thank u I now understand stonks

Nice clean little video here, it's got everything I could have asked for!

Trippy.

The equation you write at 5:50 is in fact not correct as stated. The expectation you write must be *under the risk-neutral probability measure* (!) rather than the actual probability measure. This is a crucial point. So, to make it correct, you’d have to write an index at the bottom of the expectation: \E_{Q} where Q is this risk neutral measure (which is unique in the Black-Scholes model).

Very beautiful proof. I liked the combinatorial nature underlying the argument. Excellent exposition on this important and fascinating, most fundamental result in all of probability theory & statistics.

Amazing job! You really explain this well. Keep it up!

I’m a researcher in math. Great effort, good and clear explanation. One thing that perhaps could be added (it might clarify it more for some students, but perhaps could also confuse some others): Given a rv (that is, a measurable function on \Omega), say X, we get a distribution *of X with respect to* some given probability measure P; if we were to consider an other, different probability measure say Q, then the rv X is (obviously!) still exactly the same as it was before, but its *distribution* (considered wrt Q now) is different from before (when it was under P). So, the concept of rv is in some sense more fundamental (or “primitive”/rudimentary)-indeed it is defined simply by a measurability condition, and has a priori nothing to do with any probability measure. Keep up the good work!

Thank you !

Poll: What looks cooler in your opinion? Zooming in (beginning of video), or Zooming out (middle of video)? Like the comments below to vote.

Lovely!

I loved it thanks a lot

Extremely insightful. Thanks a lot for sharing!

Question: If Delta draws from [0,1], does the X-Delta not count {1,2,3,4,5} twice? Thus it´s not exactly the same as nU? And if you simply use Delta draws from (0,1] or [0,1), then you exclude 0 or 6. Also, if you round up nU, and you happened to draw 0, then you round to 0, which is not in the ranga of X. Does this matter for the proof at all?

not sure if i "discovered" this or not, but i think i found the formula for the average maximum of an N sided die with M advantage(m+1 rolls): Sum(from X=1 to N): (X*Sum(from A=1 to M+1): (((X-1)/N)^(A-1) * (1/N) * (X/N)^(M+1-A)) sorry for the HORRIBLE formmating of the comment, its hard to write in plaintext because the formula i made up uses sigma notation INSIDE a sigma notation. So please, if someone who understands math can tell me if its right/ if theres a mistake in it id be glad :) P.S im not sure if it can even qualify as a formula since it uses sigma notatoin inside of a sigma notation, either way its O(n^2) complexity in computer terms so its not too bad for a computer to calculate. also i just think its neat. EDIT: found an easier way to calculate it, heres the formula: N = amount of sides M = amount of rolls Sum(from x=1 to n): ((x/n)*(x^m * (x-1)^m))

Isn't the P(N_maxers) dependent on the value of the maxiumum? E.g. if the max of 3 dice is 1 then the probability that all three dice have the maximum value of 1 is 100%. It's probably a higher order term, but I think this isn't considered in your exact formula

A formula that is exact is 1 + n - sum_{k=1}^{n-1} (k/n)^m. I found this by defining a matrix A such that pi' = A*pi takes the probability distribution of the max value of rolling m n-sided die to the probability distribution of the max value of rolling m+1 n-sided die. The Eigendecomposition of A is easy to compute by hand as it has a nice form. Using the Eigendecomposition you can compute pi_m = Q*(diag(v)^(m-1))*(Q^-1)*(1/n, ..., 1/n) efficiently by exponentiating the Eigenvalues. If you just want the expected value of the max, you can simplify further to the formula provided.

I feel like this doesnt pass the simplest sniff test. As m approaches infinity the expected max should approach n as it becomes extremely likely you roll the highest value. This equation instead approaches n +1/2 - infinity which is very wrong.

Hello Dr. Nica, If, per chance, you are looking for an interesting problem to solve and post to YT, may I suggest a puzzle I have long wondered about, but have not been able to make progress on-the Strong Birthday Puzzle. Rather than asking how many people you need in a room to have a 50% chance that there is _some_ match (yeah, making all the uniform assumptions and 365 days), I am interested in how many people you need in a room to be 50% sure that _everyone_ has at least one match. Such a simple change, but other than simulation, I can’t make progress on it. Thanks again.

Hi! I randomly came across your video and spent the whole day exploring this topic-I've found it fascinating! I think I’m starting to grasp the coin flip problems, but I’m still struggling with some aspects and would love your insight. Here’s what I’ve been working on: For coin flips, I think I understand the idea of focusing on the last few flips for overlaps. For example: HTTTH = 2⁵ + 2⁴ + 2² + 2¹ = 54 I feel pretty confident about this one, though I’d like to know if my interpretation is correct. However, when I try to apply the same framework to other problems like dice rolls or words, I keep running into issues. Here are a few examples: 1. HTHTHHTTTH = 2¹⁰ + 2⁴ + 2² + 2¹ = 1046 I think this is right, but I’m unsure how to check if I handled the overlaps correctly (e.g., for the final TT). 2. Dice Sequence: 62453621 = 6⁸ = 1,679,616 Here, I treated it the same as the coin flips but with a base of 6 instead of 2, since there are no repeating numbers. Is this the right approach? 3. Word: circumference = 26¹² = ~9.54e+16 For this, I’m unsure if repeating letters (e.g., c, r, e) should change the calculation or if I’m overthinking it. I’m having trouble understanding how repeating elements (like letters or numbers) affect the total and whether my method for handling overlaps is correct. If anything I wrote here is wrong or unclear, I’d really appreciate your feedback-I’ve never worked on anything like this before but find it really interesting. Thank you so much for the video! (Reworded with AI for readability reasons.) (If anything I wrote is wrong, unclear, or confusing, please let me know!)

I think the exact solution to the average maximum should be exactly n+1 minus the average minimum, so n+1 - (1^m + 2^m + 3^m + ... + (n-1)^m + n^m)/(n^m), but I don't have an elegant way to prove why yet, so I'm going full Fermat just laying out the claim.

This is my attempt before watching the video: X1,X2,…Xm are IID from X ~ Unif{1,…,n}. For constant a taking values in {1,…, n} ==> P(min(Xi) >= a) = P(X1 >= a, … , Xm >= a) = Π[i = 1 to m] P(X >= a) by independence = [ P(X = a) + P(X = a+1) + .. + P(X = n) ] ^ m = [ 1/n + 1/n + … + 1/n ] ^ m = ((n - a + 1)/ n)^m ==> P(min(Xi) <= a) = 1 - P(min(Xi) >= a+1) = 1 - ((n - a)/ n)^m = 1 - (1 - a/n)^m (Cumulative distribution F(a) of minX) ==> discrete pmf given by: P(minXi = a) = F(a) - F(a-1) = 1 - (1 - a/n)^m - 1 + (1 - (a-1)/n)^m = (1 - (a-1)/n)^m - (1 - a/n)^m ==> E[minXi] = Σ[a = 1 to n] a(1 - (a-1)/n)^m - a(1 - a/n)^m It’s horrible but it actually turns out to be the same as your solution (probably just using binomial), but far less elegant

Hello Dr could you please make a video explain the intuition behind the concept of expected values, there are so many way of explaining it but each one doesn't always fit perfectly when it comes to different contexts.

Cool video, but the thing that catches the eye about 2term vs 3rd term approximations is: if we fix the N and let M get arbitrary large, with only 2 terms E[max] approaches N, which sounds correct since no matter the size N, if we throw the dice many times (M >> N) then surelly we will get the biggest possible value sooner or later, but with 3rd turm E[max] goes to negative inf, maybe the possitive O(N^2) term could fix this? idk but the concept of better approximation which only works on the N ~ M interval seems strange... surely the simple (M * N)/(M+1) +0.5 formula cant be the best for every pair (N, M)?

I wrote a comment to Matt Parker's video with my own attempt at proving this 2 years ago after watching it. I'll copy it here, since I think it works but want to know where I've erred if it doesn't. For n,k >= 1 and m <= n, let D(n, k; m) := # ways that the max value of k rolls of n-sided dice is m. (m > n obviously gives 0, so that's not very interesting) Equivalently, D(n, k; m) = #{ s in {1, . . ., n}^k | max(s) = m}. An interesting thing happens here, though: Since we're taking m <= n, we can actually restrict the domain to {1, . . ., m}^k without losing any elements. This is because any set of k rolls where some roll > m occurs, we won't be counting it. Therefore, D(n, k; m) = #{s in {1, . . ., m}^k | max(s) = m} Now, let's enumerate all the possibilities. By definition, we have to roll at least one m, and nothing larger than m. Suppose m shows up as the first roll. Then, the other (k-1) rolls can be anything between 1 and m, so this accounts for m^(k-1) elements. If m doesn't show up as the first roll, but is the second roll, that means we had (m-1) possible values for the first roll, 1 value for the second (this is where m is first rolled), and then m^(k-2) possibilities for the last k-2 rolls. This accounts for m^(k-2) * (m-1) elements. This process continues until the only case is where the final roll is m, leaving (m-1)^(k-1) possibilities for the first (k-1) rolls. This gives us the sum: Sum( m^(k-j) * (m-1)^(j-1), j = 1 to k ). I'll leave it as an exercise to the reader to show that this sum comes out to m^k - (m-1)^k. Therefore, D(n, k; m) = m^k - (m-1)^k. Since there are n^k possible sets of k rolls, the probability that the max value of k rolls of n-sided dice is m = (m^k - (m-1)^k)/n^k. Now, we can compute the expected value. Let E(n, k) denote the expected value for the k-roll experiment with n-sided dice. E(n, k) = Sum( m * (m^k - (m-1)^k)/n^k, m = 1 to n ) = (1 / n^k) * Sum( m * (m^k - (m-1)^k), m = 1 to n ) = (1 / n^k) * (n^(k+1) - Sum( m^k, m = 1 to n-1 )). As before, I'll leave the final equality as an exercise to the reader. For k = 1,2,3, it's easy to simplify this down further using the well-known formulas for the sum of the first (n-1) integers, squares, and cubes (resp.): E(n, 1) = (1/n) * (n^2 - [n * (n - 1)] / 2) = (n + 1) / (2) -> For a single roll of a D6, we get (6 + 1) / (2) = 3.5, as expected. E(n, 2) = (1/n^2) * (n^3 - [n * (n - 1) * (2n - 1)] / 6) = (4n^2 + 3n - 1) / (6n) -> For 2 rolls of a D20, we get (1600 + 60 - 1)/(120) = 13.825, which agrees with the 2 rolls of D20 simulations in the video. E(n, 3) = (1/n^3) * ([n^2 * (n-1)^2] / 4) = (3n^2 + 2n - 1) / (4n) -> Matches the value computed in the video. Taking limits as n->infinity also work (using the ~ asymptotic notation): E(n, k) ~ kn/(k+1) and so E(n, k) / n -> k/(k+1) as n->infinity. For the full expression of E(n, k) / n, we'd need to make use of Faulhaber's formula, which is a big complicated expression involving Bernoulli numbers. This means that your conjecture isn't quite true, but it is "true" in an asymptotic sense. Using Faulhaber's formula, E(n, k) = kn/(k+1) + 1/2 + o(1) as n->infinity. [Note: Little-o notation - f(n) = o(g(n)) means f(n)/g(n) -> 0 as n -> infinity. In this case, f(n) = o(1) means f(n) -> 0] For rolling with disadvantage (formulas provided without justification; it's essentially the same as above): d(n, k; m) := # ways that the min value of k rolls of n-sided dice is m = (n-m+1)^k - (n-m)^k probability of min value after k rolls of n-sided dice is m = ( (n-m+1)^k - (n-m)^k ) / n^k e(n,k) := expected value of taking the min of k rolls of n-sided dice = (1/n^k) * Sum( m^k, m = 1 to n ) Using Faulhaber's formula for the sum of the first n kth powers, we find: e(n, k) = n/(k+1) + 1/2 + o(1) as n->infinity where that o(1) packs in a bunch of complicated expressions involving Bernoulli numbers.

Another way to get to the maximum of 3 points being ¾ is to think of finding 1 minus the minimum, so 1 - S1 and since S1 is ¼ you get ¾

This is one of the best descriptions of series approximations that I've seen on YT. It's great that you've given each term a qualitatively different explanation, which definitely has parallels with perturbation theory for me (while being simple enough for a general audience!). I particularly liked the graphs at the start, they really show what you gain at each step

When you say “k-th largest” I think you mean to say “k-th smallest” 18:17

At 38:15 why is it that it's multiplied by (1-2/n)? Shouldn't it be (1-1/n) ? Because the first two dices have the same values so there is only one unique value that the next dice should be different from?

1:30 yo the 1/12n reminds me of the adjustment in stirlings formula

really cool video man! funny as it is, the bit that was most confusing to me was the part you subtracted and added the 1/2. took me a few good seconds to realize what has happened there.

For this problem i'd probably have done with descriptive shorthands, like s for sides, and n for number, or d for dice

so cool >:3 nice animations btw

In the calculation of P(All Different) around 37:40 shouldn't the error term be of order O(m^3/n^2), which is not negligible when m^3 is comparable to n^2?

Yahtzee!

As a mathy person who struggles with probability beyond just crunching examples, I'm loving this! I'm at 28:07, and almost shivering with antici...pation to discover where that 12th is gonna emerge from.

With the subtraction of a uniform distribution, you introduced 0 as an outcome. The only way that would be true is if it never could happen, but it has an infinitely small probability rather than a 0 probability. You claimed an equal distribution that is not the same. It still requires an approximation. The delta at 1 can also tie the maximum as 6 - 1 = 5 - 0. This would be exceptionally rare, but not impossible. This also means that delta is ambiguous when you claim it to be a constant as two terms will have deltas and the delta for one is 1 and the delta for the other is 0. From a statistics standpoint, this is still a 0 for the probability, but it is not a 0 exactly. Yes, it is so small that it probably would be covered by the O(1/n^3), but one could argue with the equal sign.

Excellent explanation, thank you!

Why are you pointing? We can see what's on the screen. You don't need to tell us what to look at.

Could you do more on the general strategy of Ninja Proofs? I really like what you showed here.

so, we can add more terms, and they will have n^2, n^3, etc..., upto n^m terms.

18:56 correction: it's the "kth smallest", not the "kth largest", but we can just use the min-max trick from earlier. The kth largest would be (m-k)/(m+1).

10:53 Why not 1+(5*rand)? It should be impossible to ever get zero with any n-sided die.

Does the 0 to 1 range of the uniform random variables include both 0 and 1? When you "round up", you use the ceiling notation. If your uniform random number happens to be 0 and you multiply that by the number of sides on a die, then you'd get 0, which is not a valid roll. Likewise, if you got a 1 but the delta that you subtract is also 1, you'd also get a 0. Rounding 0 up to 1 to make it a valid roll adds a minuscule bias, making a roll of 1 ever so slightly more likely than any other value. Since the goal is to improve an approximation that's already ignoring very small terms, perhaps this doesn't matter. But I'm not a mathematician, so maybe I'm wrong about how that endpoints of the uniform random variables affect the reasoning.

Didn't you mean: _X_ ≈ _floor(nU +1)_ _nU_ is going to be a value [ 0, _n_ ), or is this why you say approximately equal? Edit: hmm, I should have just watched further.

Really thoughtful explanation of fundamental ideas. The reason to use of equations involving random variables is totally clear. It is clear that random variables are just functions, not themselves random, but some probability over omega is *pushed through* the random variable function, leading to the possibility of calculating expected values, probability mass/density vs values, and, one is led to presume, calculation of statistics in general. What a treat!

Great video! As always! On a side note, here's a link to a video that also proves the general formula for Matt Parker's max-of-dice conjecture. This video shows the exact probability distribution for any n and m, which an exact mean value can be calculated. kzbin.info/www/bejne/aJTLhaeZord0qbcsi=ZiyYxadGX2axbuJZ

Great presentation, that was surprisingly easy to follow.