Really lovely video, lots of good intuition, and yet just enough details that you can see how a rigorous argument proving the same thing would shake out, without bogging people down in tedium. Thank you for taking the time and effort to explain this, and the "reveal" at the end that the 1/n² term vanishes was quite the surprise! Just as a heads-up, whatever greenscreen technology you are using seems to not quite be able to mask the very top right corner of your camera feed correctly, and so there is a small artifact more or less in the middle of the screen for most of the video. Not a big deal at all, but probably easy to correct.
Thank you so much! I'm so glad you liked it and the "plot" was followablr. Yes indeed the correction I need to make is to not bump into the green screen next time :)
So, the P(Nmax=2) perspective on the third term explains why the third correction term doesn't make sense when m=1 - intuitively you should anticipate that m=1 should still be a valid input, but Nmax>=2 being the cause of the third term means m needs to be >1 (you can't have two dice being maximum out of a set of one 😂)
Indeed the P(Nmax=2)=0 for a single dice (not 1/2n as the formula would predict!). Another way to see it is to notice that in our derivation we actually had a factor of (m-1) which cancelled out from the numerator and denominator so this is a 0/0 dovision when m=1!
Yes, its exactly because of the 0 there. But I don't have a nice probability explanation as to why! Probabilitistically its something like an overestimate for the P(N_max = 2) and an underestimate for P(N_max = 3) being equal or something like that
The fourth term is zero, because the sequence follows a similar, if not the same process as finding successive polynomial equations for sums from 1 to n of i^kth power. i.e 1^k + 2^k + 3^k ... + n^k. k=3 => n^4/4 + n^3/2 + (0)n^2 + n/4 As an amateur I've worked on a method to take the equation for k = x to k = x + 1, and disregarding the effects of integration and other trivial things, the underlying structure of how the equations coefficients change looks exactly the same as this formula from the video
You make a "probability preserving map": it's a function phi that rotates the points like in the animation. Then you argue that the probability of any event is unchanged under phi (this is because the dots are all uniformly random, so rotating them doesnt change any probabilities). Then showing the expected values are all equal is true since E[S_1]=E[phi(S_1)]=E[S_2]
Thanks for the answer! Then I question the choice of circle, couldn't you make the same argument using a regular polygon (or any other funny shapes for that matter)? How would you make the argument if the underlying distributions are unknown, but you have prior beliefs that it's uniform, could you make a similar argument?
@@elonitram more technically the operation is just shifting on the interval [0,1] with the convention that what goes off the left reappeares on the right side: but as a picture this is a circle! I'm not quite sure what it would mean to do a shape other than a circle...like in what way are the vertices/edges of it being used? If there aren't any it's the same as a circle! The argument does not work for non-uniform things unfortunately
3 minutes into the video : I don't know why but the ½ by itself and -1/12 coefficient are screaming out Bernoulli to me... I guess I'll have to watch and see!
i remember a couple years ago i tried to work on this myself, I managed to prove that asymtotically the +1/2 would work and was quite chuffed with myself
Ya a limit is just proving the error from the first to term is "o(1)" (which just means "the limit is 0"). But in this case since there are no fractional powers, o(1) is the same as O(1/n) :)
I proved this in a more pedestrian way in a paper a couple of years ago: Does Gaia Play Dice? : Simple Models of non-Darwinian Selection. eqn 9. There are some other dice games and related problems in there you might find interesting 🙂
Yes! I didn't want to bring this up in the video but on some sense the -1/12 is the *same* -1/12 that appears in the "1+2+3+...=-1/12" thing from that one Numberphile video!
Well hopefully you saw the bonus 4th term at the end of the video! After that you have to read en.m.wikipedia.org/wiki/Faulhaber%27s_formula and en.m.wikipedia.org/wiki/Bernoulli_number to get more terms...they are truly very weird and I have no intuition on what they mean!
@@MihaiNicaMath Yes I saw the fourth term in the video and I looked up that formula but am a bit confused because the terms given don't correlated exactly to the formula terms given, or at least require the correct name subsiitutions and applying some formulas. It does appear that the the 'fifth' term has both a small polynomial term and a small constant term.
@@BramCohen looking at it quickly so I might have a mistake but I think the next term is 1/30*m*(m-1)*(m-2)/n^3 and then therm after that is -1/42*m*(m-1)*(m-2)*(m-3)*(m-4)/n^5 . (Note that these are for the maximum: for the sum of powers the plus/minus signs are reversed)
People comment this on all of my videos that involve dice (which is a lot of them!), but singular "dice" can be the more common usage depending on where you live (particularly different in North American English vs British English). See e.g. www.oxfordlearnersdictionaries.com/us/definition/english/dice_1 . I personally find singular "dice" more clear since "die"/"dye" has several other meanings, while "dice" is unambiguous.
I think there's something wrong with your green screen: there's a little sliver of something in the middle of the video just about level with the top of your head and it's really distracting, sorry!
Ya I only noticed this after I uploaded: what happened was my green screen very slightly moved and what your seeing is tiny corner of it. I'm so sorry! Will try to fix for next time (coming at this from the math side of things so this is all new to me!)
Ooooh this is very nice - the perfect math christmas present:D
Haha only a few days late! Even knowing I always underestimate how long animating takes, I *always* underestimate how long animating takes
The subject area is mathematics. Short version, maths.
Thanks!
Wow thank you so much! That is very generous of you!!!
Really lovely video, lots of good intuition, and yet just enough details that you can see how a rigorous argument proving the same thing would shake out, without bogging people down in tedium. Thank you for taking the time and effort to explain this, and the "reveal" at the end that the 1/n² term vanishes was quite the surprise!
Just as a heads-up, whatever greenscreen technology you are using seems to not quite be able to mask the very top right corner of your camera feed correctly, and so there is a small artifact more or less in the middle of the screen for most of the video. Not a big deal at all, but probably easy to correct.
Thank you so much! I'm so glad you liked it and the "plot" was followablr. Yes indeed the correction I need to make is to not bump into the green screen next time :)
Very cool math tricks, and I appreciate that you show us exactly where the approximation differs from the real value and by what terms
a great video, such an underrated channel
Thank you so much, that is very kind! Hoping to make more videos in 2025!
So, the P(Nmax=2) perspective on the third term explains why the third correction term doesn't make sense when m=1 - intuitively you should anticipate that m=1 should still be a valid input, but Nmax>=2 being the cause of the third term means m needs to be >1 (you can't have two dice being maximum out of a set of one 😂)
Indeed the P(Nmax=2)=0 for a single dice (not 1/2n as the formula would predict!). Another way to see it is to notice that in our derivation we actually had a factor of (m-1) which cancelled out from the numerator and denominator so this is a 0/0 dovision when m=1!
I bet if we had a machine get every term, we could plug it in and get the right answer. If it isn’t an infinite and unpredictable sequence.
@IsaacDickinson-tf8sf There is an exact formula called Faulhabers formula! But it's complicated and I don't know a nice intuitive reason for it!
This is a great exposition!
Thank you! Much appreciated :)
Is the 0 term related to half the Bernoulli numbers being 0, since they appear in Faulhaber's formula?
Yes, its exactly because of the 0 there. But I don't have a nice probability explanation as to why! Probabilitistically its something like an overestimate for the P(N_max = 2) and an underestimate for P(N_max = 3) being equal or something like that
For m rolls of n sided die of consecutive positive integer faces the expected maximum value of a singular die is
n - n^{-m} \sum_{i=1}^{n-1} i^m
Ya this formula is mentioned at 7:47 :) the rest of the video gives a simple approximation for this sum
@@MihaiNicaMaththanks for the reply, that's correct, i must've dozed off for a moment there
Incredible description of your logic!!
Thank you!
thanks!! very nice and instructive video.
Glad it was helpful!
The fourth term is zero, because the sequence follows a similar, if not the same process as finding successive polynomial equations for sums from 1 to n of i^kth power. i.e 1^k + 2^k + 3^k ... + n^k.
k=3 => n^4/4 + n^3/2 + (0)n^2 + n/4
As an amateur I've worked on a method to take the equation for k = x to k = x + 1, and disregarding the effects of integration and other trivial things, the underlying structure of how the equations coefficients change looks exactly the same as this formula from the video
great video. Thank you!
You are welcome!
A very nice explanation. At 28:31 you misspoke, it is m=2
Thanks! Yes I completely agree: it's the max of two things so it should be "m = 2".
The circle argument makes sense intuitively! How would you show it rigorously?
You make a "probability preserving map": it's a function phi that rotates the points like in the animation. Then you argue that the probability of any event is unchanged under phi (this is because the dots are all uniformly random, so rotating them doesnt change any probabilities). Then showing the expected values are all equal is true since E[S_1]=E[phi(S_1)]=E[S_2]
Thanks for the answer! Then I question the choice of circle, couldn't you make the same argument using a regular polygon (or any other funny shapes for that matter)?
How would you make the argument if the underlying distributions are unknown, but you have prior beliefs that it's uniform, could you make a similar argument?
@@elonitram more technically the operation is just shifting on the interval [0,1] with the convention that what goes off the left reappeares on the right side: but as a picture this is a circle! I'm not quite sure what it would mean to do a shape other than a circle...like in what way are the vertices/edges of it being used? If there aren't any it's the same as a circle! The argument does not work for non-uniform things unfortunately
3 minutes into the video : I don't know why but the ½ by itself and -1/12 coefficient are screaming out Bernoulli to me... I guess I'll have to watch and see!
Yes it's Bernoulli!!!
I had to stop the video to check for this comment after blurting out "why is -1/12 in there?!"
i remember a couple years ago i tried to work on this myself, I managed to prove that asymtotically the +1/2 would work and was quite chuffed with myself
possibly if i was happier with big O i would've gotten a better result, back then my approach to avoiding lower order terms was to take a limit
Ya a limit is just proving the error from the first to term is "o(1)" (which just means "the limit is 0"). But in this case since there are no fractional powers, o(1) is the same as O(1/n) :)
I proved this in a more pedestrian way in a paper a couple of years ago: Does Gaia Play Dice? : Simple Models of non-Darwinian Selection. eqn 9. There are some other dice games and related problems in there you might find interesting 🙂
Cool! Here's the link for anyone else reading this that wants to check it out arxiv.org/pdf/2301.02623
For m and n = 1 your special term is equal to -1/12 = zeta(-1).
Yes! I didn't want to bring this up in the video but on some sense the -1/12 is the *same* -1/12 that appears in the "1+2+3+...=-1/12" thing from that one Numberphile video!
It's everywhere.
I'm very curious what the next few terms are
Well hopefully you saw the bonus 4th term at the end of the video! After that you have to read en.m.wikipedia.org/wiki/Faulhaber%27s_formula and en.m.wikipedia.org/wiki/Bernoulli_number to get more terms...they are truly very weird and I have no intuition on what they mean!
@@MihaiNicaMath Yes I saw the fourth term in the video and I looked up that formula but am a bit confused because the terms given don't correlated exactly to the formula terms given, or at least require the correct name subsiitutions and applying some formulas. It does appear that the the 'fifth' term has both a small polynomial term and a small constant term.
@@BramCohen looking at it quickly so I might have a mistake but I think the next term is 1/30*m*(m-1)*(m-2)/n^3 and then therm after that is -1/42*m*(m-1)*(m-2)*(m-3)*(m-4)/n^5 . (Note that these are for the maximum: for the sum of powers the plus/minus signs are reversed)
@MihaiNicaMath why is the coefficient of the first term 1/12 instead of 1/6?
The singular of dice is a die.
People comment this on all of my videos that involve dice (which is a lot of them!), but singular "dice" can be the more common usage depending on where you live (particularly different in North American English vs British English). See e.g. www.oxfordlearnersdictionaries.com/us/definition/english/dice_1 . I personally find singular "dice" more clear since "die"/"dye" has several other meanings, while "dice" is unambiguous.
Hello, your topics are highly original and insightful, but if you can work a bit on the verbosity :)
Noted!
ah yess yhe classic mat parker formula case
Is it not nᵐ√(¹/ₙ)
_Roll a n-sided dice m times_
You'd think anyone explaining with a formula would use _s-sided_ and _t times._
I thought about this but I wanted to match the notation from Matt Parker's video so people could go back/forth more easily
@@MihaiNicaMath
Fair enough, cheers. :)
I think there's something wrong with your green screen: there's a little sliver of something in the middle of the video just about level with the top of your head and it's really distracting, sorry!
Ya I only noticed this after I uploaded: what happened was my green screen very slightly moved and what your seeing is tiny corner of it. I'm so sorry! Will try to fix for next time (coming at this from the math side of things so this is all new to me!)
I'm not watching this because of the background drumbeat.
I made a version for you with the background music removed: ruclips.net/video/ZSG-QpFcOag/видео.html Thanks for the feedback!
@@MihaiNicaMathwhat an amazing response to that comment!
Thanks for the kind words...I'm genuinely unsure how to proceed with the background music so happy to let people have options if possible!
@@MihaiNicaMath Your videos are fantastic. I barely noticed there was music.
@@MihaiNicaMath If you have no background music, a viewer can turn on their own if they want.