Correction: the optimal number for the original 20-sided die problem is actually 35 (this is slightly better than 34). James has been fed to the dragon as punishment for this error.
it's actually kinda interesting, because with numbers this large you end up with the problem that you cannot inch towards such a number in this universe by adding small enough numbers. There's just not enough information storage space in the universe to represent all the intermediate steps. edit. so, peculiarly, you can throw the die enough times to reach graham's number, but you'd never in this universe know whether you actually did or not
What surprises me is that 21 is so unlikely even though it is exactly double the average, and it is the most likely to come up if you just have 2 throws.
You are correct to be surprised, as he seems to be calculating only the odds of hitting the number in a single throw after you get to within 20 of that number. He is not considering the chances of reaching the number in 2 or more additional throws once you get to within 20. Unless I missed something. So he is leaving out the possibility, for example of 1+1+1+1......+1+1 = 21.
@@GreylanderTV The recurrence relation for 21 includes the probability that you were previously at 20 and rolled one, which already includes the probability that you were at 19 and rolled one, etc.
@@GreylanderTV When he talks about "recurrence relations" he means that the computer will take his 1-roll equation and keep iterating it for an arbitrarily large number of rolls. This is why the other ways of reaching 21(or anything else) are included. If you didn't know this, then I get why it looked like he wasn't considering large numbers of rolls.
I thought along the same lines - How many numbers are "double evil" where summing the whole as well as decimals ends up evil where the Nth decimal position number arriving at 666 is the same number as the whole?
I've never seen this James Munro guy but please use him more! He has such a good, conversable tone, you can hear the excitement in his voice for maths as he speaks and he has the ability - seemingly - to speak to a range of ages and abilities and foster an enjoyment in maths. In fact this guy's probably what Rishi wants from all his teachers lol. Great video Brady. Interesting subject topic. I've literally just got home from a tutoring session with a student regarding probability so this was a nice continuation from that for me.
This problem actually is a fantastic application of dynamic programming. If you do it without it, the time complexity of your algorithm is O(20^n). With DP it is O(n)
@@tehdarkneswithin I guess that requires finding the 20 exact roots of a 20th order polynomial first? Please show us the way... 😉 (And it seems to assume we can do exponentiation in O(1), which I kind of doubt when we want to be exact...)
@@tehdarkneswithinwell. i was thinking of a different problem, the finding of the max for any kind of die. which was O(s^3). Yes finding the answer to your problem however is probably like O(s*log(n))
Pi being evil is another reason to use Tau instead - It reaches 663 on the 139th digit and 668 on the 140th, nicely skipping over 666 as any well behaved number should do!
Update: I have verified the first 3000 values for N. Was able to compute each result in O(N^2) cause of knowing what place to start looking and an improved algorithm that does not compute things based on number of rolls. The answers to the question of what the best guess > N is seems to follow f(N) = round((e - 1)N + 4 - pi) for N=2...3000 EXCEPT N=232 and N=1233 both in which case the actual answer is f(N)+1 but thats not something im completely sure about since the difference in probability for N=232 is only 6.182648143449043e-12 and for N=1233 is a meager 5.937702586555904e-13. Edit: I am sure now. have tried with 112 bit percision floating point numbers.
The issue with picking the closest value to the peak is that doing that only works for specific kinds of functions. (They at least require that, around a peak x, f(x + k) = f(x - k).) It doesn't work for generalised functions, but if the function can be approximated by one that fulfills this property, it will usually get you the right answer. Any quadratic function meets this rule, so if the function has a very small third derivative, you can get away with it.
Watching hardcore mathmeticians trying to do simple mental arithmatic is always amusing, a friend is doing a phd is maths and we always find it funny when he stuggles to work out his portion of the bill 😂
Interesting that the 0 is ignored in the average of the digits in pi. If you include it to get an average digit of 4.5, then the value 1/4.5 represents the expected number of times that a specific very high total will be hit, since it can be more than once if one or more zeros occur immediately after hitting the total.
@@slo3337 Any number can appear on any die. In fact, most 10-sided die are numberd from 0-9 (as well as some numbered from 00 to 90, that when paired with a 0-9 allow you to roll 0-99 for percentages). This would also perfectly simulate rolling a 9-sided die in this application (since we only care about the sum, not the number or rolls).
@@phiefer3 Well, in a literal sense, the only numbers which can appear on a die... are the numbers which are marked on the faces of said die. A way to have a guaranteed victory against the video's dragon is by putting your chosen number (or a factor of it) on all the faces of the dragon's die.
If you want the probability that a given number will be hit, then zero must be ignored But if you want the expected number of digits, then zero should be included (He was only doing the former, not the latter)
Remember the recurrence relation? p(k) = 1/n p(k - 1) + 1/n p(k - 2) + ... + 1/n p(k - n). (In the digits example, n = 9.) If we allow zero, the probability becomes 1/(n + 1) (because there's now an additional possible outcome), and you get the extra addend for zero: p(k) = 1/(n + 1) p(k - 0) + 1/(n + 1) p(k - 1) + ... + 1/(n + 1) p(k - n). But p(k - 0) is obviously p(k), so you can move that term to the other side, and you get p(k) - 1/(n + 1) p(k) = n/(n + 1) p(k) on the left-hand side, meaning you have to divide all terms by n/(n + 1) - which gets you back to the original equation without zero.
The calculation for the optimal dragon dungeon number can use pi too. Instead of (e-1)N, you can do the average of the die * pi to get close to the optimal guess. 6 sided die average is the sum of both sides divided by two, so 3.5, and 3.5 * pi = 11. Same for 20 sided die, but 10.5 * pi = 33, not 34
(waves) Hi! I've made about 70 episodes of an online maths club over on this channel, and I'm doing a bunch of livestreams on maths admissions test questions. ^James
Great puzzle! Tried to solve it in my head before firing up python, and the incorrect heuristic I got was roughly (2 - 1/e)*N. I figured out the probabilities for 1 through 20 to be P(x) = p*(1+p)^(x-1), where p = 1/20, which are exponentially increasing. Then P(21) would be the average of P(1) through P(20), and so on. The approximation I then made was that the maximum probability among P(21) through P(40) would be the average starting from the first number between 1 and 20 that had an above-average probability (i.e. more than P(21)), and if you work through the math on that, this would give an optimal value of N + log(e-1)/log(1+p), in the limit as N becomes large; and using series expansions this is roughly (2 - 1/e)*N. This is an approximation because the probabilities P(21) through P(40) are also (initially) increasing, so instead of finding the first x such that P(x) > P(21), we should find the first x such that P(x) > P(x+20). But that seemed too complicated to think about. Pleasantly surprised to see the actual correct heuristic also involves e, and to be more specific, a simpler expression of e that lies between 1 and 2!
The recursive rule can be seen as a Infinite Response Filter that just average the previous N values for a N sided dice. So it's a low-pass filter hence the smoothing effect.
I'm so glad that someone picked up on the smoothing effect! This was not the right video to talk about how there's a jump discontinuity then a jump in the first derivative (in the continuous version), but it's the sort of thing I see when I look at the graph :) ^James
For the values for 1 to 20, you can think of P(n) = 0 for -19 =< x =< -1 and P(0) = 1, which makes sense conceptually, since you always start with 0, and you never have negative numbers. It doesn’t make sense if you keep extending it backwards, though: you get that |P(n)| = 20 for n in [-39, 20], alternating in sign. And then you get P(-40) = 800.
The Biblical reference relates 666 to “the number of the beast (or antichrist).” The Antichrist mimics the true Christ with stolen power & knowledge. So naturally the original 666 from the true Christ is not evil.
That is a bit meta. The dragon in the game is asking the characters to choose numbers for dice rolls which the players then perform on behalf of the characters.
There is a fragrance called Pi, by Givenchy (and created by one of the most prolific and successful perfumers: Alberto Morillas.) Pi smells really great, warm sweet, suitable for cold weather season, so this the best time to wear it in the northern hemisphere . I'd encourage all math/Pi enthusiasts to check it out.
Seeing the same by just rolling a lot in python: number of success for 32: 96891 number of success for 33: 97156 number of success for 34: 97212 number of success for 35: 97605 number of success for 36: 97127 1 million rolls each time.
Ah I wondered if anyone would spot this. Sorry - I have no idea why I said 34 on the day. In my defense, 34 and 35 have very very similar probabilities. ^James
Perhaps because (e-1)*20 is closer to 34 than 35 the intuition would be that 34 is slightly better than 35. However, a bit of experimentation shows that 35 is favoured.
If the dragon let you choose 2 target numbers upfront, what would the best choices be for a 20-sided die? Will they maybe be 34 & 35, or will they be spread out?
My intuition tells me that since the average roll of a d20 is 10.5, choosing 33 and 34 is the best option. To explain why spread out numbers wouldn't work, imagine the dragon gave you 10 numbers to choose. You could choose any 10 random numbers, or you could choose 10 numbers around 34. Since we know from this video that 34 is the best choice, this gives us 4/5 numbers to select as a sort of "early exit", and the other 5/4 numbers as a "late exit", essentially allowing our number to "fall through" a "bigger hole"
Nice problem for a computer simulation. You have to be careful when you simulate a dice because what you get may not be evenly divisible by the number of sodes. The rng I use fills an integer with random bits which doesn’t work well if you want to simulate a dice with, say, 17 sides. So you have to discard some of the values close to the max value.
Re: Evil I tried doing my own and wasn't getting 666, but then saw that you stuck a note in the animation about ignoring the leading number. Once I started from the decimal, we get matching results,. I went position by position for the first 200 are here are the counts of different end points (where I stopped adding up digits when adding the next one puts the total over 666): 666: 45 665: 42 663: 25 664: 24 662: 20 660: 17 661: 15 659: 10 658: 2
If you had a full set of dice (D6 D8 D10 D12 D20) and you can choose a different one for each roll, what number is best to pick then? Is there a strategy? What if to have to plan the sequence of rolls before you start?
Could you please include some reference links in the description so people who are interested in these problems can look more into them? I'm interested in the continuous generalization he mentioned, but I don't know how to search for this problem or the related work on it.
Let me know if you find anything - I'm the person in the video and I don't have a reference for this! Just a small bit of maths I did on the back of an envelope. Might have to write it up now it's on numberphile... ^James
@@jamesmunro I figured out continuous functions for (1) the segment for S between 0 and n: P_1(S) = (1 / n + 1)^S / (n + 1), with a maximum probability at the point (n, (n + 1)^(n - 1) / n^n), and (2) the segment for S between n and 2 n: P_2(S) = n^(-S) (n + 1)^(S - 1) - S n^(n - S) (n + 1)^(S - n - 2), with a maximum probability at the point ((n + 1)^(n + 1) / n^n + 1 / ln(n / (n + 1)), -(n / (n + 1))^(n - n^(-n) (n + 1)^(n + 1)) / (e (n + 1)^2 ln(n / (n + 1)))).
Player 1 has a standard 6 sided die, sides labelled 1-6. Player 2 has a unique 6 sided die, with 5 sides labelled "0" and 1 side labelled "21" (Sum:1-6). The goal of the game is to reach a number, "x", or exceed that number, in the least rolls. What dice should you pick? Does it matter? Does the value of "x" have an impact on the dice you should choose?
Combinatorics comes to mind. Partitions in particular. With the added complexity of a limited number of faces in the dice. The discontinuities are borne in the denominator I believe… (There’s a scene about these partitions in the Ramanujen movie)
I'm seeing the long term average strategy for huge values... sort of... however: You will always have a last roll - which either hits your value or exceeds it. (If it does neither, it isn't your last roll). When you are thus within range, there is ALWAYS a 1 in 20 chance of getting the right number. To put it another way, there is never a situation where more than one value on the d20 will end the game in a victory... so... all numbers are equal. Please explain how this is wrong, since it clearly seems to be.
12:35 You can make an "any-sided" die by taking a pointy prism and create as many faces on the side as you need. Within reason, from 3 to 10 works fine.
Mathematicians: here's the most statistically probable number you can get Me: the dragon never said I had to say what the number was out loud, I'll just the roll the dice 3 times and say the sum of the 3 rolls is the number I've chosen
Excellent video. I don't believe I've seen a video with James before, but that was really an excellent topic, and made entertaining! Would live to see more. My first intuition was that this had to do with partitions, but it doesn't seem needed (or maybe you can use them but it's overkill for this particular topic as you can use more traditional methods?). Cheers,
Great Video! I stumbled over two things: 5:50 What does he mean by "You have to make 21 in two rolls"? That the highest probability is achieved with two rolls? Because of course there is also 5+5+11, 5+5+5+6, 21*1 etc. 11:30 He is talking about the average one digit makes in the summation of the digits, shouldn't zero be included here? Zero occurs with the same likeliness in Pi as all other digits, so the average should be 0+1+2+3+4+5+6+7+8+9/10 (Mind the divided by 10 instead of 9!). In effect, only 1/4.5 = 22.23 % of numbers are 3v1l, on average.
I don't like, "the odds that you see [a billion and three] is one in 10.5 because the average of a d20 is 10.5." This doesn't make sense to me. If you had a die that rolled 0, 1, or 2, its average is 1. But you wouldn't say the odds of landing on a specific high number with that die are 1 in 1.
The question is about landing on a specific total _sum_ of all rolls. So after rolling the D20 (tenns or hundreds) of millions of times there is a 1/10.5 chance that the sum will ever be exactly one-billion-three
@@daxaferCheck out what I wrote again. If I had a 0/1/2 die, the odds of reaching any arbitrary large number _aren't_ 100% merely because the average roll is 1. Similarly, if I had a 0/1 die, the chances of reaching a given large number aren't 200%.
I wrote a python script to simulate the empirical answer. For a 0/1/2 die, it levels off to ~66%, and not 100%, as expected. So I think the better way to think about why for a d20 it levels off to 1/10.5 is to consider, for the positions ahead, the chances of _overshooting_. 20 or more spaces ahead, you have 0/20 chance of overshooting. 19 spaces ahead, you have a 1/20 chance of overshooting. 18 ahead, you have 2/20 chance of overshooting. etc. Since the probability levels off, you can take each of these prior positions to be equally likely, so you can average over these to reach an overshoot probability of Σ[0 to 19] = 9.5 (out of 20), which of course leaves 10.5 out of 20 chance of landing. So while James wasn't _wrong_ per se, I think his explanation was a tad idealized - that is, it probably only applies to dice that are integers 1 to N.
Out of curiosity I calculated the probability of P(2), and got an answer that was quite surprising and, as it turns out, wrong. Can anyone point to my error? In order to NOT hit a total of 2 using a d20, you must: 1) Not land on 2 with your first roll (19/20); and 2) Not land on 1 with your second roll (19/20); and 3) Not land on 1 with your third roll. So P(2) = 1 - (19/20)^3 =~ 0.14. I expected it to be lower and, indeed, from the video at around 6:06 the correct answer is slightly over 0.05. I'm not sure where I'm going wrong, though.
its interesting to hear that Brady's intuition was something that had to be small because my mind immediately went that extremely large numbers could be better because there are more and more ways to reach that number. My intuition was telling me that it should be a very large multiple of 21 since it would be a multiple of the average roll (10.5) of a 20 sided die
a formula would really be best for comprehension, extension, and application.this is A start: with r rolls of an s sided dice the probability that the rolls' result sum above total t is given by ... this would be very helpful on a type on functor category scheme i'm looking into.
There are 668 comments right now ... Thanks to two persons, I can write without being sad about it I don't understant why 34 is the best number when 20 seems to have an higher probability when I look at the graph. Also, it's an interesting math topic, liked it
PLEASE SOMEONE CORRECT ME. I send my 3 ideas here (A, B, C). A. ACCEPTING 1 ROLL RESULTS: I would argue that for a fair 6-sided die, a good bet would be on number 6, as it has SIX possible combinations (6, 5+1, 1+5, 3+3, 4+2, 2+4), better than 8 and 5 which has FIVE combinations, 4 7 and 9 which have FOUR combinations, and 10 and 3 and which have THREE combinations. Their probability for 6 should be: 1/6 + 5/36 = 11/36 instead of lets say 5 which would have 1/6+4/36 = 10/36 or 8 which would have 5/36. B. REQUIRING >= 2 ROLLS: With the same logic, I would bet on 8, not 10 as you suggest here 8:35 . As p8=5/36 and p10 = 3/36 C. MEDIAN SUM IDEA: 1. The more the rolls -> more possible combinations for the median sum of x rolls + the actual dices configurations would follow closer the normal distribution = gets more possible to hit the most probable sum for more number of rolls. I think I am talking about the median. My intuition says that betting on the most probable sum that occurs out of 1 million rolls (top allowed), is more secure than betting on the most probable sum of only 100 (top allowed) rolls. As 100 rolls' best pick would be examined more times. So I would bet on a relatively high number. Is that right? 2. So say we have an upper limit of 500 rolls. To bruteforce it in my computer, I would run thousands of 500-rolls experiments to get the most probable sum (median sum?) of 500 rolls (including the intermediate sums created before reaching the 500th roll). I think it would be more secure than the median sum of 250 rolls, as the best pick of 250 rolls is better examined in the 500 rolls experiments. Maybe I am wrong? I would suspect that it would lie a little higher than the average sum of x rolls.
Not a mathematician, and no experience with advance calculation. What occurred to me was adding the integers from 1 to 20. 1+2+3+4+......20 and that got me 210. 210 then halved to get the average of 105, which when divided by 3 = 35(also noticed the 10.5 average you mention is a factor of 105{10.5x10}). So out of curiosity : Total of integers on all faces/sides of die lets call "T" divided by 2 to get T/2 then divided by 3, something like (T/2)/3? The question is does this hold up on a greater sided die and or even on a twenty sided die with random integers?
Pi is 1 digit off of bein a hitchhiker number, without the leadin 3; 41 For the Tau fans, it also isnt a hithchiker number with or without its leadin 6; with tho it ends up one over at 43... Which feels quite befittin for pi to be one under and tau to be one over heh (without leadin 6, it gets to 38 and 45; bcuz of an unlucky 7 after a 1)
had a massive 42 syncro, jester-die. 101010 if I rember correctly. Also 2*3*7. Endless wheel. It being very "bithish", it´s value is also 2A (TwoAlity). Never mind. It is what it is.
Wow! I estimated the best number would be either 22 or 33. Based on what I know from backgammon, the average roll would be approximately half the highest possible roll plus one (11). Then look at the multiples of 11 higher than 20 and less than 40.
Update: I have checked the first 10000 values of N. g(N) = round((e - 1)N + q) where 0.8591145934369706 < q < 0.8592248809409284 which is a very narrow interval of just 0.00011. There are 8 values of g(N) outside the formula round((e - 1)N + 4 - pi) which disproves my earlier thought! i wonder if this constant, q, has a name?
I haven't calculated this, but if you asked for a googolplex to the power of a googolplex, I'm pretty sure you'll still be rolling at the heat death of the universe.
Do it for blackjack! What if you get to pick the number, so it doesn't have to be 21. That is going to get stuck in my head now for a very long time. So many variables.
It got cute there at the end. Your intuition at the start was so good! Exactly the right reasoning that led you to the right conclusion and you got there quickly while I was still pondering. Now I'll never know if my intuition was gonna be that good! haha cheers
The best number is 58008. Iykyk, but also eventually the dragon will fall asleep before you get to your number, and you just walk out. Just make sure to bring cocaine.
By the logic used in the video, 2/3 is infinitely evil. Not only because eventually you’ll hit 666 when adding up the digits, but because the decimal expansion is only 6s.
I have news about the evilness of e, by the power of REGULAR EXPRESSIONS IN EXCEL! So we know e is NOT evil, but you need to understand it comes VERY close. It hits 665 after summing the first 140 digits, then the 141st digit is a 0, then the 142nd digit is a 3, bringing the total to 668.
Correction: the optimal number for the original 20-sided die problem is actually 35 (this is slightly better than 34). James has been fed to the dragon as punishment for this error.
Thanks! I've been searching for an off-by-one error in my code :D
Haha, hope the Dragon liked fried James. 🙂
It says James still works at Oxford in outreach. Is he doing this from within the dragon's belly?
@@Corwin256 It is dark in here, but on the plus side I've got more time to think about maths now! ^James
No graphic on screen for that?
The best number you can pick is Graham's Number. That way you can stall long enough for a rescue party.
it's actually kinda interesting, because with numbers this large you end up with the problem that you cannot inch towards such a number in this universe by adding small enough numbers. There's just not enough information storage space in the universe to represent all the intermediate steps.
edit. so, peculiarly, you can throw the die enough times to reach graham's number, but you'd never in this universe know whether you actually did or not
>don't want to be in a dungeon forever
>choose a number that keeps you there forever
@@Sopel997 Would it be achievable in some different number base?
Can't see the dragon for the TREE(3)?
Too bad the dragon has a time machine which defeats any time problem
What surprises me is that 21 is so unlikely even though it is exactly double the average, and it is the most likely to come up if you just have 2 throws.
If you have exactly 2 throws, that is. The extra chance for numbers 1 to 20 comes precisely from the option of stopping after the first throw.
You are correct to be surprised, as he seems to be calculating only the odds of hitting the number in a single throw after you get to within 20 of that number. He is not considering the chances of reaching the number in 2 or more additional throws once you get to within 20. Unless I missed something. So he is leaving out the possibility, for example of 1+1+1+1......+1+1 = 21.
@@GreylanderTV No, any way of reaching a number is included in the calculation, even 1+1+1+1+...+1+1 = 21.
@@GreylanderTV The recurrence relation for 21 includes the probability that you were previously at 20 and rolled one, which already includes the probability that you were at 19 and rolled one, etc.
@@GreylanderTV When he talks about "recurrence relations" he means that the computer will take his 1-roll equation and keep iterating it for an arbitrarily large number of rolls. This is why the other ways of reaching 21(or anything else) are included. If you didn't know this, then I get why it looked like he wasn't considering large numbers of rolls.
Brady you are such an amazing interviewer. Every single time I am in awe of your abilities.
Yes I didn't get his intuition about it being most likely to be in between 20 and 40, but it turned out to be correct!
Is a number even more evil if it hits 666 twice because there is a zero in that spot?
I like your thinking.
@@numberphile Then sqrt(90) would be triple evil, only counting decimal places
Obviously 666 zeroes at that spot is the most evil.
664 Neighbors of the beast.
I thought along the same lines - How many numbers are "double evil" where summing the whole as well as decimals ends up evil where the Nth decimal position number arriving at 666 is the same number as the whole?
I've never seen this James Munro guy but please use him more! He has such a good, conversable tone, you can hear the excitement in his voice for maths as he speaks and he has the ability - seemingly - to speak to a range of ages and abilities and foster an enjoyment in maths. In fact this guy's probably what Rishi wants from all his teachers lol.
Great video Brady. Interesting subject topic. I've literally just got home from a tutoring session with a student regarding probability so this was a nice continuation from that for me.
I bet the editor had real fun with that wilhelm scream lol
1:43
I can't believe I missed that! It was quiet in my defence though 😂
This problem actually is a fantastic application of dynamic programming. If you do it without it, the time complexity of your algorithm is O(20^n). With DP it is O(n)
Even better, this is a (homogenous) linear recurrence (with constant coefficients) so has a closed form that you can compute in O(1)
@@tehdarkneswithin I guess that requires finding the 20 exact roots of a 20th order polynomial first?
Please show us the way... 😉
(And it seems to assume we can do exponentiation in O(1), which I kind of doubt when we want to be exact...)
@@tehdarkneswithinwell. i was thinking of a different problem, the finding of the max for any kind of die. which was O(s^3). Yes finding the answer to your problem however is probably like O(s*log(n))
Based comment I've been grinding dp problems, when I close my eyes all I see is `if dp[i] >-1 return dp[i]`
Brady's intuition is amazing. Every time.
Pi being evil is another reason to use Tau instead - It reaches 663 on the 139th digit and 668 on the 140th, nicely skipping over 666 as any well behaved number should do!
Pi being evil is another reason to keep using pi. Tau is not evil and therefore boring.
Ten points to G̶r̶i̶f̶f̶i̶n̶d̶o̶r̶ Steve Mould
I'm OK with any additional reason to use Tau.
Would you rather eat pi or a tau (towel)?
But I have no umbra formas why would I go tau
If you can choose any number, just choose 0 and don't roll at all. Instant win.
choose -1 so that you would roll forever knowing that you will never reach it. bore the dragon until it succumbs and frees you
The problem raises when it would demand at least one roll from you.
@@drenz1523 that would be a instant lose, as it is exact. If above, you die.
@@thomas-carter so it's result as a instant lost as well.
Best would use X
boom, guarantied win here!
@@claudetheclaudeqc6600 oh right... darn it!
Pi is evil?! Welp, looks like you gotta change your profile picture, Numberphile!
😂
Long live Tau
@@imveryangryitsnotbutterTau is 2*pi so does that mean it’s twice as evil?
Update: I have verified the first 3000 values for N. Was able to compute each result in O(N^2) cause of knowing what place to start looking and an improved algorithm that does not compute things based on number of rolls. The answers to the question of what the best guess > N is seems to follow f(N) = round((e - 1)N + 4 - pi) for N=2...3000 EXCEPT N=232 and N=1233 both in which case the actual answer is f(N)+1 but thats not something im completely sure about since the difference in probability for N=232 is only 6.182648143449043e-12 and for N=1233 is a meager 5.937702586555904e-13.
Edit: I am sure now. have tried with 112 bit percision floating point numbers.
The issue with picking the closest value to the peak is that doing that only works for specific kinds of functions. (They at least require that, around a peak x, f(x + k) = f(x - k).) It doesn't work for generalised functions, but if the function can be approximated by one that fulfills this property, it will usually get you the right answer. Any quadratic function meets this rule, so if the function has a very small third derivative, you can get away with it.
You are wrong.
I love that you can screw around with numbers and get something meaningful out of sensless goofing around.
Pi might be evil, but Numberphile is a great good.
Watching hardcore mathmeticians trying to do simple mental arithmatic is always amusing, a friend is doing a phd is maths and we always find it funny when he stuggles to work out his portion of the bill 😂
I enjoyed differential equations in college… But had to use my fingers or scratch paper to add and subtract. 😂
Interesting that the 0 is ignored in the average of the digits in pi. If you include it to get an average digit of 4.5, then the value 1/4.5 represents the expected number of times that a specific very high total will be hit, since it can be more than once if one or more zeros occur immediately after hitting the total.
Ya but there is no 0 on a die
@@slo3337 Any number can appear on any die.
In fact, most 10-sided die are numberd from 0-9 (as well as some numbered from 00 to 90, that when paired with a 0-9 allow you to roll 0-99 for percentages). This would also perfectly simulate rolling a 9-sided die in this application (since we only care about the sum, not the number or rolls).
@@phiefer3 Well, in a literal sense, the only numbers which can appear on a die... are the numbers which are marked on the faces of said die.
A way to have a guaranteed victory against the video's dragon is by putting your chosen number (or a factor of it) on all the faces of the dragon's die.
If you want the probability that a given number will be hit, then zero must be ignored
But if you want the expected number of digits, then zero should be included
(He was only doing the former, not the latter)
Remember the recurrence relation? p(k) = 1/n p(k - 1) + 1/n p(k - 2) + ... + 1/n p(k - n).
(In the digits example, n = 9.)
If we allow zero, the probability becomes 1/(n + 1) (because there's now an additional possible outcome), and you get the extra addend for zero:
p(k) = 1/(n + 1) p(k - 0) + 1/(n + 1) p(k - 1) + ... + 1/(n + 1) p(k - n).
But p(k - 0) is obviously p(k), so you can move that term to the other side, and you get p(k) - 1/(n + 1) p(k) = n/(n + 1) p(k) on the left-hand side, meaning you have to divide all terms by n/(n + 1) - which gets you back to the original equation without zero.
I'm gonna call this rule34 of dice 😈
The calculation for the optimal dragon dungeon number can use pi too. Instead of (e-1)N, you can do the average of the die * pi to get close to the optimal guess. 6 sided die average is the sum of both sides divided by two, so 3.5, and 3.5 * pi = 11. Same for 20 sided die, but 10.5 * pi = 33, not 34
so many occult numbers appearing in this video lol, 33, 42, 144, 666
More videos from Mr. Munro please! I like this guy!
yes he's great
(waves) Hi! I've made about 70 episodes of an online maths club over on this channel, and I'm doing a bunch of livestreams on maths admissions test questions. ^James
Fun fact: 666 is a mistranslation; the actual number is 616. Is pi still evil by the more accurate texts?
Neat. Do you have a citation for that?
its interesting how the first 144 decimals of pi add up to 666, when the number 144,000 is mentioned not long after 666
Great puzzle! Tried to solve it in my head before firing up python, and the incorrect heuristic I got was roughly (2 - 1/e)*N. I figured out the probabilities for 1 through 20 to be P(x) = p*(1+p)^(x-1), where p = 1/20, which are exponentially increasing. Then P(21) would be the average of P(1) through P(20), and so on. The approximation I then made was that the maximum probability among P(21) through P(40) would be the average starting from the first number between 1 and 20 that had an above-average probability (i.e. more than P(21)), and if you work through the math on that, this would give an optimal value of N + log(e-1)/log(1+p), in the limit as N becomes large; and using series expansions this is roughly (2 - 1/e)*N. This is an approximation because the probabilities P(21) through P(40) are also (initially) increasing, so instead of finding the first x such that P(x) > P(21), we should find the first x such that P(x) > P(x+20). But that seemed too complicated to think about.
Pleasantly surprised to see the actual correct heuristic also involves e, and to be more specific, a simpler expression of e that lies between 1 and 2!
The recursive rule can be seen as a Infinite Response Filter that just average the previous N values for a N sided dice. So it's a low-pass filter hence the smoothing effect.
*an N-sided die
I'm so glad that someone picked up on the smoothing effect! This was not the right video to talk about how there's a jump discontinuity then a jump in the first derivative (in the continuous version), but it's the sort of thing I see when I look at the graph :) ^James
@@TheGreatAtario Yes - the singular noun is die, and its' plural is dice. "The dice are loaded", not "the dice is loaded".
This is actually a fantastic observation, and could merit an entire video just exploring this topic itself.
For the values for 1 to 20, you can think of P(n) = 0 for -19 =< x =< -1 and P(0) = 1, which makes sense conceptually, since you always start with 0, and you never have negative numbers.
It doesn’t make sense if you keep extending it backwards, though: you get that |P(n)| = 20 for n in [-39, 20], alternating in sign. And then you get P(-40) = 800.
When the "many more sides die" was suggested, my mind immediately went to 600 - one of the 4D platonic solids.
I like that under the digit definition of evil 666 itself is not an evil number
It's a lucky number.
That's the best trick of evil
The worst evildoer thinks they are righteous.
@@BobStein It's like humanity in general, it mostly does evil on this planet and thinks it's all right.
The Biblical reference relates 666 to “the number of the beast (or antichrist).” The Antichrist mimics the true Christ with stolen power & knowledge. So naturally the original 666 from the true Christ is not evil.
Putting this challenge in my D&D game!
That is a bit meta. The dragon in the game is asking the characters to choose numbers for dice rolls which the players then perform on behalf of the characters.
There is a fragrance called Pi, by Givenchy (and created by one of the most prolific and successful perfumers: Alberto Morillas.) Pi smells really great, warm sweet, suitable for cold weather season, so this the best time to wear it in the northern hemisphere . I'd encourage all math/Pi enthusiasts to check it out.
Actually, 35 is the most likely for a 20-sided die, not 34.
Thanks for the sanity check. Just wrote code to solve this myself and it was giving me 35.
Seeing the same by just rolling a lot in python:
number of success for 32: 96891
number of success for 33: 97156
number of success for 34: 97212
number of success for 35: 97605
number of success for 36: 97127
1 million rolls each time.
Ah I wondered if anyone would spot this. Sorry - I have no idea why I said 34 on the day. In my defense, 34 and 35 have very very similar probabilities. ^James
Thought I messed up the Desmos graph when I saw 35.
34 - 0.09751
35 - 0.09767
Perhaps because (e-1)*20 is closer to 34 than 35 the intuition would be that 34 is slightly better than 35. However, a bit of experimentation shows that 35 is favoured.
I really like this James, he reminds me of classic numberphile
oh crazy, James was my guide at a cambridge summer school back in 2014!
If the dragon let you choose 2 target numbers upfront, what would the best choices be for a 20-sided die? Will they maybe be 34 & 35, or will they be spread out?
That's a really good question. I thought it would obviously be 34 and 35 but the more I think about it the more I'm unsure.
My intuition tells me that since the average roll of a d20 is 10.5, choosing 33 and 34 is the best option. To explain why spread out numbers wouldn't work, imagine the dragon gave you 10 numbers to choose. You could choose any 10 random numbers, or you could choose 10 numbers around 34. Since we know from this video that 34 is the best choice, this gives us 4/5 numbers to select as a sort of "early exit", and the other 5/4 numbers as a "late exit", essentially allowing our number to "fall through" a "bigger hole"
That’s an interesting question because you want to pick two that are both likely but cover different sequences of rolls.
Right. P1 + (P2 | !P1). Interesting.
This is an excellent follow-up question. ^James
Nice problem for a computer simulation. You have to be careful when you simulate a dice because what you get may not be evenly divisible by the number of sodes. The rng I use fills an integer with random bits which doesn’t work well if you want to simulate a dice with, say, 17 sides. So you have to discard some of the values close to the max value.
That dragon dice problem feels like convolution is involved, but I cant really put my finger on it
Not discussed in the video, but a nice result anyway: the probability of hitting N with an N-sided die approaches e/(N+3/2) for large N.
Re: Evil I tried doing my own and wasn't getting 666, but then saw that you stuck a note in the animation about ignoring the leading number.
Once I started from the decimal, we get matching results,. I went position by position for the first 200 are here are the counts of different end points (where I stopped adding up digits when adding the next one puts the total over 666):
666: 45
665: 42
663: 25
664: 24
662: 20
660: 17
661: 15
659: 10
658: 2
If you had a full set of dice (D6 D8 D10 D12 D20) and you can choose a different one for each roll, what number is best to pick then? Is there a strategy? What if to have to plan the sequence of rolls before you start?
p=0.30013020833333337 when guess is 8 and your plan is to use: 8, 6, 10, 12, 20. (this is the optimum)
I heard the Wilhelm scream. You didn't slip it past us.
Could you please include some reference links in the description so people who are interested in these problems can look more into them? I'm interested in the continuous generalization he mentioned, but I don't know how to search for this problem or the related work on it.
Let me know if you find anything - I'm the person in the video and I don't have a reference for this! Just a small bit of maths I did on the back of an envelope. Might have to write it up now it's on numberphile... ^James
@@jamesmunroAh, so this is something you came up with yourself? That's cool. How did you approach the continuous case?
@@jamesmunro I figured out continuous functions for
(1) the segment for S between 0 and n: P_1(S) = (1 / n + 1)^S / (n + 1), with a maximum probability at the point (n, (n + 1)^(n - 1) / n^n), and
(2) the segment for S between n and 2 n: P_2(S) = n^(-S) (n + 1)^(S - 1) - S n^(n - S) (n + 1)^(S - n - 2), with a maximum probability at the point ((n + 1)^(n + 1) / n^n + 1 / ln(n / (n + 1)), -(n / (n + 1))^(n - n^(-n) (n + 1)^(n + 1)) / (e (n + 1)^2 ln(n / (n + 1)))).
Player 1 has a standard 6 sided die, sides labelled 1-6. Player 2 has a unique 6 sided die, with 5 sides labelled "0" and 1 side labelled "21" (Sum:1-6). The goal of the game is to reach a number, "x", or exceed that number, in the least rolls. What dice should you pick? Does it matter? Does the value of "x" have an impact on the dice you should choose?
Missed a great chance for a Matt Parker cameo with whatever highest number fair die he created a while ago.
LOL the wilhelm scream at 1:43
Since (e-1)N leads to an approximation, just use √3N
James: "I don't know how to make a nine-sided die"
Most recent Curiosity Box: "hold my light-year of water"
You summarize every signals intelligence intuition I've felt throughout my life very well
I'd pick something like 9239328574395749 and leave once the dragon gets bored.
12:34 The easiest way to make a 9-sided die is with a 9-sided prism and doming the ends so it can't land on the nonagonal faces.
Brady choosing 42 then there’s the Wilhelm’s scream during animation, welcome to nerdphile ;)
11:04 I find it interesting that the corresponding decimal place is 12 squared
Yes, In Analog Clock, Hour Needle passes 2 times from 12... And there are 1440 minutes in a Day..
Another level of irrelevance is to consider whether the number of the beast was actually 616.
Numberphile: Pi is evil
Also, Numberphile (9 years ago): Pi is beautiful 😂
Combinatorics comes to mind. Partitions in particular. With the added complexity of a limited number of faces in the dice. The discontinuities are borne in the denominator I believe…
(There’s a scene about these partitions in the Ramanujen movie)
Tried telling my maths teacher that maths was evil. She was unimpressed.
1:43 obligatory Wilhelm scream
I'm seeing the long term average strategy for huge values... sort of... however:
You will always have a last roll - which either hits your value or exceeds it. (If it does neither, it isn't your last roll). When you are thus within range, there is ALWAYS a 1 in 20 chance of getting the right number. To put it another way, there is never a situation where more than one value on the d20 will end the game in a victory... so... all numbers are equal. Please explain how this is wrong, since it clearly seems to be.
Oh, if you're allowing an interval now, I choose [1,20] inclusive.
12:35 You can make an "any-sided" die by taking a pointy prism and create as many faces on the side as you need. Within reason, from 3 to 10 works fine.
A Parker Number is where the digits do not add up to ANY literary sum. Not 42, not 666, not 451...
Thank you for using an accurate dragon cartoon
Mathematicians: here's the most statistically probable number you can get
Me: the dragon never said I had to say what the number was out loud, I'll just the roll the dice 3 times and say the sum of the 3 rolls is the number I've chosen
more evidence for tau to replace pi
Are you sure that-for this definition of evil-tau is not also evil?
@@PhilBoswell I haven’t check but I remain in hope
What's even better than doing the math is just refusing the dragon's game. If he then tries to keep you there, it's wrongful imprisonment.
I appreciate the Wilhelm scream.
And that's why we should be using tau instead. Periodicity of the circle for the win!
Excellent video. I don't believe I've seen a video with James before, but that was really an excellent topic, and made entertaining! Would live to see more. My first intuition was that this had to do with partitions, but it doesn't seem needed (or maybe you can use them but it's overkill for this particular topic as you can use more traditional methods?).
Cheers,
I'm pretty sure that expanding that recurrence to a non-recurring version requires solving partitions along the way, but that's just my intuition.
It is worth noting that tau is not evil (with or without the initial 6). τ > π
Great Video! I stumbled over two things:
5:50 What does he mean by "You have to make 21 in two rolls"? That the highest probability is achieved with two rolls? Because of course there is also 5+5+11, 5+5+5+6, 21*1 etc.
11:30 He is talking about the average one digit makes in the summation of the digits, shouldn't zero be included here? Zero occurs with the same likeliness in Pi as all other digits, so the average should be 0+1+2+3+4+5+6+7+8+9/10 (Mind the divided by 10 instead of 9!). In effect, only 1/4.5 = 22.23 % of numbers are 3v1l, on average.
I don't like, "the odds that you see [a billion and three] is one in 10.5 because the average of a d20 is 10.5." This doesn't make sense to me.
If you had a die that rolled 0, 1, or 2, its average is 1. But you wouldn't say the odds of landing on a specific high number with that die are 1 in 1.
The question is about landing on a specific total _sum_ of all rolls. So after rolling the D20 (tenns or hundreds) of millions of times there is a 1/10.5 chance that the sum will ever be exactly one-billion-three
@@daxaferCheck out what I wrote again. If I had a 0/1/2 die, the odds of reaching any arbitrary large number _aren't_ 100% merely because the average roll is 1. Similarly, if I had a 0/1 die, the chances of reaching a given large number aren't 200%.
I wrote a python script to simulate the empirical answer. For a 0/1/2 die, it levels off to ~66%, and not 100%, as expected.
So I think the better way to think about why for a d20 it levels off to 1/10.5 is to consider, for the positions ahead, the chances of _overshooting_.
20 or more spaces ahead, you have 0/20 chance of overshooting.
19 spaces ahead, you have a 1/20 chance of overshooting.
18 ahead, you have 2/20 chance of overshooting.
etc.
Since the probability levels off, you can take each of these prior positions to be equally likely, so you can average over these to reach an overshoot probability of Σ[0 to 19] = 9.5 (out of 20), which of course leaves 10.5 out of 20 chance of landing.
So while James wasn't _wrong_ per se, I think his explanation was a tad idealized - that is, it probably only applies to dice that are integers 1 to N.
Gonna implement this into my next DnD game.
okay but bold of you to assume I wouldn't want to stay with the dragon forever
"The fractional part of Pi is evil, maybe the 3 saves it."
There's a theology joke in there somewhere.
Out of curiosity I calculated the probability of P(2), and got an answer that was quite surprising and, as it turns out, wrong. Can anyone point to my error?
In order to NOT hit a total of 2 using a d20, you must:
1) Not land on 2 with your first roll (19/20); and
2) Not land on 1 with your second roll (19/20); and
3) Not land on 1 with your third roll.
So P(2) = 1 - (19/20)^3 =~ 0.14. I expected it to be lower and, indeed, from the video at around 6:06 the correct answer is slightly over 0.05. I'm not sure where I'm going wrong, though.
its interesting to hear that Brady's intuition was something that had to be small because my mind immediately went that extremely large numbers could be better because there are more and more ways to reach that number. My intuition was telling me that it should be a very large multiple of 21 since it would be a multiple of the average roll (10.5) of a 20 sided die
Short bit of (python) code to calculate and print the probabilities given an N=20 sided die:
N, P = 20, [ ]
for i in range(43):
P.append((i == 0) + sum(P[i-k-1] for k in range(min(N, i))) / N)
print(f"{i:2}:{P[-1] * 100:7.3f} %")
Results in:
0:100.000 %
1: 5.000 %
2: 5.250 %
3: 5.513 %
4: 5.788 %
5: 6.078 %
6: 6.381 %
7: 6.700 %
8: 7.036 %
9: 7.387 %
10: 7.757 %
11: 8.144 %
12: 8.552 %
13: 8.979 %
14: 9.428 %
15: 9.900 %
16: 10.395 %
17: 10.914 %
18: 11.460 %
19: 12.033 %
20: 12.635 %
21: 8.266 %
22: 8.430 %
23: 8.589 %
24: 8.743 %
25: 8.890 %
26: 9.031 %
27: 9.163 %
28: 9.287 %
29: 9.399 %
30: 9.500 %
31: 9.587 %
32: 9.659 %
33: 9.714 %
34: 9.751 %
35: 9.767 %
36: 9.761 %
37: 9.729 %
38: 9.670 %
39: 9.580 %
40: 9.458 %
41: 9.299 %
42: 9.350 %
Nice code.
a formula would really be best for comprehension, extension, and application.this is A start: with r rolls of an s sided dice the probability that the rolls' result sum above total t is given by ... this would be very helpful on a type on functor category scheme i'm looking into.
Really fun little problem and approach!
There are 668 comments right now ... Thanks to two persons, I can write without being sad about it
I don't understant why 34 is the best number when 20 seems to have an higher probability when I look at the graph.
Also, it's an interesting math topic, liked it
PLEASE SOMEONE CORRECT ME. I send my 3 ideas here (A, B, C).
A. ACCEPTING 1 ROLL RESULTS:
I would argue that for a fair 6-sided die, a good bet would be on number 6, as it has SIX possible combinations (6, 5+1, 1+5, 3+3, 4+2, 2+4), better than 8 and 5 which has FIVE combinations, 4 7 and 9 which have FOUR combinations, and 10 and 3 and which have THREE combinations.
Their probability for 6 should be: 1/6 + 5/36 = 11/36 instead of lets say 5 which would have 1/6+4/36 = 10/36 or 8 which would have 5/36.
B. REQUIRING >= 2 ROLLS:
With the same logic, I would bet on 8, not 10 as you suggest here 8:35 . As p8=5/36 and p10 = 3/36
C. MEDIAN SUM IDEA:
1. The more the rolls -> more possible combinations for the median sum of x rolls + the actual dices configurations would follow closer the normal distribution = gets more possible to hit the most probable sum for more number of rolls. I think I am talking about the median. My intuition says that betting on the most probable sum that occurs out of 1 million rolls (top allowed), is more secure than betting on the most probable sum of only 100 (top allowed) rolls. As 100 rolls' best pick would be examined more times. So I would bet on a relatively high number. Is that right?
2. So say we have an upper limit of 500 rolls. To bruteforce it in my computer, I would run thousands of 500-rolls experiments to get the most probable sum (median sum?) of 500 rolls (including the intermediate sums created before reaching the 500th roll). I think it would be more secure than the median sum of 250 rolls, as the best pick of 250 rolls is better examined in the 500 rolls experiments. Maybe I am wrong? I would suspect that it would lie a little higher than the average sum of x rolls.
I love that the golden ratio is evil. Take that hippies
Not a mathematician, and no experience with advance calculation. What occurred to me was adding the integers from 1 to 20. 1+2+3+4+......20 and that got me 210. 210 then halved to get the average of 105, which when divided by 3 = 35(also noticed the 10.5 average you mention is a factor of 105{10.5x10}). So out of curiosity : Total of integers on all faces/sides of die lets call "T" divided by 2 to get T/2 then divided by 3, something like (T/2)/3? The question is does this hold up on a greater sided die and or even on a twenty sided die with random integers?
Pi is 1 digit off of bein a hitchhiker number, without the leadin 3; 41
For the Tau fans, it also isnt a hithchiker number with or without its leadin 6; with tho it ends up one over at 43... Which feels quite befittin for pi to be one under and tau to be one over heh (without leadin 6, it gets to 38 and 45; bcuz of an unlucky 7 after a 1)
What a great video for Spooktober! Thanks Brady
had a massive 42 syncro, jester-die. 101010 if I rember correctly. Also 2*3*7. Endless wheel. It being very "bithish", it´s value is also 2A (TwoAlity). Never mind. It is what it is.
Wow! I estimated the best number would be either 22 or 33. Based on what I know from backgammon, the average roll would be approximately half the highest possible roll plus one (11). Then look at the multiples of 11 higher than 20 and less than 40.
Coming soon on StandUpMaths... Rolling 20-sided dice!
Right on. Thanks for sharing.
Update: I have checked the first 10000 values of N. g(N) = round((e - 1)N + q) where 0.8591145934369706 < q < 0.8592248809409284 which is a very narrow interval of just 0.00011. There are 8 values of g(N) outside the formula round((e - 1)N + 4 - pi) which disproves my earlier thought! i wonder if this constant, q, has a name?
"because the digits of pi are evil, I vote we cut them off and just make pi = 3"
- engineers probably
I haven't calculated this, but if you asked for a googolplex to the power of a googolplex, I'm pretty sure you'll still be rolling at the heat death of the universe.
This is a cool bit of maths, I will try to remember it the next time I am trapped in a dungeon with an angry dragon and a dice.
12:47 it’s a ten sided die with zero. This is a common die.
Do it for blackjack! What if you get to pick the number, so it doesn't have to be 21. That is going to get stuck in my head now for a very long time. So many variables.
Whole numbers themselves are also looking pretty shady. Add up 1 through 35 and you get 630...
I'd choose 700 quintillion, and hope the dragon would either get bored and go away or just die of old age before I'd get close to the number
The Wilhelm scream is a nice easter egg.
It got cute there at the end. Your intuition at the start was so good! Exactly the right reasoning that led you to the right conclusion and you got there quickly while I was still pondering. Now I'll never know if my intuition was gonna be that good! haha cheers
The best number is 58008. Iykyk, but also eventually the dragon will fall asleep before you get to your number, and you just walk out. Just make sure to bring cocaine.
No, pi is delicious, it's the calories that are evil 😂
Graham's number - Because you (or the dragon) would die of old age before hitting it.
What if you allow 1-20, but disallow “wins” on the first roll? (So it always has to be a proper “sum” of 2 or more numbers)
By the logic used in the video, 2/3 is infinitely evil. Not only because eventually you’ll hit 666 when adding up the digits, but because the decimal expansion is only 6s.
I have news about the evilness of e, by the power of REGULAR EXPRESSIONS IN EXCEL!
So we know e is NOT evil, but you need to understand it comes VERY close. It hits 665 after summing the first 140 digits, then the 141st digit is a 0, then the 142nd digit is a 3, bringing the total to 668.