion give a good approximation for k=10: 0.6280...
@@Bruno_Haible I didn't do it for bigger cases, but for k=1000 the 10th guest still eats the largest portion, with the 33rd already eating less than 1%
I really like your solution. I found that 10 was the largest slice by ignoring how much was left, and simply pairing numbers in a sequence. So I assumed there was 100% of whatever remained left, then took 10% out( I jumped there because the math was easy). After taking 11% of what remained( 90%), I realized that the 10th guest gets more than the 11th. After comparing it to the 9th guest in the same manner, I arrived at the solution.
@@stigcc That's not what they said. They're using 100% as 100% of the pie remaining after the first nine guests. After taking 10% of what remains, 11% of the new remainder is less than the 10% of the initial value.
@@JoeThomas-lu6fy Right, thank you. So after the 8th person is done, the 9th person takes 9% and returns 91 which the 10th person takes 10% of which is 9.1% of the original. Nice
Amazing idea! thanks for sharing. This directly proves that Guest 10 gets more than Guest 11 as well as Guest 9. Also it is not hard to see that Guest 10 is indeed the maxima. I was just going for the rough numbers until guest 2, 3, 4 and I realized soon that Guest 10-13 would be the tipping point. I really liked the problem.
this problem can be solved slightly more elegantly by instead of calculating the exact amount for any given person k, set the amount of pie left when person k to be x which means person k gets xk/100 % of the pie and person k+1 gets x*((100-k)/100)*((k+1)/100) now if person k+1 gets more pie than person k; the following inequality will hold xk/100 < x*((100-k)/100)*((k+1)/100) this simplifies to 100k < (100-k)(k+1) which simplifies to k(k+1) < 100 the last integer this is true is 9 therefore person 10 gets the most pie
Now just for fun, I've calculated the share of the entire pie (using the formula given after 4:31), for all numbers of guests from 1 to 171. (My spreadsheet can't handle the calculations for any greater number.) For some numbers of guests, there will be 2 guests who share the largest proportion. (With 2 guests both #1 and #2 will get 50%. With 6 guests both #2 and #3 will get 27.7recurring%.) There will be 2 guests on the equal-largest share when the number of guests is 2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156. An obvious progression, and I can't be bothered pursuing the reason for it, but I thought some of you might be interested.
I know I’m late to the party but this pattern is because when looking at the ratio between successive portions we got the inequality n > k^2 + k so when n = k^2 + k, the portion size remains exactly the same for one iteration. Indeed, we see 2 = 1^2 + 1, 6 = 2^2 + 2, 12 = 3^2 + 3, etc
@@zinc_magnesium Wow. I don't remember this at all. Thanks for the reply. When I've got time I might come back and find out what it was all about. I'll rewatch the video, work out what I was calculating and then go through what you wrote to find out why. (I have read your post, but out of context it's all a bit baffling.)
@@zinc_magnesium or simply every consecutive multiplication. 1x2=2, 2x3=6, 3x4=12, 4x5=20, ..., 12x13=156. It makes it very easy to figure out every guest value where this occurs. For example, 1,001,000 would work.
Wonderful explanation. Thanks! I did a excel manipulation and got the answer at Guest 10. Create three columns. In Column 1, put 100 in row 1 and a formula from row 2 onwards. Fornula is to get difference between cellAbove and cell3_Above (column3-rowAbove). Repeat it till 100 rows. In column 2, put percentages from 1% till 100%. in column 3, put multiplication of two cells (cell1 * cell2). Observe that it increases till Row10 (share of Guest10 and then decreases.
One of my intuitive guesses was that it would probably be around 10 because that's the geometric mean of 1 and 100 and this seemed like a multiplication-heavy problem. Seeing the formulas was neat.
I did it by a much more brute force method. I started out multiplying .99 by .02, adding that to .01, multiplying that by .03, adding that, multiplying by .04, but I quickly saw that that would get too difficult. So I started from the other end. I realized that whatever 99 got, 100 would get only 1% of that, and realized you could just treat every guest as the first and multiply the next guest's percentage by whatever was left from the previous guest. So 51 was going to get 51% of whatever 50 got, clearly much less, and then I just had to multiply .41 by .6, .31 by .7, .21 by .8, and then .11 by .90, and since that kept producing smaller products than .4, .3, .2, and .1, respectively, I realized that that the tipping point had to be somewhere between 1 and 10. So I kept working my way backwards, multiplying .91 by .10, realizing that that was obviously larger than .9, and realized that the tenth guest must have gotten the largest piece.
Guest n.10 gets the largest piece: 6,28/100 ... instead of the formal complete evaluation as showed by Presh you can notice that each person gets a piece that is approximated by -2/99 -3/99 -4/99. As we substract a growing term, and from this term we reduce the total amount available, there must exist a certain number where the next term is less than the one before. This number can be found by approximation in the same way we find the intersection of two curves on the XY plane. For example get n1=6 and n2= 20 you see that for 5.15 (n1) > 2.60 (n2) so you pick n3 close to n1 and so on ... till you find the max.
Can be resolved mentally, with no algebra. Look at what happens with guest 2: gets 2% instead of 1%, but the pie has barely shrunk since guest 1, so of course guest 2 gets more. At some point, the pie starts shrinking faster than the percentage increase between guests. Look at guest 9: they take 9%, leaving 91%. Then guest 10 takes 10% of that 90%, or 9.1%. Guest 10 gets more. Look at guest 10: they take 10%, leaving 90%. Then guest 11 takes 11% of that 90%, or 9.9%. Guest 10 gets more.
Sir, at 1:52, when you mentioned that “after Second guest has the pie there will be 99%•98% left” I think there is something off. See, the second guest gets 99%•2% pie is apparently right. However, when I calculate the left over, which generated by (100%-99%•2%), I get the result 9802/10000, which is by 1% off from your result 99%•98%=9702/10000. 1% might be a small difference, but this makes the actually result unable to be present in that form, and approximating it to another number could lead to tremendous difference in a large series. (I’m sorry English is not my first language, If You don’t get my idea Id love to explain it more detail.)
Nope. The first guest takes 1% of the whole pie, so there is 99% left for the second guest. The second guest takes 2% of 99% of pie, so there is (100% - 2%) * 99% = 98% * 99% left for the third guest. The third guest takes 3% of (98% * 99%) of pie, so there is (100% - 3%) * (98% * 99%) = 97% * 98% * 99% left for the fourth guest. The fourth guest takes 4% of (97% * 98% * 99%) of pie, so there is (100% - 4%) * (97% * 98% * 99%) = 96% * 97% * 98% * 99% left for the fifth guest. etcetera. The "1% of pie" difference that you're finding, is the 1% that the first guest took. The correct calculation of the left over after the second guest, is 100% - 1%*100% - 2%*99% = = 100/100 - (1/100)*(100/100) - (2/100)*(99/100) = 100/100 - 100/10000 - 198/10000 = 10000/10000 - 100/10000 - 198/10000 = 9702/10000 I hope that helps.
Well I did it in python: #init guestAmount = [0, 0.01] pieLeft = [1, 0.99] #calc for i in range(2, 101): guestAmount.append(pieLeft[i-1] * (i/100)) pieLeft.append(pieLeft[i-1] - guestAmount[i]) #result biggestSlice = max(guestAmount) # 0.062815... guestNumber = guestAmount.index(biggestSlice) # 10
"Pretty neat" is very well said ... as my brain got fried by this one :)) ... I finally found the right solution (at least to the question, not the exact amount of the largest piece) by finding a similar formula (not the one you mentioned) and then by trying and approaching - thanks for posting and explaining so well!
if you change the number of people to any square number it'll be the root-th piece (assuming using fractions 1/m, 2/m, 3/m... with m people rather than percentages or u end up eating more than one pie.)
Not true. I like pie, but I don't want a slice of pie the size of a small moon. And if the 100th guest is to have at least an atom, then the pie will need to be that big. I wonder who the actual lucky guest is that gets enough to enjoy but not so much that it's impossible to eat in one sitting?
Minute 1:53 How did you know that the remaining pie was 99/100*98/100?? If you do the problem you get 99/100 (the amount of pie left from previos persona) minus 2% of 99/100 (the amount given to the second guest) and you get 99/100 - 198/10000. What was the mind process that you did to know that, that subtraction was equal to that multiplication??
Hello my friend, I totally agree with you! I think the solution presented is wrong, because it assumes that the remaining pie is exactly 98/100, 97/100, 96/100 and so on, which is not accurate: it doesn't take into consideration the exact portions the previous guests have taken! I don't know whether the right function has a maximum for k=10, but I think almost everyone has missed this detail!
(98/100)x=(100/100)x-(2/100)x When the problem asks "take 98% of what remains," either approach is a mathematically accurate interpretation, but the approach on the left is more useful for further simplifying and getting a solution.
So the answer is 2% of π? 🤨 Dammit π, stop popping up everywhere. ¬_¬ (Yes it's just a coincidence, the question could just as easily have been a pizza or cake or pile of money being split up.)
I came to the same conclusion, but without all the factorial stuff I just considered how the values change so going from the first piece, which is 0.01, to the second one 0.99*0.02 = 0.99 * (0.01*2), we multiply by 0.99*2. Going from the second to the third, we multiply the second value by 0.98*3/2 The pattern is that the first factor always decreases by 1%, while the second one is the current element number divided by the previous one Those 2 factors can be written as (1-n/100)*(n+1)/n now I'm just interested when this is becoming
Wrong. Error (1): It's not the education system that sucks the fun out of math. What sucks the fun out of math is all the other kids not having the slightest interest in math. Therefore, error (2): The reason for having fun with math after school is actually not the fact that *you* are out of school, but the fact that *those other kids* are out of school as well.
I totally agree french man. Learning out of passion is way more effective than by an autority. Though we need people to inspire them and give them the materials needed.
That second inequality at 3:45 100 > k^2 + k is also the same as asking which is the greatest triangle number under 50. I wonder if there is an alternate way to express this in terms of those simple consecutive sums.
Did you use simple arithmetic also? So easy to just round the numbers as they don't have a HUGE effect on a problem of this sort... Of course you could have used Euler's constant and limits and came to the same answer... funny how flexible Math can be for finding solutions.
I solved this with a slightly modified compound interest formula and calculus. I set up the formula y = P(1-r/n)^tn where P is the principle amount (100), r is the "interest" rate (-1% or -0.01, multiplied by the number of times the pie has been taken from), n is the compounds per year (which I set to 1), and t is the number of years (the number of times the pie has been taken from) as: y = 100(1-0.01t)^t I then took the derivative: dy/dx = 100(-0.01)ln(1-0.01t)(1-0.01t)^t The 100 and the 0.01 cancel to get: dy/dx = -ln(1-0.01t)(1-0.01t)^t Setting this equal to 0, ln(1-0.01t) will never equal 0 for positive integer t, but (1-0.01*10)^10 is 0, so that's either a local minimum or maximum. Testing the value of dy/dx just below and above ten, you can see that it switches from positive to negative, so t = 10 is a local maximum on the original curve, so guest 10 gets the biggest piece.
I started a different way but ended up with in the same spot: The remaining Size of the pie is (1 - 1/100) * (1 - 2/100) * ... * (1 - n/100) So the pie, that the next guest Nr. "n" takes from, is always (1 - n/100) smaller than the piece of his predecessor. But Guest Nr. "n" takes (n+1)/n more than the one before him. That means if both factors together are above 1, then the piece he takes is bigger than than the ones taken before. We just have to find out the single piece after which the sizes of the slices are not increasing anymore (which means both factors combined are roughly 1) so we have now: ((n+1)/n) * (1 - n/100) = 1 which can be simplified into: 1/n - (n+1)/100 = 0 after some rearranging we get: n^2 + n = 100 For which I would have needed a calculator so I guessed something between 9 and 10 and failed, cause I choose 9 XD
+MindYourDecisions Presh, you really dropped the ball on this one. If the kth of N guests gets k/N of the remaining pie, you can show that k=sqrt(N) will get the largest piece.
@@mattrogers6646 I have difficulty doing equations in comments, but I'll give you a verbal head start. Follow Presh's method, but substitute the variable N for the given constant 100. When you simplify the inequality for the ratio of k+1 and k, you should get k(k+1)
define a function for the size of the piece of the pie for the x‘th person. p(x) say p‘(x)=0 and you get x1 if p‘‘(x1) < 0 you know person x1 has largest piece
I didn’t understand this leap from k=9 with the fraction greater than 1, to “size increases to guest 10.” Wouldn’t that be k=10, so that would make the fraction less than 1? 4:27
To calculate what 3rd guest gets, u need to subtract guest 2 and guest 1's share from the total and then multiply 3/100 ..U did not subtract guest one's share!!
My guess was correct. I figured 10 th to 15 th guest will get the biggest Piece because the remaining amount of the Pie gets smaller and smaller even though the percentage increases.
I could do the mental math, but I think the point is to think of a more elegant solution, which I can't. My intuition is that the largest piece will be somewhere between 8 and 12% (of the total pie, that is). I was right about it involving factorials, yay.
There’s a much easier way of thinking about it. Guest 2 gets 0.99 x 0.02/0.01 > 1x as much as guest 1, because when the pie gets to the 2nd person there is 99% as much pie as when it got to the first person, and guest 2 gets 2% of that whereas guest 1 got 1% of theirs (so 0.99x as much pie left, but 2x as much of the remaining pie gets taken) Similarly, guest 3 gets 0.98 x 0.03/0.02 > 1x as much as guest 2 Generalizing, guest n+1 gets (100-n)/100 x (n+1)/n times as much as guest n. So, the guest who gets the most pie is the last guest who gets more pie than the guest before them, or the highest value n+1 where the above equation is still greater than 1. Now we have a simple algebraic equation: -> (100-n) x (n+1) > 100n -> n^2 -99n - 100 < -100n -> n^2 + n < 100 -> n = 9 -> n+1 = 10 QED
How I got the answer: I came to the conclusion that the pie will become small really quicly, so higher numbers are out of the question. I thought that after about 10 guests the cake would have lost a major portion, so more % would make too little difference then. So I approximated it to be 10. Turns out I was right. Sometimes "close enough" is close enough, I suppose. The actual answer in the video was very beautiful though.
I came to same conclusion on the logic then when there are two factors in a situation where their isn't one extreme that's clearly better it's usually the value where they are the same (making a square) so it was probably somwhere around 10 because you then have 10%. (It's a handy rule I think that can come in hand
Oh! At 4:40, you say "... and just as a funny coincidence...," then you observe that the largest pie piece is roughly 2π/100. Well... It's not a coincidence! So long as your group size (number of people wanting some pie) is an even power of 10, then the size of the largest piece will approximate the digits of pi. For larger even powers of 10, the approximation improves significantly. Try cutting the pie for 10,000 people or 1 million people -- you will get more and more of the correct digits of pi (in base 10). This _must_ be so. Why? I'm not telling you! I prefer to drive you insane by trying to figure out WHY that factorial-rational expression (from 3:30 in your video) _must_ converge to a whole-fraction of pi. (Hehehe...)
@@pgoeds7420 Niiiiice! Of course, there are other means to have known this interesting property... but that clickity-clackity square thing IS pretty awesome! Mind bending...
2:34 I wanted to see how could that function look so I typed it in Desmos with inserting 'x' for 'k's and got very weird function existing only for X from (-153; 198.41) and (198.977; 199.019). Have I broke Desmos or is that due to domains or something like that?
Even though that equation simplifies to small numbers the exponential and factorials both produce ridiculously large numbers, so at those values of x you’re exceeding Desmos computational thresholds... at least that’s my guess. I’m also thinking desmos might be broken in some way because at values k>100 you’re taking the factorial of a number < 0, which is impossible... so the domain should be -infinity < k < 100. Edit: Yeah just plugged it in myself. Desmos is definitely not happy computing with numbers of this magnitude. It’s busted.
It's easier to think about it as the expression k(100-(k-1)) which simplifies to 101k-k^2. Then find the maximum value of this which is √101 which is around 10.
I've come to a weird conclusion taking this a step further. What if you devide the piece in to "n" persons? Well the first steps are easy: in 1 person, the first person gets ... all the pie in 2 persons, both the first and the second get 1/2 a pie, so they are both "winners" in 3, 4 or 5 persons, only the second person is the winner, in 6 persons both the second and the third person are winners, in 7 to 11 persons he third person wins, in 12 it will be the 3th and 4th that win, from 13 to 19 it is the 4th,... If n= persons and k= winner than: if n=k*(k-1) (exept for k=1, this is when n=1) than both k and k-1 are winners if k*(k-1)
If you solve the quadratic equation at 3:48. And change the 100 to N, so that each person gets their position / N of the pie, you get the equation. k = root [ (N + 1/4) ] - 1/2. For N = 100 (this problem), the person after the ≈9.5 spot (i.e. #10) get the most. For N = 10, the person after the ≈2.7 spot (i.e. #3) gets the most. For N = 20, the person after the ≈4.0000 spot gets the most - meaning #4 and #5 get the same.
So you started with one pie and wound up with two pi.
Lol
The bakers were Banach and Tarski.
@@pjabrony8280 Funded by vsauce
This deserves the pin.
pjabrony woooow, learning maths from a joke!
Funnily enough, the tenth guest eats roughly 6.7 duodecillion times more pie than the last one. Assuming that the pie is made entirely of elemental carbon, if the last guest gets to eat only a single atom, the tenth guest eats 1.3 trillion tons of pie.
Geek NERD ASIAN
@Rohit Iyengar I am not. My parents lived there
@Rohit Iyengar not your fault. Raj Singh is an Indian name. And I wouldn't deny my geekness. 😉
?
Based on Mr Sanjay's information (and what I found on the interwebs), this pie weighs almost as much as Lake Baikal (imagine Lake Baikal's rift filled with a pie!) or about 128 times more than Mt. Everest. That's 2^7 Mt. Everests made of pie and maybe even more significant than the pi within the pie!
At first 'this seems quite easy but I just cant get it'
Watches solution 'no, I was very wrong'
Hahaha same here.
Did you think it was the 100th guest too?..that's what the pattern suggests at first..
@@leif1075
It can't really suggest that. #1 gets 1% and the next few get more than 1%. Yet the amounts have to AVERAGE 1%. So clearly they have to start falling.
There is no point in you working it out. It has no practical application. You will never see it again, whether you drop out of high school, or whether you become the next Einstein. It is utterly menaingless. You will not encounter it if you go to one of the top 5 universities for maths in the world. So, don't worry.
Ikr, I thought I nearly had it and then the solution was way to hard for me
"Don't use calculator"
- Mmmmkay..
”..and this fraction: 62815650955529472/100^9"
I get the joke, but that operation of Presh's solution wasn't asked initially: we were only asked to find the person with the largest pie piece
Well this is very easy, 100^9 = 10^18, so you just need to count 18 numbers from the right, and put the decimal here ^^
or the dot, or comma, or whatever the correct formulation is in English
@@tryphonunzouave8384 he is talking about (99 factorial upon 90!) Times (10/100^10) simplification
@@tryphonunzouave8384 Most (predominantly) English speaking countries use the decimal point as opposed to the decimal comma, but it's more regional driven than it is language. Some English speaking countries do use the decimal comma and Britain itself was almost one of them.
I'm impressed that Presh showed us the simplified calculation of (99!/90!)*(10/100 exponent 10) without using a calculator in the end.
Yeah... same here...
Since ancient Pythagoreans considered 10 to be a sacred number, I guessed it was the tenth guest. I was a Classics minor, not a math major.
I have no idea if they thought that because it had to do with a mathematical principle that somehow affects the answer here but I'm not surprised that even a cursory knowledge of the history of math could be helpful in some situations since there are so many mathematical patterns everywhere. Or its just coincidence
I have a feeling that finding pi here is more significant
Devan Burke> I have a feeling that finding pi here is more significant
π has a habit of showing up in the strangest places. 3b1b has some videos of it popping up in odd places.
pavel goor> It's logical as it's a pie problem :P
Ugh, hardly. 😒 It could just as easily have been about splitting a sandwich or a pile of money. 🙄
i thought it might be too, was hoping to find it turn up as a limit as the number of people increases but alas that takes the slice size to zero :/
@Devan Burke Actually he found tau
There is not pi, just a number "near" pi for a human brain prespective. Just a fun fact
@@vasilischatzipanagiotou9051 Someone pointed out on other part of the comment but when you try this same problem with 10^4, 10^6, 10^8, 10^12, etc. those numbers will approximate the digits of 2*pi more and more closely. Not coincidence; the reasoning is rather complicated but I think 3Blue1Brown kinda makes an explanation to a not so similar but probably very parallel problem with the bouncing squares.
It’s a very easy problem to solve with an Excel spreadsheet. So I used that, then clicked the video and unsurprisingly see the “no computers allowed.” Oh well
I tried to solve this problem on paper and found it too difficult for me, but on Excel this is child's play as you say… …funny how the first 20 or so guests eat almost all the cake and for the last of the guests there's hardly atoms left...
Allan Boone, yeah that would just leave the part with finding the equation.
Oops, sorry, I used a computer to see that part of the problem.
Haha, did the same thing, using Excel I found out the 10th guy gets the biggest slice in like 1 minute, and it's enough for me. Saved 4 minutes of my time. :D Don't give a damn about the equation, lol.
float sum = 0, x = 0.01;
int i = 1;
while(i < 101) {
cout
FUN FACT: The 100th guest will get less than 10^-40 percent of pie...bechaara😂😂
Wow
So a piece skinnier than my bicept
Actually, not really. That is because there are approximately 10^27 atoms in a pie. Don't ask me why that information is available on the InterWeb, but it is: ( answers.yahoo.com/question/index?qid=20111004110759AAolUCP ) So it turns out that the 90th person gets a single atom of pie, and the rest cannot actually receive even one atom.
It doesn't matter what percent I get as long as the pie is large. Plus - I wants some cream anyway.
@@JoeHarpo wow,I haven't thought about it!
When you say ,"To build confidence of the students around the world..."
You mean it sir!
I will do this at my next dinner party. It will probably be my last dinner party.
Yes. Because if you serve a cake whose volume falls below that of Lake Baikal, then you would have to split atoms before the last guest receives his share. And doing so usually results in some serious explosion.
Just make sure you like the first 10 persons, the rest won't like you that much, the ones around the 20th upwards will hate you...
You go to dinner party's?!
@@chrisschmeitz1139 Not only that, having 100 guests also
There's a much quicker way of solving this than the method in the video.
First realise that the answer, n, is greater than 1. (easy to verify)
Second realise that the answer is n (x - x·n/N)·(n+1)/N
Simplifying gives
n² + n - N > 0
and the solution is the first integer greater than
[√(4N + 1) - 1 ]/2
For N = 100 this gives n =10
No need for factorials and general terms for the nth portion.
I realized the first part just by reading the description without even watching the video xD I didn't bothered tho to procede further (and to watch the video xDD)
It's the same amount of work actually. He just took ratio of successive terms and you took difference.
@@purplewine7362 I'd say Presh's method took considerably more work, because he needed the full formula for the total amount of pie available to the n'th guest. Whereas this solution observes that it doesn't really matter how much you get in total because you know the fraction of the remainder you and the next person are taking so you can calculate which of you is getting more and that's enough. Ultimately Presh's solution simplifies a lot (hence all the cancellations) but you didn't really even need to include those factors if you had set up your inequality better.
Hey don't you think the 50th person gets to eat the most pie....
Because he is getting 50% of the remaining 50% pie so that is 25% of the original pie..
Please Explain where do I got wrong...
@@harshnair7126 You misread what Ged Langosz
said, he doesn't mean that the 50th gets the most he means that it is easy to exclude anyone after as candidates.
Me *see 6.28*: oh, it’s tau
Video: it’s two pi
Me: Here we go
*Pure mathematics is in its way the poetry of logical ideas* 🕊️🕊️
👏👏👏🤓
Einstein
This is applied mathematics. This is a constraint convex optimization problem.
No, don't compare math to poetry.I suck at poetry. Doesn't mean I'm any good at math
@@randomdude9135 Do you even know what poetry is? Think carefully, as examples get very obscure...
So what do you do with logical ideas to get maths?
One may not be good with poetry of letters and words, but may be great with poetry of logical ideas.
A simpler way is to find the exact tipping point, where slices do not become any bigger anymore:
x = (1 - x) * (x + 0.01) with x between 0 and 1.
Solving the quadratic equation yields a result obviously above 9% but below 10%.
So as long as x is below 10%, the slices will become bigger. Thus the tenth guest will get the biggest slice.
Yep, this is the way I did it. You don't need to figure out the exact amounts, just when they stop getting larger.
Would you mind explaining how you got the equation x=(1-x)*(x+0.01) ?
@@jessstuart7495 It is obvious that the pieces are getting bigger in the beginning while they are shrinking towards the end. So there must be a percentage in between (not necessarily an integer) where the current piece has the exact same size as the next one.
The x on the left is the size of the current piece.
(1 - x) is the part that remains for the next piece and (x + 0.01) is the part of the rest that the next piece would be.
Thus the equation defines the exact percentage a piece needs to be in order for the next piece to be of the same size.
It is important to notice that is does not matter how much of the cake is actually left when comparing two adjacent pieces.
@@jessstuart7495 The only assumption is that it's not a trick question; i.e. there is a local maximum which is also a global maximum, as asked. The starting percentage and increment (indirectly) are given, corresponding to a convenient %_n=n. How close the starting amount of cake to the amount of cake at the nth guest in question is irrelevant as each step only concerns the amount of cake left after the previous. Algebraically, the fraction of the original cake at any step is a common factor to the steps after it, so it doesn't change the size comparison of two subsequent pieces.
But the number of the guest issue not the same as the percent number, so how can this be right?
Actually, the occurrence of 2pi is a misleading coincidence. One would hope that with increasing number of guests, the largest share would approximate 2pi better and better, i.e. that the largest piece approximately equals to 2pi/m -- where m is the number of guests. This is not so.
But other interesting mathematical constant appears!
Generally, for m guests, the formula for the n-th guest piece is: x_n= n(m-1)!/m^n/(m-n)! -- the rules are the same, the first guest takes 1/m of the pie, the second 2/m of the remaining and so on.
If the number of guests is a square number, i.e. m=k^2, the largest piece goes to the k-th guest. His share is thus x_k= k(k^2-1)!/k^(2k)/(k^2-k)! which -- for large k -- approximately behaves like 1/k e^(-1/2). Thus the inverse of the square root of e is the correct constant!
@Girisha Polzin It does. If there are m guests, first takes 1/m pice, the second 2/m pice of the remaining, and so on until the m-th guest takes m/m= 100 % of the remaining pie. There would be nothing left.
Moreover if you express in terms of m, person k gets ~.6/sqrt(m) of the pie - so adding more people means they quickly get more than a fair portion of 1/m. The last person gets factorially little of the pie
True. You can see it converge:
k=10 → 0.6281...
k=100 → 0.6085...
k=1000 → 0.60673...
k=10000 → 0.606550...
e^(-1/2) = 0.606530...
Now you can even do the Taylor expansion in 1/k: k → e^(-1/2) * (1 + 1/3 1/k + 2/9 1/k² + ...)
Already the first three terms of this Taylor expansion give a good approximation for k=10: 0.6280...
@@Bruno_Haible I didn't do it for bigger cases, but for k=1000 the 10th guest still eats the largest portion, with the 33rd already eating less than 1%
@@m4riel DId you change the denomination of the pie cutting to 1000^2?
I really like your solution. I found that 10 was the largest slice by ignoring how much was left, and simply pairing numbers in a sequence.
So I assumed there was 100% of whatever remained left, then took 10% out( I jumped there because the math was easy). After taking 11% of what remained( 90%), I realized that the 10th guest gets more than the 11th. After comparing it to the 9th guest in the same manner, I arrived at the solution.
Even I have a different approach to this problem here : th-cam.com/video/CkFZvr24xB0/w-d-xo.html
Why did you assume that the 10th guest had 90% to take from?
@@stigcc That's not what they said. They're using 100% as 100% of the pie remaining after the first nine guests. After taking 10% of what remains, 11% of the new remainder is less than the 10% of the initial value.
@@JoeThomas-lu6fy Right, thank you. So after the 8th person is done, the 9th person takes 9% and returns 91 which the 10th person takes 10% of which is 9.1% of the original.
Nice
Amazing idea! thanks for sharing. This directly proves that Guest 10 gets more than Guest 11 as well as Guest 9. Also it is not hard to see that Guest 10 is indeed the maxima.
I was just going for the rough numbers until guest 2, 3, 4 and I realized soon that Guest 10-13 would be the tipping point. I really liked the problem.
this problem can be solved slightly more elegantly by instead of calculating the exact amount for any given person k, set the amount of pie left when person k to be x
which means person k gets xk/100 % of the pie and person k+1 gets x*((100-k)/100)*((k+1)/100)
now if person k+1 gets more pie than person k; the following inequality will hold
xk/100 < x*((100-k)/100)*((k+1)/100)
this simplifies to
100k < (100-k)(k+1)
which simplifies to
k(k+1) < 100
the last integer this is true is 9 therefore person 10 gets the most pie
I did this is exaxt way
I thought exactly the same. It was much more easy
Me too!
How Did you think of that? Why the hell didn't I think of that??
Yeah exactly
Now just for fun, I've calculated the share of the entire pie (using the formula given after 4:31), for all numbers of guests from 1 to 171. (My spreadsheet can't handle the calculations for any greater number.)
For some numbers of guests, there will be 2 guests who share the largest proportion. (With 2 guests both #1 and #2 will get 50%. With 6 guests both #2 and #3 will get 27.7recurring%.)
There will be 2 guests on the equal-largest share when the number of guests is 2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156.
An obvious progression, and I can't be bothered pursuing the reason for it, but I thought some of you might be interested.
I know I’m late to the party but this pattern is because when looking at the ratio between successive portions we got the inequality n > k^2 + k so when n = k^2 + k, the portion size remains exactly the same for one iteration. Indeed, we see 2 = 1^2 + 1, 6 = 2^2 + 2, 12 = 3^2 + 3, etc
@@zinc_magnesium
Wow. I don't remember this at all.
Thanks for the reply. When I've got time I might come back and find out what it was all about. I'll rewatch the video, work out what I was calculating and then go through what you wrote to find out why. (I have read your post, but out of context it's all a bit baffling.)
@@zinc_magnesium or simply every consecutive multiplication. 1x2=2, 2x3=6, 3x4=12, 4x5=20, ..., 12x13=156. It makes it very easy to figure out every guest value where this occurs. For example, 1,001,000 would work.
Wonderful explanation. Thanks! I did a excel manipulation and got the answer at Guest 10. Create three columns. In Column 1, put 100 in row 1 and a formula from row 2 onwards. Fornula is to get difference between cellAbove and cell3_Above (column3-rowAbove). Repeat it till 100 rows. In column 2, put percentages from 1% till 100%. in column 3, put multiplication of two cells (cell1 * cell2). Observe that it increases till Row10 (share of Guest10 and then decreases.
One of my intuitive guesses was that it would probably be around 10 because that's the geometric mean of 1 and 100 and this seemed like a multiplication-heavy problem. Seeing the formulas was neat.
great insight! thanks
I did it by a much more brute force method. I started out multiplying .99 by .02, adding that to .01, multiplying that by .03, adding that, multiplying by .04, but I quickly saw that that would get too difficult. So I started from the other end. I realized that whatever 99 got, 100 would get only 1% of that, and realized you could just treat every guest as the first and multiply the next guest's percentage by whatever was left from the previous guest. So 51 was going to get 51% of whatever 50 got, clearly much less, and then I just had to multiply .41 by .6, .31 by .7, .21 by .8, and then .11 by .90, and since that kept producing smaller products than .4, .3, .2, and .1, respectively, I realized that that the tipping point had to be somewhere between 1 and 10. So I kept working my way backwards, multiplying .91 by .10, realizing that that was obviously larger than .9, and realized that the tenth guest must have gotten the largest piece.
But 2π/100 ist only 2% of one pie... so it’s not equal to 6.28%.
Took me a moment to figure that one out. :)
Underrated comment is underrated
2 * pi is approximately 6.28, so you get 6.28/100, or 6.28%
@@surjo8095 2π/100 = 2/100*π = 2/100 of π. Now read it out loud ^^.
Aaaaaaaand best comment award goes to
My eyes glazed over 30 seconds into the solution. I'd like to see 100 mathematicians share a single pie. Smart asses!
*LOVE* that Presh *didn't* say "Did you figure it out?"
👍👍👍
Yeah... he knew we didn't.
Guest n.10 gets the largest piece: 6,28/100 ... instead of the formal complete evaluation as showed by Presh you can notice that each person gets a piece that is approximated by -2/99 -3/99 -4/99. As we substract a growing term, and from this term we reduce the total amount available, there must exist a certain number where the next term is less than the one before. This number can be found by approximation in the same way we find the intersection of two curves on the XY plane. For example get n1=6 and n2= 20 you see that for 5.15 (n1) > 2.60 (n2) so you pick n3 close to n1 and so on ... till you find the max.
Can be resolved mentally, with no algebra.
Look at what happens with guest 2: gets 2% instead of 1%, but the pie has barely shrunk since guest 1, so of course guest 2 gets more.
At some point, the pie starts shrinking faster than the percentage increase between guests.
Look at guest 9: they take 9%, leaving 91%. Then guest 10 takes 10% of that 90%, or 9.1%. Guest 10 gets more.
Look at guest 10: they take 10%, leaving 90%. Then guest 11 takes 11% of that 90%, or 9.9%. Guest 10 gets more.
Best solution, congratulations🎉
You find the solution more easily by simultaneously solving G(k)/G(k-1) > 1 and G(k+1)/G(k) < 1
I get Presh's Notification, I click!!😀
if the last guest got 1 pound of pie and I was one of the other guests then I would not be disappointed.
A plot with the fraction that each guest had would have been great!
Sir, at 1:52, when you mentioned that “after Second guest has the pie there will be 99%•98% left” I think there is something off. See, the second guest gets 99%•2% pie is apparently right. However, when I calculate the left over, which generated by (100%-99%•2%), I get the result 9802/10000, which is by 1% off from your result 99%•98%=9702/10000. 1% might be a small difference, but this makes the actually result unable to be present in that form, and approximating it to another number could lead to tremendous difference in a large series. (I’m sorry English is not my first language, If You don’t get my idea Id love to explain it more detail.)
I totally agree
Nope.
The first guest takes 1% of the whole pie, so there is 99% left for the second guest.
The second guest takes 2% of 99% of pie, so there is (100% - 2%) * 99% = 98% * 99% left for the third guest.
The third guest takes 3% of (98% * 99%) of pie, so there is (100% - 3%) * (98% * 99%) = 97% * 98% * 99% left for the fourth guest.
The fourth guest takes 4% of (97% * 98% * 99%) of pie, so there is (100% - 4%) * (97% * 98% * 99%) = 96% * 97% * 98% * 99% left for the fifth guest.
etcetera.
The "1% of pie" difference that you're finding, is the 1% that the first guest took. The correct calculation of the left over after the second guest, is
100% - 1%*100% - 2%*99% =
= 100/100 - (1/100)*(100/100) - (2/100)*(99/100)
= 100/100 - 100/10000 - 198/10000
= 10000/10000 - 100/10000 - 198/10000
= 9702/10000
I hope that helps.
Well I did it in python:
#init
guestAmount = [0, 0.01]
pieLeft = [1, 0.99]
#calc
for i in range(2, 101):
guestAmount.append(pieLeft[i-1] * (i/100))
pieLeft.append(pieLeft[i-1] - guestAmount[i])
#result
biggestSlice = max(guestAmount)
# 0.062815...
guestNumber = guestAmount.index(biggestSlice)
# 10
The early bird gets the worm.
But the second mouse gets the cheese.
Ah but the 10th person gets the biggest piece of pie.
-Typo: --2:47-- The denominator of the top fraction should have "k + 1" instead of "k - 1".- I made a mistake, not Presh.
[100-(k+1)]! = (100-k-1)! You have to distribute the minus sign
@@TastyBagel I did not catch that; Presh did not make a mistake there.
There was only 1 pie then how would the 10th guest get 2 pie he he jokin
Can this be shown as a graph? Curious as to exactly why person no.10 gets the largest slice, and thinking a graph can show why.
"Pretty neat" is very well said ... as my brain got fried by this one :)) ... I finally found the right solution (at least to the question, not the exact amount of the largest piece) by finding a similar formula (not the one you mentioned) and then by trying and approaching - thanks for posting and explaining so well!
Square root of 100
That was my intuition too. It turns out to be true, but i had no idea how to prove it
if you change the number of people to any square number it'll be the root-th piece (assuming using fractions 1/m, 2/m, 3/m... with m people rather than percentages or u end up eating more than one pie.)
The lucky guest will get largest piece of pie!
One pie equals two pi.
Not true. I like pie, but I don't want a slice of pie the size of a small moon. And if the 100th guest is to have at least an atom, then the pie will need to be that big.
I wonder who the actual lucky guest is that gets enough to enjoy but not so much that it's impossible to eat in one sitting?
Minute 1:53 How did you know that the remaining pie was 99/100*98/100?? If you do the problem you get 99/100 (the amount of pie left from previos persona) minus 2% of 99/100 (the amount given to the second guest) and you get 99/100 - 198/10000. What was the mind process that you did to know that, that subtraction was equal to that multiplication??
Hello my friend, I totally agree with you! I think the solution presented is wrong, because it assumes that the remaining pie is exactly 98/100, 97/100, 96/100 and so on, which is not accurate: it doesn't take into consideration the exact portions the previous guests have taken! I don't know whether the right function has a maximum for k=10, but I think almost everyone has missed this detail!
(98/100)x=(100/100)x-(2/100)x
When the problem asks "take 98% of what remains," either approach is a mathematically accurate interpretation, but the approach on the left is more useful for further simplifying and getting a solution.
Presh: No calculators allowed
also Presh: In order to solve this problem, we must first calculate...
D’oh! I’m a calculator.
So the answer is 2% of π? 🤨 Dammit π, stop popping up everywhere. ¬_¬
(Yes it's just a coincidence, the question could just as easily have been a pizza or cake or pile of money being split up.)
Actually, it's not tau. It's the inverse of the root of e
:^)
@@Seb135-e1i the limit is 1/(n*sqrt(e)) if there are n^2 guests as n goes to infinity. In this case n is pretty small so the answer is a bit larger.
It is not a coincidence God put pi there
I came to the same conclusion, but without all the factorial stuff
I just considered how the values change
so going from the first piece, which is 0.01, to the second one 0.99*0.02 = 0.99 * (0.01*2), we multiply by 0.99*2. Going from the second to the third, we multiply the second value by 0.98*3/2
The pattern is that the first factor always decreases by 1%, while the second one is the current element number divided by the previous one
Those 2 factors can be written as (1-n/100)*(n+1)/n
now I'm just interested when this is becoming
2:17 I dont understand how you derive this formula
Sadiq Abbaszade sane here. Anyone can help?
Yup me too. I am under impression 2% of 99/100
99 is the 1st term, and k represents the term we're looking for, say k = 1. You'll get 99-(-1) which will be 100% which is the original pie size
This is the quality content I subscribe for.
I always start Mind your decision videos with a like, there after I start to watch the video🤗
After a long time of fondling with numbers, i was like, let's see how Presh is saving the world, again.
2pi??. You mean tau
Well yeah, thats what he said
*t w o p i*
I took this problem without knowing the answer and told it to my friends to act like I’m smarter than I am.
Honestly, when you're out of school, learning math is 100x more fun. The education system sucks the fun out of math, reading and science
Wrong.
Error (1): It's not the education system that sucks the fun out of math. What sucks the fun out of math is all the other kids not having the slightest interest in math.
Therefore, error (2): The reason for having fun with math after school is actually not the fact that *you* are out of school, but the fact that *those other kids* are out of school as well.
I totally agree french man. Learning out of passion is way more effective than by an autority. Though we need people to inspire them and give them the materials needed.
That second inequality at 3:45 100 > k^2 + k is also the same as asking which is the greatest triangle number under 50. I wonder if there is an alternate way to express this in terms of those simple consecutive sums.
Me: I think I got it
"Watches explanation"
Me: This is way more complicated than what i did I have to be wrong
"The answer is 10"
Me: oh, I was right
Did you use simple arithmetic also? So easy to just round the numbers as they don't have a HUGE effect on a problem of this sort... Of course you could have used Euler's constant and limits and came to the same answer... funny how flexible Math can be for finding solutions.
I solved this with a slightly modified compound interest formula and calculus.
I set up the formula y = P(1-r/n)^tn where P is the principle amount (100), r is the "interest" rate (-1% or -0.01, multiplied by the number of times the pie has been taken from), n is the compounds per year (which I set to 1), and t is the number of years (the number of times the pie has been taken from) as:
y = 100(1-0.01t)^t
I then took the derivative:
dy/dx = 100(-0.01)ln(1-0.01t)(1-0.01t)^t
The 100 and the 0.01 cancel to get:
dy/dx = -ln(1-0.01t)(1-0.01t)^t
Setting this equal to 0, ln(1-0.01t) will never equal 0 for positive integer t, but (1-0.01*10)^10 is 0, so that's either a local minimum or maximum. Testing the value of dy/dx just below and above ten, you can see that it switches from positive to negative, so t = 10 is a local maximum on the original curve, so guest 10 gets the biggest piece.
A great puzzle !!! I solve it in my mind this morning, stuck in traffic jams (1 hour long !).
:-)
Pythagoras invitation 100 people for a pie party.
People after watching this video : fighting for the 10th seat.😆
coding solution:
fraction
Sebastian Lol
Which programming language is this?
When you are a 13 year old student seeking joy...
I thought it would be somewhere around the 10th-15th guest logically but I wouldn't be able to calculate it.
"simplify this into *something even more confusing*"
Fun Fact: Guest 44 gets 0.001% and everyone after that gets 0.000%
I started a different way but ended up with in the same spot:
The remaining Size of the pie is (1 - 1/100) * (1 - 2/100) * ... * (1 - n/100)
So the pie, that the next guest Nr. "n" takes from, is always (1 - n/100) smaller than the piece of his predecessor.
But Guest Nr. "n" takes (n+1)/n more than the one before him.
That means if both factors together are above 1, then the piece he takes is bigger than than the ones taken before. We just have to find out the single piece after which the sizes of the slices are not increasing anymore (which means both factors combined are roughly 1) so we have now:
((n+1)/n) * (1 - n/100) = 1
which can be simplified into:
1/n - (n+1)/100 = 0
after some rearranging we get:
n^2 + n = 100
For which I would have needed a calculator so I guessed something between 9 and 10 and failed, cause I choose 9 XD
Who's that host that wants to divide a pie to a hundred guests.
Bro in 4th class a dude had a 100 watermelons
Rmon Velsco a cheap baster lol 😆
+MindYourDecisions Presh, you really dropped the ball on this one. If the kth of N guests gets k/N of the remaining pie, you can show that k=sqrt(N) will get the largest piece.
Could you further explain how you deduced that? I am interested in understanding how sqrt(N) applies.
@@mattrogers6646 I have difficulty doing equations in comments, but I'll give you a verbal head start. Follow Presh's method, but substitute the variable N for the given constant 100. When you simplify the inequality for the ratio of k+1 and k, you should get k(k+1)
1. 0,01 = 1%
2. 0,02 * 0,99 = 1,98 %
3. 0,03 * 0,9702 = 2,91 %...
2:23 how did you make that simplification
thats basic arithmetic
define a function for the size of the piece of the pie for the x‘th person. p(x)
say p‘(x)=0
and you get x1
if p‘‘(x1) < 0
you know person x1 has largest piece
I didn’t understand this leap from k=9 with the fraction greater than 1, to “size increases to guest 10.” Wouldn’t that be k=10, so that would make the fraction less than 1? 4:27
That one was a hard one, without a computer/calculator.
The 10th guest would gets the largest piece of the pie.
To calculate what 3rd guest gets, u need to subtract guest 2 and guest 1's share from the total and then multiply 3/100 ..U did not subtract guest one's share!!
My guess was correct. I figured 10 th to 15 th guest will get the biggest Piece because the remaining amount of the Pie gets smaller and smaller even though the percentage increases.
Oops I used a spreadsheet to get the answer.
10. guest gets 6.28% of the whole cake.
I could do the mental math, but I think the point is to think of a more elegant solution, which I can't. My intuition is that the largest piece will be somewhere between 8 and 12% (of the total pie, that is).
I was right about it involving factorials, yay.
There’s a much easier way of thinking about it.
Guest 2 gets 0.99 x 0.02/0.01 > 1x as much as guest 1, because when the pie gets to the 2nd person there is 99% as much pie as when it got to the first person, and guest 2 gets 2% of that whereas guest 1 got 1% of theirs (so 0.99x as much pie left, but 2x as much of the remaining pie gets taken)
Similarly, guest 3 gets 0.98 x 0.03/0.02 > 1x as much as guest 2
Generalizing, guest n+1 gets (100-n)/100 x (n+1)/n times as much as guest n.
So, the guest who gets the most pie is the last guest who gets more pie than the guest before them, or the highest value n+1 where the above equation is still greater than 1.
Now we have a simple algebraic equation:
-> (100-n) x (n+1) > 100n
-> n^2 -99n - 100 < -100n
-> n^2 + n < 100
-> n = 9
-> n+1 = 10
QED
intuitively i figured it should be half way. I was right on a log scale
Lesson to be learned : Don't be late to the party.
No. Be the 10th person to the party.
take the entire pie
What's the difference between jam and jelly???
Lol they are completely different things. 😂😂😂
You can't jelly a slice pie in your mouth.
(Rated-G version of actual joke)
This is one of the best math tricks
How I got the answer: I came to the conclusion that the pie will become small really quicly, so higher numbers are out of the question. I thought that after about 10 guests the cake would have lost a major portion, so more % would make too little difference then. So I approximated it to be 10. Turns out I was right. Sometimes "close enough" is close enough, I suppose. The actual answer in the video was very beautiful though.
I came to same conclusion on the logic then when there are two factors in a situation where their isn't one extreme that's clearly better it's usually the value where they are the same (making a square) so it was probably somwhere around 10 because you then have 10%. (It's a handy rule I think that can come in hand
It is not surprising at all that π appears in the solution... After all, we are solving for a gigantic π
But what does this have to do with pythagoras?? xD
Pythagoras was the name of the magazine the puzzle first appeared in.
Pythagoras was the 10th guest
@@arnavgoyal7952 Lol he was a sneaky bastard. Just shows you math can be useful irl.
Pi showing up is never a coincidence.
Why is it called Pythagoras pie puzzle?
cause pythagoras was the 10th guest
Pythagoras was the name of the magazine the puzzle first appeared in.
Thank you for the video! All of you friends are super awesome!
Oh! At 4:40, you say "... and just as a funny coincidence...," then you observe that the largest pie piece is roughly 2π/100.
Well... It's not a coincidence!
So long as your group size (number of people wanting some pie) is an even power of 10, then the size of the largest piece will approximate the digits of pi.
For larger even powers of 10, the approximation improves significantly. Try cutting the pie for 10,000 people or 1 million people -- you will get more and more of the correct digits of pi (in base 10).
This _must_ be so. Why? I'm not telling you! I prefer to drive you insane by trying to figure out WHY that factorial-rational expression (from 3:30 in your video) _must_ converge to a whole-fraction of pi.
(Hehehe...)
I tried it for 1000 people and I got 0.0193839 for guest 32. Doesn't work unless I messed up.
Adrian K No, you are wrong. It works. The comment specified even powers of 10. 1000 is not an even power of 10. You messed up.
I know why because of some blues and browns.
@@pgoeds7420 Niiiiice!
Of course, there are other means to have known this interesting property... but that clickity-clackity square thing IS pretty awesome! Mind bending...
@@pgoeds7420 are there three and one of them too?
me, an intellectual: uses trial & error
What a world, I'm watching maths videos for fun now
2:34 I wanted to see how could that function look so I typed it in Desmos with inserting 'x' for 'k's and got very weird function existing only for X from (-153; 198.41) and (198.977; 199.019).
Have I broke Desmos or is that due to domains or something like that?
Even though that equation simplifies to small numbers the exponential and factorials both produce ridiculously large numbers, so at those values of x you’re exceeding Desmos computational thresholds... at least that’s my guess.
I’m also thinking desmos might be broken in some way because at values k>100 you’re taking the factorial of a number < 0, which is impossible... so the domain should be -infinity < k < 100.
Edit:
Yeah just plugged it in myself. Desmos is definitely not happy computing with numbers of this magnitude. It’s busted.
4:50 did you mean: π/50
Same
I figured it out on my computer, then started watching the video and was told I'm not allowed to use a computer. Oops, sorry.
It's easier to think about it as the expression k(100-(k-1)) which simplifies to 101k-k^2. Then find the maximum value of this which is √101 which is around 10.
I've come to a weird conclusion taking this a step further. What if you devide the piece in to "n" persons?
Well the first steps are easy:
in 1 person, the first person gets ... all the pie
in 2 persons, both the first and the second get 1/2 a pie, so they are both "winners"
in 3, 4 or 5 persons, only the second person is the winner,
in 6 persons both the second and the third person are winners,
in 7 to 11 persons he third person wins,
in 12 it will be the 3th and 4th that win,
from 13 to 19 it is the 4th,...
If n= persons and k= winner than:
if n=k*(k-1) (exept for k=1, this is when n=1) than both k and k-1 are winners
if k*(k-1)
I doubt Pythagoras used percents :) Extensive use of hundreds came after French revolution (metric system).
You lost me at 98/100 so you can imagine how much I understood of the rest of this problem.
50th is my guess
damn that wasnt even close
To be fair, it is a reasonable guess since the question conjures images of the binomial distribution.
An example of the crazy problems you are given in Math.
Like αν καμαρώνεις και εσύ που είσαι Έλληνας...
If you solve the quadratic equation at 3:48. And change the 100 to N, so that each person gets their position / N of the pie, you get the equation.
k = root [ (N + 1/4) ] - 1/2.
For N = 100 (this problem), the person after the ≈9.5 spot (i.e. #10) get the most.
For N = 10, the person after the ≈2.7 spot (i.e. #3) gets the most.
For N = 20, the person after the ≈4.0000 spot gets the most - meaning #4 and #5 get the same.