Two Candles, One Cake (Part 2) - Numberphile

Ben Sparks continuing from Part 1, which is at: • Two Candles, One Cake ...
Go deeper with Ben's explanation at • The Cake and Candles -...
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[@numberphile]

Part 1 is here: kzbin.info/www/bejne/fJy5lmunp7ljgqM Go deeper with Ben's Geogebra explanation at: kzbin.info/www/bejne/ZpbUc3psod9lo6c

[@PhilBoswell]

Your playlist has triggered one of my pet peeves: this "Part 2" is after the associated "Part 1" 🤦‍♂️ Sorry!

[@doctorscoot]

What if you add the restriction that the cut must pass through the centre, which is normally the way one would 'cut' a cake: a random number between 0 and 2π is all you need, (i.e. the angle, in radians)?

[@MathNerd1729]

The fact that it's way more complicated in 2-dimensions reminds me of a couch problem which is simple enough for a Calculus class if we use a 1-dimensional couch, but is unsolved as of now if we use any 2-dimensional couch. Nice vid! :)

[@andrewyoumans2161]

Now bring it into 3 dimensions! I would be interested to see the distribution of two points on the surface of a three dimensional object intersected by a plane.

[@fabrizio3408]

Hello @Numberphile, you guys are excelent at math so I was woundering if you know if/how I can solve a function like f(x)= a*3rd root(x^c-1) +0.538 when I have 5+ points and it‘s not the third Root but any other uneven root? I am trying to find this out for school for a programming project in which I want to put a function. I only got close with geogebra. I would be very greatful hearing from you. :) Yours faithfully Fabrizio

[@jaopredoramires]

The "what are the chances?!?" after he cut between the candles really got me

[@omgawesomeomg]

Never tell me the odds!

[@vigilantcosmicpenguin8721]

If only someone could determine the chances...

[@majuss06]

"Don't answer that" is what got me :)

[@tuckergymnest0]

@@majuss06 math jokes are always the funniest

[@Triantalex]

??.

[@EchosTackyTiki]

"My main conclusion of this is that I didn't know what's going on." I felt that statement on a spiritual level.

[@RichardWinskill]

I feel like the random point and random angle method more closely matches how I would approach cutting a physical cake into 2 random pieces

[@prdoyle]

I thought the same. I'd turn the knife to some angle and then whack the cake in a random place.

[@jursamaj]

Problem is, "how I would do it" doesn't make it actually random.

[@RichardWinskill]

@@jursamaj True, but It was more about expressing a preference for one of the presented methods; as in "this one feels like the realistic one, and I like that"

[@jursamaj]

@@RichardWinskill Yes, but my point is that sort of 'gut feeling' is often wrong. A good analysis shows that in this case, it is correct. And also shows that the resulting probability is more like 28.8%.

[@dowesschule]

I‘d say choosing a random point on the circumference and then a random angle would be the most realistic representation so it would answer the question for 2 candles on a 2D-cake. What is actually random for a line and two points on circle is probably unknowable and always depends on the context.

[@steffahn]

5:08 So here are exact values for all 4 cases: random end points: 1/3 - 5/(4 π²) ≈ 0.206682 random mid point: 1/8 + 2/(3 π²) ≈ 0.192547 random radial point: 128/(45 π²) ≈ 0.288202 random point random angle: 1/3 ≈ 0.333333 And as a bonus, here's a distribution of cuts that wasn't in the video: Take two random points inside of the cake and make a line through them. results in: 1/3 + 35/(72 π²) ≈ 0.382587 Exercise for the interested reader: Verify these results (or show where I might be wrong).

[@Doeniz1]

I worked out the four cases as well and came to the same results. But I had no idea how to handle the bonus case with two candles inside the cake. Can you tell me how you've done it?

[@gmalivuk]

I believe based on your result for two random points and the cut through them, it implies that the probability that four random points in a circle will *not* form a convex quadrilateral is exactly 35/(12 π²)

[@steffahn]

@@gmalivuk I know, that's a fun “roundabout” way of getting that result. Not that I would know of any more direct approach to do it... (I, too, actually made the same kind of observation myself even before first posting here.) To be clear: I did not proceed the other way, determining the probability for the convex hull of 4 points in a circle forming a quadrilateral as opposed to a triangle was not a step of my approach.

[@gmalivuk]

@@steffahn Out of curiosity how did you approach the random interior segment calculation? I got the same result with a numerical integral in Mathematica, but the expression I was integrating was far too awful for it to give me a nice closed form result like you got.

[@steffahn]

@@gmalivuk Which one is "random interior segment"?

[@JNCressey]

7:13 I think if you asked someone, without briefing them, to randomly cut a cake that they would pick a random angle or point of the edge, but then have the cut going through the centre.

[@romaindubray2325]

There are a lot of biases to consider here, all of which seem really interesting in their own right but complicate the problem beyond probability. If you casually ask in a "normal" context, I doubt people would even consider not cutting "fair"/equal parts. If you explicitly mention randomness in your request, I'd expect a lot of people to instead avoid the center, but still creating two comparably sized parts. I'd also imagine the majority of people to cut from where they stand and without changing the alignment of the knife as they would usually hold it. And lastly, I'd expect the vast majority of people wouldn't consider anything but a vertical cut, or a straight cut.

[@tristanridley1601]

Depends on the person probably

[@davidwebb2318]

@@tristanridley1601 Yes, if you ask mathematicians I expect you would get a completely different distribution of cuts than if you asked artists.

[@Qermaq]

The candles are random too. What if they were clumped way over to one side? Then it would be difficult to cut between them.

[@tristanridley1601]

@@davidwebb2318 Or just two different mathematicians or artists even.

[@andriypredmyrskyy7791]

"what are the chances?! Don't answer that"

[@Marconius6]

As a programmer I'm watching this and just screaming internally... Any time someone asks me "oh yeah, just randomly generate it", I now how to be like "but which KIND of random do you mean... they're all different... there's no good answer... what do I do..."

[@UsernameXOXO]

You generate it based on a bunch of parameters at once and then get complaints when there is a streak of similarities because it's "not random enough" 😝 Iirc apple had this issue when they released the iPod shuffle because songs would play twice in a row or more, something that could absolutely randomly happen... So you add more parameters to ruin the randomness to make it actually useful outside of analytics.

[@faastex]

8:12 "What are the chances!? Don't answer that" lol

[@Cliff86]

I like the method where you pick 4 random points in the circle. Then pick 2 of those randomly to define the line of the chord, and take the other 2 as candles That makes the answer one third

[@timseguine2]

It definitely seems like the most natural generalization to me. The method would even generalize to candles and hypercuts on a hypercake.

[@Shiandow]

The method where you pick a point and a random angle for the line also gives you a third by a similar argument to the linear cake if you rotate the cut such that it is horizontal (a change of coordinates which leaves the distribution of the two candles invariant).

[@danielyuan9862]

Nope. It's greater than 1/3. If one point happens to be within the triangle formed by the other three, then 3 of the 6 chords split the other two points instead of 2.

[@nathanp6928]

Isn’t this the same as choosing 2 random points for the chord and then 2 random points for the candles instead of randomly picking 2 of the 4, because the events are independent?

[@RollcageTV]

This was my idea as well. I would have liked to see them cover this method.

[@erickehr4475]

It feels to me like picking two points at random and cutting along the line that joins them is the most intuitive way to cut a 2D shape at random. And it easily transfers to cutting a 3D shape at random by choosing 3 points and cutting along the plane they define.

[@goininXIV]

That one is pretty much guaranteed to be non-uniform though, and it's easy to see: Use a concave shape, let's say the outline of the letter c. You know have cuts that hit the edge in 4 places and are thus defined in 6 point combinations rather than just 1. Why should that cut be more likely than another? And picking a random point and angle transfers to 3d just as well, there you just pick the angle of the normal. (which of course leads to the question of how to uniformly randomly pick a 3d angle ...)

[@lynk_1240]

@@goininXIV I fail to see how Eric's method creates non-uniformity of line distribution. If we are talking about a concave shape, it is obvious that there will be some lines that cross the shape more than once, but how does that make the lines non-uniform? Who stated we were limited to slicing a shape into only two pieces? Placing that kind of limitation is what would cause a non-uniformity.

[@victorribera5796]

The reason I may see it as non uniform is that: for example, take a square that surrounds the most tightly possible your shape, then take two points near a corner of the square that does not contain the shape, those cases would be always rejected which makes me feel like distributes them more towards passing through the center, but I'm not sure and it is only intuition, I haven't made any calculation

[@truefiasco2637]

you could argue it's just as intuitive to pick 2 random vec3s for planes centre position and Euler rotation

[@lynk_1240]

@@victorribera5796 But we are talking about points inside the circle, not outside of it.

[@OrangeC7]

My favourite part was when he said, "So, in conclusion, I have no idea." (5:14)

[@3snoW_]

You could make the cut always be vertical, because for every angled cut you could just rotate the entire cake until the cut was vertical, and the candles would move but since their positions are random anyways you shouldn't be affecting the overall distribution. This way you don't have to deal with the "how do you cut a circle randomly" problem, you just assign a point randomly between 0 and 1 and do a vertical cut there.

[@jursamaj]

Exactly this. Always look to simply a problem without loss of generality.

[@3snoW_]

@@jursamaj actually, ive though about this and it isn't quite the same as what he's doing. If you do as i suggested and pick a random value between 0 and 1 to be its X coordinate, then the chances of it being between 0 and 0.1 are the same as the chances of being between 0.2 and 0.3 or 0.4 and 0.5, it's always 10%. However, the other 2 random points don't have uniform distributions for their X value, they're more likely to be near the center because of the process he uses to generate the points: random (X, Y) pair, if outside the circle try again. So if X = 0.1 then Y has to be between like 0.4 and 0.6 for the point to be accepted, however if X = 0.5 then it will be accepted for any Y value.

[@jursamaj]

@@3snoW_ That's not a problem. His method for the points makes any point in the circle equally likely to be chosen.

[@michal_wieczorek]

I was thinking of a same thing. You can make a cut first, always vertical, then choose the points. The points will have normal distr in the circle, however, as you said in answer, points will not have norm distr if you kinda place them all on horizontal diameter of the cirlce, because of the fact that circle has more area in the middle. So what have i done is I did the math but probabilities of points were based on the area of the circle. Using some geometry you can figure out area of the cut part of the cake based on the distance between the horizontal end of a circle and the cut. Than when i had the function of an area, you can calculate the probability that poinst will end up in different parts, but using an integral (a nasty one). Soo, I coded something to calculate it somehow closely and I got ~0.3 :)

[@jursamaj]

@@michal_wieczorek I did the same, but basically did numerical integration, because I knew the formula would be nasty. I did it in a spreadsheet with 1000 uniformly distributed cuts, and got .2882…

[@rachidvanheyningen]

This paradox is part of the reasons why I keep telling people "science is óur way of making óur reality comprehensive and pragmatic for ús; nót a way to actually describe reality"

[@user-gc8cy5tx1w]

At 5:13, it's possible to calculate explicit values of these probabilities. Random end points : 1/3-5/4/pi^2 ~ 0.20668185378 Random mid point : 1/8+2/3/pi^2 ~ 0.19254745576 Random radial point : 128/45/pi^2 ~ 0.28820247796 Random point random angle : 1/3 ~ 0.3333333333 (surprisingly it's a rational !)

[@DumbMuscle]

Random point random angle I think reduces to the 1d case, if you take the projection perpendicular to that angle (this assumes that for a random point on a plane with uniformly distributed x,y, it also has a uniform distribution with the projection of those points onto any line of the plane). Which would be why that ends up rational (exactly where the cut is will depend on the angle, but for any choice of angle the probability is 1/3, so the distribution of the angle doesn't make a difference so long as it's independent of the distribution of the points)

[@steffahn]

@@DumbMuscle The distribution after projection is not uniform, however the important point here is that the distribution of each of the two candles and the distribution of the cut are *the same* distribution, after projection onto the line perpendicular to the cut. When those three points (projected two candles, projected one cut) are distributed (independently) with the same distribution, then all 6 possible permutations for their order (ignoring probability zero cases where points coincide) have the same probability, and 2 of the 6 permutations have the cut between he candles. So while it doesn't actually reduce to the original 1d case, it does reduce to *some* 1d case that can be solved using the same kind of permutation argument.

[@rdtim-bx1nq]

I’m not even remotely good at maths, but I’ve always love the ways and creativity you present these problems and theories. For some reason, watching these videos really bring peace to me. Thank you for making these videos.

[@frederf3227]

Your placement strategy affects the density distribution. E.g. Two endpoint chord gives a donut density. As long as your cut and candle placement densities are the same thenyou get ⅓ regardless. It's only when you mix and match density characters that you get deviation from ⅓ which is a measure of interaction of the different density maps. Uniform is only objective if you privilege one coordinate space over another, e.g. random R, random Theta is uniform in R, Theta space but center heavy as a disk in X,Y space. And conversely circular disk X,Y uniform is center thin in R,Theta. Which is considered uniform depends which is your favorite space.

[@OliverJennrich]

Late to the party, but: The rotational symmetry here is a blessing. Imagine placing the candles at random and chose a chord for cutting the cake. After you picked the candle positions and where to cut the cake, you can always rotate the cake so that the cut is parallel to the y-axis of a coordinate system. Rotating the whole setup cannot change the outcome of the experiment, so it cannot change the overall probability. So all you have to figure out is the areas of the two segments of the circle that are formed by the cut, as the probability of a candle to be in either part is proportional to the area. If we denote the fractional area (ie the area of the segment divided by the total area of the cake) as A, the probability for candle 1 and candle 2 falling into different segments is p=2*A*(1-A). If h is the 'height' of the segment, then A is arccos(1-h)-2sqrt(1-(1-h)^2). With that p(h) can be integrated over h=0..2 and yields 128/(45 pi^2) = ~0.288.

[@xk445g]

SO basically for "every" point from 0 to 1 along the diameter of a circle, the area of each section defines the probability of a candle landing there. , which almost brings us back to a 1 dimensional problem. Only where the distribution is focused towards the middle.

[@thenoobalmighty8790]

The rotational symmetry helps yes, so if we put the cut line as the vertical and work from the extreme left of the cake its the same problem. 1/3 chance we come to candle 1 2 or the cut first etc etc so its 1/3. No need for any trig 😅

[@thenoobalmighty8790]

0.288 is wrong 😂 its 1/3

[@Jiffy_Park]

growing up I had a lot of arguments with friends about probability, so it's nice seeing that even the probability sometimes argues with itself

[@Roopanini]

8:10 "What are the chances!? Don't answer that.." haha so perfectly said

[@Bunchachis]

Ben: Ah, we're between the candles! What are the chances! Me: I thought you would tell me!

[@domzi]

Cuts the cake and creams "what are the chances?!" THAT's what this video is all about 😂😂😂 love it!

[@wayfinder1882]

hearty chuckle at "what are the chances!"

[@martinepstein9826]

I just had a thought about the method where you define the chord by picking two random points on the circumference. It looks reasonable if we do this on a circle but imagine doing it on a square. Each chord would have a 1/4 chance of being an edge!

[@gmalivuk]

Yeah, in this way the only two options that seem to generalize nicely to other shapes and dimensions are either picking a random point and a random angle, or picking two random points and considering the line between them. In 3d that's equivalent to picking a point and a normal direction (though be careful that direction is spherically uniform) or else picking three points and drawing the plane that includes all of them. Unfortunately, I believe these two methods still give different results from each other, so while it eliminates some "bad" options it doesn't give a satisfactory final answer to the question.

[@cytotoxic6247]

5:14 Relatable "My main conclusion of this is --> I don't know what's going on."

[@luisfonseca2299]

For the one with a random angle for the cut, here is my "intuitive" explanation for it to be 1/3: Given a random angle and 3 non collinear points (here we have probability 0 of having the 3 points in the same line), there is always only one point that divides the other two when cut in that angle. Since we have 1/3 probability that point is the knife, the answer would also be 1/3.

[@johnmulhall5625]

“I wasn’t sure, and didn’t think about it.” Clearly was thinking about it…a lot lol. Love these videos

[@isavedtheuniverse]

"the point candle model was nearly not sufficient" OMG!! I'm dying.

[@allisonmarston9777]

Choosing a random midpoint makes the most sense. Cuts through the middle contain more points than cuts closer to the edge. So if you choose a random point and angle, cuts through the middle are more likely than cuts closer to the edge. The reason the simulation of random midpoint cuts is sparse through the middle is because all possible cuts cut near the edges, but only cuts through the middle, well, cut through the middle. We want each cut to be equally likely, but there's a wide variety of length to those cuts, so a simulation of random cuts should be thinner through the middle.

[@johnchessant3012]

It's also worth noting that in real life, if you were cutting a cake for two people, you'd probably cut it in half so each person gets the same amount of cake. This would give an answer of 1/2. (Place the cut first; the angle of the cut doesn't matter, and the probability that the two candles are on different halves is 1/2.)

[@Ovetupp]

This is really interesting and an important lesson in itself. This communicates so clearly how even unintentional biases ends up affecting the end result.

[@Kwauhn.]

8:10 "What are the chances!?... Don't answer that." lol

[@justarandomanimegirlpassin5341]

''real cakes ,not linear'' yeah thats the way my brain think after my third math lesson in a row

[@esce69]

Loving this one more than part 1. Questioning one's bias and coming to the conclusion that random isn't random.

[@Sgt-Gravy]

With a round cake, as long as the cuts are completely vertical, you can use just half of a cake, because no matter the angle the longest cut is cutting the cake in half. The cuts can be plotted by picking the intersection of the halving cut & the outer edge of the cake, & randomly choosing any other point.

[@andrewkepert923]

Distribution #4 wasn't considered by 3B1B's vid on Bertrand paradox, but it is the one with the simplest answer to the CCK problem: it will give probability of 1/3. The difference between distributions #3 and #4 for the position of a chord can be understood intuitively. In both you can choose position and direction independently, and the measured chord length (in the Bertrand paradox) or the relative size of the two circle segments (in this candle-candles-kut problem) doesn't depend on angle, so you can assume the line is (say) vertical. So it's really about the choice of the position you're going to put the vertical line. In #3 the position is a uniform distribution between -r and r. In #4 you are choosing a point in the circle randomly and taking the x coordinate from that as the x coordinate of the chord. This skews the distribution so you are "more likely" to choose one of the longer chords, as there are more points on it to choose. Obviously this isn't literally true, as all these probabilities are 0, and all chords have the same number of points. However it can be formalised in a way that carries this intuitive sense of being more or less likely. When you use distribution #4 in the Bertrand paradox (this one wasn't done in 3B1B vid, from memory) you get a probability of 1/3 + √3/(2π) ≈ 0.609 - you are more likely to choose longer chords than the other distributions. What this means for candle-candle-kut is that #4 you can remove the angle from the problem, and you're just choosing three points uniformly on the circle, and then comparing x coordinates. By the simplest ordering argument covered in the first part of the first video (i.e. symmetry under permutations), #4 should give probability of CKC being exactly 1/3. By thinking about how the other distributions are more likely to give shorter chords, hence unequal regions, you can deduce that they're more likely to give CCK or KCC than distribution #4.

[@andrewkepert923]

Simpler argument for 1/3 in the case of distribution #4: to choose a kut using method 4 and two candles, you are indepently choosing 3 points uniformly from the circle's area, and one direction for the cut. You then project the three points parallel to the cut, say onto a line perp to the cut. You want the probability the projection of the third point lies between the projection of the other two points. Since all 3 points are chosen from the same distribution, the probability is 1/3.

[@anononomous]

Trying to slice blind at random: asks the guy looking at the cake to say _when_ 🙂 8:06

[@skyscraperfan]

If you have a circular cake, the likelihood is smaller than with a square cake, because if you draw the largest possible circle on the square cake and the do the random cuts and candle positions, it is more likely that at least one of the candles misses the circle than that the cut misses the circle. So the average distance between the candles on a circle shrinks more compared to a square than the distance of the cut from the center shrinks. You could see that by doing a simulation with the square and the circle at the same time with the same cuts and candles and than look at the cases where either the cuts or at least one candle is outside the circle. If you choose two random points on the square, the likelihood that both are on the circle is pi^2/16, which is about 0.61685. However if you do a random cut of your square by first picking a random point and then a random angle, the likelihood is much closer to one. If that random point already is on the circle, the line will also cut the circle. That already gives us a likelihood of pi/4=0,78539..., but even if that random point is not on the circle, there still is a high chance that the cut will also cut the circle. I have to think about a formula for that likelihood, but even in the most extrem case (if the point is one of the corners of the square), half of all possible angles will create a line that also cuts the circle. That brings the overall likelihood to at least pi/8+0,5=0,8926....

[@hwangsaessi2335]

The important thing to realize here is that it's not the maths that is unclear, it's language (the word "randomly") and the physical process of cutting the cake. If you properly define the distribution of cuts and candles, you can always get a clear answer mathematically. Wether or not that answer is "useful" is another question altogether, and is sort of outside of maths.

[@tonekjansen5597]

Several people have shown that the result for the Random Point Random Angle method should be 1/3. I like it that their arguments show that the result does not depend on the shape of the cake. You can also extend it to higher dimensions. Suppose you have a 3D cake with two cherries hidden in it. You make the cut by randomly choosing a point in the cake and randomly choosing the normal of the plane of the cut. The probability that there is one cherry in each piece will be 1/3. And if you are worried about how to actually choose the plane randomly, it doesn't matter.

[@mattofbum]

I think the fundamental issue with methods for choosing random chords is that different methods will have different frequencies of duplicating chords. Take the point and angle method. There are many point-angle combinations for each possible chord. If you were to choose every single possible point-angle combination, some chords would appear more often than others. Each method probably selects each chord with differing frequencies, leading to different solutions for the problem.

[@AndrewTyberg]

This was a super intriguing series of videos! I like how (as of writing) part 1 has 166k views and part 2 has 92k views. Over 50% watch rate for a follow up video is very high.

[@justdata3650]

The fourth one isn't influenced by the shape of the circle (random point and angle) and so truly treats the entire probability space evenly so I see that one is being the more correct but as you say it comes down to definitions.

[@topilinkala1594]

The complication is that no-one cuts cakes randomly. When you try to imagine what happens when you start cutting randomly you realise first that all deliberate cuts have randomness in them and if you up that you get random cut but that is not uniform distribution it's normal. So trying to figure how to get a uniform distribution we have the same problem as in Bertrand's paradox.

[@danielbranscombe6662]

One way you could approach random cuts on the square cake is in 2 steps 1) pick 2 different sides of the cake at random 2) pick a random point on each of those 2 sides then the cut is the line connecting those two points. The idea is that every cut of the cake hits the perimeter at 2 different points and because it is straight line those 2 points can't be on the same side.

[@riadsouissi]

Again using integration, we can solve some of the random cases. For instance, the knife cut is a uniform distribution of the distance r between the line and the center, so it ranges between -1 and +1. Because of symmetry, we can consider the cut as vertical and r becomes the x-coordinate of the line intersecting with the x-axis. Let (x1, y1) and (x2, y2) be the coordinates of candle 1 and 2 respectively. We want (x1>r and x2

[@penteractgaming]

its 1/2 -128/45pi^2

[@Dysiode]

I'm glad you touched on at the end of how a person would cut a cake. I'm curious how picking a random point along the circumference and a random angle would play out. That's my estimation of how I'd cut "randomly"

[@radonato]

Random point on edge of circle, then random WALK until the other side is reached

[@goseigentwitch3105]

What I'd like to see is a batch of pre-made cuts something like 30 cuts all parallel evenly distributed along the cake, then all 30 rotated 1/30 around the center of the cake 30 times to make 900 total cuts then just pick randomly from among those 900 cuts and see which of our random distributions that curated set of cuts approximates something like that might give us a better understanding of what each way of making a random cut represents

[@KipIngram]

If you pick a random point in a circle and let it be the mid-point of a chord, of course you are going to get points more likely at larger radius, because more area of the cake is out there. On the other hand, if you pick random angles around the cake and take the end points on the periphery at those angles, that's entirely different. I don't see this as "not well defined." You just have to decide WHAT YOU WANT, and then you can do that in a perfectly well defined way. I don't know if I think of that as "philosophical" or not - the difference between the two schemes is really a matter of GEOMETRY.

[@ReynaSingh]

The desire for neat mathematical solutions comes from wanting to tie up the loose ends of existence. So many answerless problems, let math be the haven

[@bushka2000]

There's a third way to do this, which is to calculate the probability curve and then integrate. If we call the position of the cut x from 0 to 1, then the chance for both candles to be on one side is x*x, and the chance for them to both be on the far side is (1-x)*(1-x). So the total chance is x^2+(1-x)^2. Simplified, that comes to 2x^2-2x+1. Integrated over 0 to 1, that comes out to 2/3, which leaves the chance to not be on the same side as 1/3. For a note on the random cut on a circle thing, I think the key thing to think about here is to first imagine what it would look like to make every cut on a circle. That's easier to figure out, as you could start with some angle theta, and then start with a cut on a tangent line using that angle. Then move an infinitesimal distance over and make the same cut again, repeating until you reach the other tangent line. Once that's done, you rotate theta by an infinitesimal amount, and do the whole thing again over and over until you've rotated just short of 180 degrees (at which point you would start repeating). The point of this is that for a distribution to be uniformly random, it should be equally likely to give you any of those cuts. The simplest way to accomplish that would be to make it so that it could only produce any given cut in exactly one way, and then make it so that it has an equal chance to create any cut it's possible of creating. Generating a point and then an angle doesn't work for this, because it could (in theory) produce the same line many different ways, by placing a point anywhere along that line and then randomly getting the angle for that line. Getting two random points on the circumference seems promising at first, as it does uniquely create every single cut only one way (well two ways technically, but that works fine too), however the it's not equally likely to generate each outcome. As for why, imagine taking all the results from this method where the angle is some value theta. We can do this by generating a random point on a half-circle and then drawing a line at your given angle from that point. That reflects the probability of getting that line from this method, but is obviously not uniform as it will be more likely to generate a line on the ends where the circle is closer to vertical. As for what would create a real uniform distribution then, I would propose this method: Generate a random angle theta between 0 and 180 degrees. Then, generate a random number r between -0.5 and 0.5. Draw a line out from the center of the circle at angle theta to distance r, and then draw a perpendicular line at that point. This should be guaranteed to give a uniformly distributed cut (or chord) on the circle, because it satisfies the criteria I mentioned before. First, it can get every single cut along the circle, second, it can get each of those cuts in only one way, and third, it has an equally likely chance to get any of those cuts. The reason this works is because any change in either theta or r or both is guaranteed to give you a different, unique line from whatever one you started with. This is true for theta for obvious reasons, if it changes then the angle is different so the line must be different, regardless of what happens with r. That just leaves changes in only r, but we know that a change in only r must be a different line, because that moves the line orthogonally to the direction of the line, so it's impossible for the line to be shifted along it's own direction. We can also prove that this is able to generate any possible cut, because no matter what line you draw on a circle, it will have an angle measured relative to whatever static reference you want to use, and it will have some distance from the center of the circle to the closest point that is between 0 and 0.5 units. Finally, we know that every line is equally likely to be generated because our method of generation closely matches our method of generating any given cut along the circle. Theta is generated uniformly, so that part is covered naturally, and then to look at each chord we would simply sweep that line across the circle at a constant rate, and since r is also generated uniformly, we have an equal chance to get any of those lines that are a part of the sweep. Therefore, this method should be sufficient to generate a uniform distribution of cuts (or chords) on a circle.

[@ambassadorkees]

Wondering, is there rotational symmetry enough, to place the candles at random, and make just vertical cuts anywhere perpendicular to X-axis? Then expand to rotational symmetry. Or would that mess up with the candle randomness? Second question: *Isn't the dark area of random cuts actually the most intuitive correct option? * Reasoning: All cuts have to go through the edge of the cake twice, but there's no reason why they must go through the center of the cake. Again, using symmetry: If I'd take the bottom point of the circle as my cord's first point and any other on the circumpherence as the second, I have more lines near the edges than through the middle. Even when using a bottom point and an angle, again: Less cords through the middle (though differently distributed). Then use rotational symmetry (startpoint anywhere instead of bottom), and we'll have a "dark", empty center.

[@Kram1032]

there is no reason why it *must* go through the center, but each point ought to be "equally likely" to be part of a cut, right?

[@rmsgrey]

Rotating the cake and then making a cut perpendicular to the x-axis is one way of picking a distribution of cuts, but it's not uniquely and obviously correct.

[@ambassadorkees]

@@rmsgrey well, we're talking the nature of randomness here. If I would have vertical slices, I'd have all possible cord lengths from perimeter to perimeter. Than repeat for all possible starting points by rotation. No different than all perimeter points with all possible directions, only the order of construction changes. Anyway, my main point is: All cords go twice through the circumference, both long and short, but nothing compells randomness to go through the centre. This makes it logic, that the center is relatively empty. This calls for Matt to do a massive test: drop 10.000 sticks randomly over a drawn circle, use AI to photographically determine their positions, and make an overlay. Compare to mathematical construction methodes, and this could be closer to solving.

[@Leedramor]

@@Kram1032 is it? I thought "random" should be any line going through the circle should be equally likely to be chosen. *Line, not point.* But it feels like it's debatable, which makes me uncomfortable.

[@Kram1032]

@@Leedramor sure, but every line defines infinitely many points and if every line is equally likely, I'd expect every point to be equally likely as well.

[@jalenchrysos6226]

Here’s a neat proof that it’s 1/3 for any 2D cake given the random point random angle distribution: Select three points, one will be the cut point and the other two candles. Call this triangle ABC with C as the cut point. The probability that the cut direction falls between the candles is double the measure of angle C over 360° (double because the cut could be the opposite direction). The average angle is 60°, we can see this because any of the triangle points was equally likely to be the cut point and the angles add to 180°. Done! Or, suppose we select the random direction first, then the triangle. Exactly one of these three points will cut between the other two points using this pre-selected direction (the middle one, looking from the direction orthogonal to the selected direction).

[@musikSkool]

*Mathematician:* "Here, cut the cake randomly." *Me:* *Mathematician:* :facepalms: "That was too random."

[@MarekBorecki]

Can we reverse order of operation? 1. Make cut in X axis (cake is symmetrical so angle dont matter, only position) 2. Place candles at random. ? In this case probabilyty should be linked to areas of cake on two sides of cut.

[@jursamaj]

This. Preserve the symmetry, the rest doesn't matter.

[@MarekBorecki]

@@vpumeyyv if you tanking about finding cut - yes. If you talking about finding answer about cut and candles problem - this seems like more intuitive approach

[@jursamaj]

@@vpumeyyv Not having to worry about the angle of the line makes the model simpler, thus easier to calculate.

[@GuidoHaverkort]

7:42 you can see him starting to think😂

[@trummler4100]

5:01 and what about choosing 2 random points in the circle through which the cut passes through?

[@litigioussociety4249]

It creates the redundancy. It's the same as the random endpoints, but gives an infinite number of possibilities that lead to the same cut.

[@TheSwiftFalcon]

This sounds like the right answer to me. Simple, random, and propably works equally well regardless of the shape of the cake. Now I want to see a part three, where they simulate this, and also go up one dimension, using three points to define a plane intersecting a three dimensional cake which contains two randomly placed raisins (or something).

[@myvh773]

@@litigioussociety4249 Why is it the same?

[@adamx9065]

@@litigioussociety4249 Why is that even a problem though? Doing a line through two random points would work just fine for a random cut. The lesson here is that a "random cut" just isn't well defined. There are many ways to do it, but our choice of method is ultimately arbitrary.

[@theahmedmustafa]

Defining randomness in this case is hard because even in real life there are more than one ways to randomly cut a cake, each with its own random distribution. So either you take into account all of them (which is a new hole to dig for yourself) or otherwise stick to the most "natural" one. I think the most common way a human cuts the cake in two is along a diameter. So picking a random point uniformly on the circumference and making a cut on the line that joins that point and the center (the normal to the tangent on that point) seems like a decent approach.

[@litigioussociety4249]

Random endpoints (the two spinner option,) makes the most sense to me. It also makes sense to me that it would drop from 1:3 to something like 1:6.

[@jursamaj]

Random endpoints clearly clusters too much around the edges.

[@litigioussociety4249]

@@jursamaj I think that's just the nature of randomness on a circle.

[@jursamaj]

@@litigioussociety4249 It doesn't need to be at all. Consider all random lines in the plane, not just those that cut a given circle. For each such line, rotate the plane around so that the line is vertical (parallel to y axis). You should still have a uniform distribution of the lines along the x axis. They won't be clustered around the circumference.

[@litigioussociety4249]

@@jursamaj I can sort of see how they cluster, I'm not sure they shouldn't to properly solve the problem. Obviously, any could possibly be right.

[@jursamaj]

@@litigioussociety4249 I'd say that "a random line across the circle" should equally "paint" all points in the circle. Clustering at the edges doesn't do that.

[@fernandobernardo6324]

Mr. Ben, can we eat the cake already? This whole thing made me hungry

[@SkenonSLive]

Take the smallest square that contains the circle, pick a random linear x and y coordinate of your point, if it falls outside the circle, do the whole process again. I consider this to be a bulletproof way of mapping non-square 2D areas with random points, uniformly.

[@gmalivuk]

Yes. That is how to generate uniformly random points in a non-rectangular shape. But the question is about how to generate linear cuts through said shape.

[@niklaskoskinen123]

I feel like the best way to cut a circle would be to just choose a random line, and then limit the selection to those that intersect the circle. E.g., I might choose an angle, and then a position (obviously only position in the direction of the tangent matters). As a bonus, this way is very easy to limit to lines intersecting the circle

[@niklaskoskinen123]

Quickly thinking about this, it corresponds to choosing an angle and a distance from the origin.

[@wickmar3036]

I believe fixing the orientation of the knife, and only choosing a random x coordinate is equivalent. Because the candle points are distributed uniformly at random, the distribution is rotationally symmetric. This we can ignore the orientation of the knife when computing the probability.

[@cemafor1006]

"We're between the candles! What are the chances? ...don't answer that..."

[@supermarc]

For the random angle method it should be 1/3 (independent of how you choose the angle): Imagine choosing the angle first; then clearly any of the three points will be in the middle with respect to that angle with probability 1/3.

[@sarahferraro4641]

"what are the chances - don't answer that" XD

[@TheBobledanois]

"Ho my god between the two what are the chances?" you killed me XD thanks you for your quality content and the good vibes

[@davidheaps3336]

The problem is really about what is meant by a 'random'' cut through a circle. It can be defined several ways - each one producung a different result.

[@hughcaldwell1034]

If you do a square cake and define a random cut as the line passing through two random points (selected as in the video), then you get a probability of 1/3 with a really quick proof. Let a, b, c and d be random points on the cake. Ignoring cases with three or more colinear points (which occur with probability 0), these form a quadrilateral. There are six possible cuts (ab, ac, ad, bc, bd, cd), and no reason why one is more likely than any other. Only two choices of cut result in one candle per piece - those corresponding to the diagonals of the quadrilateral. Thus our probability is 2/6 = 1/3. Now, correct me if I'm wrong, but I think this works regardless of how we define "random point", so long as the candles and the cut-defining points are chosen the same way, since it then becomes an ordering problem (like in part 1 of the video). Edit: I forgot the rather obvious case of a, b, c forming a triangle with d in the middle, which messes this up. Not sure how to fix it, but I'll leave this here because I don't want to pretend that I don't make mistakes.

[@IllidanS4]

What still baffles me ever since from the first video with Grant is what actually triggers the Bertrand's paradox. Like what inability of ours causes the issue? The fact that it is two-dimensional, or something else? Can we reduce this problem to something minimal which is still affected by the paradox?

[@gmalivuk]

It mostly seems to come down to there being many different ways to define lines, and we don't all have the same intuition about what way is most "natural". I suspect it's made worse on a circle because we think of "chords" as lines connecting points on the circumference, which lends itself to the random circumference endpoints method, which probably wouldn't be a suggestion for a square cake. In addition, we know other things about chords like that each one can be uniquely defined by its midpoint, which then suggests the random midpoint approach that likewise probably wouldn't be anyone's first pick for different shaped cakes. However, the random point + random direction method, the two random points + connecting line method, and (a variation) of the random radius + random perpendicular line method would still all be applicable to other shapes (and higher dimensions), and still give different results from each other.

[@Qermaq]

We can orient the cut at any angle, so no angle is special. Well, aside from the two which are parallel to the line connecting the candles, but that case has no chance the cut is between the candles, and as it's just 2 out of infinite angles, we can ignore it. So just as in the 1D example, the cut can be to the left or right of the candles, or between. Thus it's 1/3.

[@kasuha]

Fun thing is, if on a round cake you choose random angle of the cut independent on all other random variables, the distribution is necessarily the same as if you always cut the cake in the same direction, e.g. vertically. I'd even guess it's the same on square/rectangular cakes but I don't want to go through the trouble of proving that.

[@Zahlenteufel1]

I choose a random point on the circumference. Without loss of generality, the tangent is orthogonal to the cut (circle is symmetric). so I choose a random point along the tangent such that an orthogonal to the tangent will still intersect the cake (uniform 0-1) that is the cut. look at it from the side now (from the tangent) makes you get the 1d problem, so 1/3 is correct.

[@edwardbarton1680]

For the 2D cake question, we can start with the "simpler" problem of an infinite cake. If we re-map the candle coordinates to coordinates based on the cut (x along, y perpendicular), the coordinates will still be random, but we can throw one out (x) and we're back at the 1D problem. The complication in moving to a finite cake is that the shape of the cake will determine the distribution of the candles. The candles will no longer have an even distribution along the "y(cut)" axis. If we choose a circle cake, we can at least simplify a bit by eliminating the rotation of the cut, by considering it as the rotation of the cake instead.

[@petertaylor4980]

The infinite 2D problem has different probabilities to the finite one, though. It gives 1/2 rather than 1/3.

[@gmalivuk]

What distribution are you going to use for picking points on the infinite cake, now that uniform is no longer possible?

[@edwardbarton1680]

@@petertaylor4980 How do you get 1/2? The logic of the ordering still holds.

[@petertaylor4980]

@@edwardbarton1680, symmetry. Cut first: then the two halves are symmetric.

[@edwardbarton1680]

​@@gmalivuk Why wouldn't uniform be possible?

[@steffahn]

"random point random angle" should obviously give exactly 1/3. If the angle is chosen first, and then the 3 points for the 2 candles and the cut are chosen, then there's an ordering between these 3 points given by how far they are in the direction perpendicular to the cut direction, and this ordering characterizes whether or not the cut lies between the candles. Since the points 3 for candles and cut are chosen independently and in an equivalent manner, all 6 possible orderings must be equally likely, and 2 of the 6 orderings have the cut between the candles (i. e. it's the same ordering argument then like in the line case).

[@Jazardly]

assuming cut angles are uniform so they can be ignored. the area isolated by a chord at distance a from the center point is arccos(a^2)-1/2 sin(2arccos(a^2)) = arccos(a^2) - a^2 sqrt(1-a^2) the percentage of the unit circle that this is is then [ arccos(a^2) - a^2 sqrt(1-a^2) ] / pi = f(a) The probability that the first candle is inside the chord is f(a), and that the next candle is outside is 1-f(a) The probability that the first candle is outside the chord is (1-f(a)) and the other is inside is f(a) so the likelihood with my definition of random chord is integral from [0,1] of 2f(a)(1-f(a))=.388959...

[@infinitivez]

I'm just here for the cake itself 😆

[@brettbreet]

"What are the chances? Don't answer that." It feels like there's a quantum superposition of probabilities that will only collapse when you say what method you're using outloud!

[@nicov1003]

Choosing a random point and random angle is effectively the same as choosing two random points and drawing a line between them, which I feel is most appropriate given his method for choosing the candle placement. Ultimately what this problem is about isn't really the circle or square, and so we should want a random metric that's shape invariant.

[@PhotonBeast]

Credit to the gameplay programmer that explained this to me: For the method where you pick an angle and then a distance away from center, you have to get that distance in a particular way. You need to generate a number between 0 and 1, take the square root, and then multiple by the radius. Otherwise, the random points will be packed near the center since most of the area of a circle is around the area near the edge not the center. Because of this, for purposes of this puzzle, I would hypothesis that this is why there is a higher chance of cutting between the candles if you don't make that adjustment. If the random point is more likely to land near the center, then it's more likely to be between any two points.

[@AdrianHereToHelp]

I feel like the "human" way to pick a random cut is to select a random point on the circumference (or half the circumference), then select an angle as long as said angle results in forming another endpoint (your cut has to be inside the circle). My intuition says that might also leave a bit of a gap towards the center, but I have no idea.

[@originalman6599]

think about this: if every cut needs to be equally likely, it makes no sense to do the bottom right option (choose a random point on the circle, then a random angle). a longer cut would mean that there are more possible places for the initial random point to land in it, so shorter cuts would occur less often. to get around this, you just need to pick a random point on the perimeter instead, and then a random angle. doing this should give every possible cut needed to satisfy this problem, and no symmetries are broken.

[@Smonserratm]

You can make a vast space full of random lines and then move the cake around.

[@jeremyhansen9197]

The way I thought about the circle cake is by thinking of the areas divided by the cord. Cutting the cake between the candles would then be the same as choosing one area for one candles and the other area for the second candle or vice-versa. For a given cord the probability we want would then be twice the probability of choosing one area times the probability of choosing the other area. Exploring the space over all possible areas, I get an overall possibility of 444/(135*pi^2) which is basically a third.

[@riccardoorlando2262]

An interesting method for picking random chords: Randomly and uniformly select a diameter (by selecting an angle uniformly), then uniformly pick a point on the diameter. Cut perpendicular to the diameter, through the point.

[@steffahn]

This method is equivalent to the "random radial point" approach. The only difference is whether the angle or the radius is chosen first (well, more precisely, a diameter leaves 2 possible angles 180 degrees apart), in both cases angles and radii are uniformly distributed and chosen independently.

[@JSLing-vv5go]

It really is amazing how tough the chord problem is. I've been trying to read about what's been done with it so far, and it doesn't appear to be solved, at least to a reasonable consensus of mathematicians.

[@chitlitlah]

I made a thread in which I explained how I solved one of the scenarios. The others would require figuring out the distribution of the distances from the midpoint of the circle to the midpoint of the chord.

[@Sinnistering]

I think the takeaway (as someone who is NOT a mathematician) is that "random" is only meaningfully defined for certain situations. Easy to define in 1D, but it requires more to define well in higher dimensions.

[@johnroberts7185]

If we have to cut a cake in half, of course the straight cuts have to go through the center, but if each piece is allowed to be smaller of larger than the other, the middle would be cut less than the rest of the cake because fewer angles would pass through that point proportionally to the area of its column/row. If we use a method of dividing a circle into a series of cubes inlaid into it, the rows near the top and bottom have the fewest cubes, and the columns near the right and left have the fewest cubes. A cut of indivisible thickness pass through more of their points, but if the cut had thickness to it, near portion of the cake would be equally likely to be cut. It's one of those "infinite area with finite paint" problems. We can't paint an infinitely thin layer, so it doesn't really make sense in real life.

[@potato4dawin1]

I would say that picking a random point and random angle is more random because if you remove the circle and just consider the question "what is a random line" then it would make the most sense to argue that a random line is 1 chosen out of the infinite possibilities of all lines in which case for determining a set of all possible lines, choosing a random point and random angle is the only system which could be applied without the constraints of the circle and so as the most general case, it seems the most fitting to a general question without context. I would very much like to see the calculations for exactly the odds on the random point, random angle method.

[@scaredscorpion]

For figuring out the "correct" randomisation method I think reversing your logic might help. Rather than how do you select points at random think, what is the range of possible options. From there it seems pretty clear that picking a random angle (0 to 180°) and random distance (0 to 2r) from left to right (at the selected angle) has equal chance of getting every possible cut, having a different angle than another cut automatically ensures they cannot be the same cut so you don't have to deal with de-duplication. That is assuming random in this case means all results are equally likely

[@jursamaj]

Assume a random line anywhere in the plane (not even limited to cutting the circle). Now, those lines which happen to pass thru the circle are truly "random cuts". We can ignore all others. If we then rotate all such cut circles so that the cut is parallel to the x axis, the cuts will be uniformly distributed along the y axis. We lose no generality by assuming the random cut is parallel to the x axis then choosing the 2 points fully randomly within the circle. So that is the best uniform distribution for the random cuts. This is also simple to model, so it makes the rest of the problem easy. Now, all we have to do is find the area of the slice. Again WLOG, give the circle radius 1 centered at the origin. Pick a point on the y axis cutting a chord parallel to the x axis. Define the distance of the chord from the top of the circle. Call it H. The proportion of area of the segment above the cut is given by A=(acos(1-H)-(1-H)*sqrt(2H-H^2))/π. The probability of the 2 points being on opposite sides of the cut is P=2A*(1-A). Averaging P for H in the range 0 to 2 yields a probability of 28.8%. This is close to 1/3, but definitely not the same.

[@atimholt]

I feel like the most “intuitive” choice for random cut is to choose from the set of all lines through all space, discarding ones that do not go through the cake. This can be done by choosing the angle *first*, then taking an even linear distribution from all the lines that go through the cake at that angle. EDIT: I was wrong, unless you choose your slice’s lateralness from a bounding circle instead of just the shape.

[@gmalivuk]

Choosing the angle first and then picking a random cut in that direction is equivalent to picking a radius and then picking a point on that radius to be the midpoint of the chord.

[@bengoodwin2141]

Here's how I would "randomly" model picking a point on the edge of a circle: Divide the circle in two equal regions and have a bit represent one or the other region, if the bit is random then you'll choose a random half of the circle. Then, divide the two new pieces in half and make a bit correspond to each of the halves, now you have 4 areas picked randomly. Repeat this process splitting the areas evenly an infinite number of times and you'll have an infinitely small segment, a "point" chosen at random.

[@davidgoffredo1738]

That's the binary representation of a number between zero and one. Then multiply by 2*pi to get the angle in radians.

[@bengoodwin2141]

@@davidgoffredo1738 kind of. It's the binary representation of a fraction, but extended infinitely. So the first step is either above or below 0.5, then the second step determines whether the number is above or below 0.25 or 0.75 depending on the result of the previous bit, and so on. Edit: wait no it's exactly that.

[@musikSkool]

I think to pick the cut you also need to try two random points on the (round) cake and draw a line between them, and then try two random points on the edge and draw a line between them. Those two interest me the most.

[@sunday87]

"Between the candles! What are the chances?" That's exactly what we are here to find out :-D

[@Yishay.S]

"Don't answer that..." Don't distract the audience from the lecture 😉

[@alexandersmith4796]

An idea I had watching this is to see what would happen in the cutting simulations if a cut was determined by two randomly chosen points in the circle, and then the line taken between them was extended to the ends of the circle. I suspect, by virtue of the points being uniformly distributed, the cuts would also tend towards that too. Edit: I also feel like the reason that it's so difficult to choose a random slice is due to it being a dimensionality problem. With just 1 dimension, if you were to randomly divide a space into two halves, there would be some sort distribution that the sizes of halves would take. With a 2D shape, you've gone up a dimension, so now you're trying to somehow randomly distribute *areas* of a region that fit some given distribution.

[@cheesebiscuits3586]

One possible route to solve this problem is to work out the different in volumes of the 2 pieces of the cake after the cut and then determine the placing of candles as you would with a waited coin. The chance of the two candles being in different pieces is 2*(v/V)(1-(v/V)) (where v is one of the piece sizes and V is the total size) the chance that the candles are in the same piece is also given: (v/V)^2+(1-(v/V))^2. This method can be generalised to arbitrarily large amounts of candles, arbitrarily many dimensions and I suspect arbitrarily many cuts: -Higher dimension only needs v to be upgraded to some hypervolume, -More candles requires you to consider more cases (eg. for the case of 4 candles the chances that you get an even split is 6*(v/V)^2(1-(v/V))^2, the case of all in one side or the other is (v/V)^4+(1-(v/V))^4, the other case 3 in 1 side and 1 in the other is also obtainable but not pretty, and can be seen by expanding out (v/V+(1-v/V))^n for n candles) -More cuts I think you just get with a more exotic expression in the bracket (a+b+c...)^n, you can think of this as rolling a waited die. The resolution of Bertrand's paradox I cannot answer though (but I have hidden it in the "work out the different volumes of the 2 pieces" component).

[@mytube001]

Part two was much better than part one! :D

[@llegaremosalasestrellas3245]

I think a way to test the methods for the random cuts is to calculate the average area of millions of cuts for each method, and double check which methods gets the area of a semicircle. The video is very interesting. Best regards.