Thanks for watching! Consider checking out the Patreon I made a couple weeks ago (the names of current supporters are in this video description) www.patreon.com/comboclass
5:50 I already knew that: Matt Parker (standupmaths) has a video about 2d6 to 1d6- this is exactly what he uses to do that transformation- as well as the transformation from 3d6 to 2d6.) In Matt's video, he even uses a 3d6 inside a cube. 6:55 sounds a lot like a clear DND reference: "What would the chart look like if you had an advantage on a saving throw that just uses 1d6?" is essentially what you're saying, but DND typically uses 1d20 for these "saving throws." DND uses all these: d4, d6, d8, d10, d12, d20, d100 (Also called "percentile dice," the d100 is just in the form of 2d10)
Been in a pretty bad place in life for a while now, and it's the little things like this that keep me going. Thanks for making these! They're silly and weird but education should be more fun in general, and your boundless enthusiasm would surely be enough to overcome the judgement of even the most ardent math-haters. The content and consistent quality is great, and I just wanted to express my appreciation. So thanks again, and to anyone who reads this, I hope you're doing well :)
@@uirwi9142 I was asking if they were Anarcho-Communist due to the flag on their PFP, Anarcho- meaning Anarchist, Communist means... well... communist.
i’ve disliked math for years, but recently things have started making sense and after watching math youtubers like yourself, i’m considering adding a math major to my degree plan. thanks for finally making it interesting and understandable
There are no people who dislike math. There are only people who dislike their bad math teachers. Or at least that's the understanding I gained from surviving through the education system.
@@btCharlie_ I think that's not really true. I agree there's lot of mediocre or bad math teachers, but it doesn't mean that nobody dislikes it. People do have preferences for certain things (particularly when it comes to just the broad duality of STEM vs non-STEM)
I'm sure you're aware of this, but in tabletop RPGs, rolling multiple dice and picking the highest result is called rolling with advantage. For 2 n-sided dice, your average result will approach 2n/3 since the equation for the average value is (n+1)(4n-1)/6n^2. If you do the limit as n approaches infinity to, you'll get 1/6 * 1 * 4 = 4/6 = 2/3.
I'm a huge fan of TTRPGs like dungeons and dragons. I actually have probably a few hundred dice from playing! I'm not very good at the game but it's a lot of fun to play with friends
I've always sucked at math, but at the same time found it interesting as hell. Your channel is right up my alley. Thanks for the great content. The only thing stopping your channel from blowing up is lack of exposure. Keep up the good work, you'll be at 100k subs in no time.
Began watching this channel a while ago and subbed pretty quick but never thought to check the subscriber count, because I always assumed it was a multi million subscriber channel; It just caught my eye though. Wdym 20k subscribers?? This is really quality content
If you treat two dice as a two-digit seximal (base six) number, you can create a number from 0 to decimal 35 if you count 6 as 0. From there, using modulo can get you equal chances for 2, 3, 4, 6, 9, 12, or 18, plus just the raw number for 36
Love your vids! Highly recommend the game Dicey Dungeons haha. It's not easy, and you'll run into so many tricky sticky probabilistic decisions to make. Seriously, top tier game!
my all time favorite dice rule is from an older DnD book. you roll a number of d6, and sum them up. then you take any dice that rolled a 6 exactly, and re roll those and add that to the sum. you repeat this process until no more 6's were rolled. the math to map n number of dice to s total is actually quite complicated. there was also a relationship where you could roll to hit. if you rolled too low you did zero damage. you increase your chance to hit by reducing the number of d6. if you reduced all but 1 of you dice, you had a way higher chance to hit, but did way less damage. if you took away no d6 then you chances to hit were lower but you did more damage if you did hit. then the question becomes, what is the relationship between how many d6 you roll and your chance to hit. if you roll a d20 to hit and you need to roll a 15, what is the optimal amount of d6 to remove to achieve the highest average damage? i have solved this using monte carlo, but the covariance solution is exceedingly complicated lol. i still don't have a good model for it that matches the monte carlo results
After thinking about it for a bit, it doesn't seem too complicated to calculate average sum of any number of d6 with 6 rerolls. For a single dice you get 5/6 * 3 (the average between 1 and 5) + 1/6 * 5/6 * 9 (single reroll) + 1/6 * 1/6 * 5/6 * 15 (two rerolls), which in the end gives you an average of 4.2. "sum 5/6 * 1/6^n * (3 + 6 * n), n=0 to infinity" And because dice are independent, for m dice you get 4.2 * m on average, 8.4 for 2 dice, 12.6 for three, etc.
It is certainly more complicated than the simple charts he shows in the vid, since you have to take weight into account rather than just counting the results (for instance, using 3 d6, even though there are 117 ways to get 20 as a result and only 6 ways to get 5, 5 is the more likely result between the two [2.78% vs 2.08%]) not to mention the fact that the results literally go to infinity. If you figure out how subtracting dice adds to hit chance, then I'm sure it wouldn't be overly difficult to figure out the mathematically determined optimal approach (you say you need at least 15 on a d20, but how many d6 is that for? and how does that number change as d6 are taken away?). Of course, the pure math can only go so far. When it comes down to actually being used, situation and play style will be the better determiners of whether one should risk missing just to land a hard hit or go for the safer weaker shot.
Very nice presentation of all the combinations, in a regular grid so you can see the histogram as well as the combinations themselves. Glad to be an early subscriber; extrapolation suggests very early! Keep up the amazing work!
While it's not strictly rolling dice, there's a tabletop game called Gloomhaven where multiples of numbered cards are drawn (functionally like a roll of the dice) and with some buffs or debuffs, the greater or lesser of the cards are discounted. So it is like rolling two dice and only counting the higher or lower of the two. It's an interesting mechanic.
Yo! Your conversion of 2D6 into 1D12 or 1D3 etc is literally my PhD. It's a combinatorial structure called Sum Systems, which asks; For two sets A and B, by taking all sums of each a in A and b in B, can you get A+B={0,1,2,...,N-1} for some N in the naturals, and each element occurring exactly once each? So, if we consider this structure as "modulo N", then you count a result of 0 as N when using these systems in dice (like how 1d100 use 2D10 and a result of 00 and 0 is 100 (almost like the d100 is a Sum System!!! (Spoiler, it is))) Both sets must have 0, as to make the resultant sum of 0. Otherwise each other element is unique. Your choice of N affects your sets too. The rule is that N = |A| x |B|, i.e. the size of your consecutive integer set is the product of the cardinality of your two "sum system component sets" Taking each element in a set to be a face on a dice gives you the desired property But it goes further! Can you have 3 dice? 4 dice? More? What's the minimum sum of these dice? What about if you include subtraction?! (Known as a sum and distance system) I lloovvee these structures hahaha Example: Let A={0,1,2,3,4,5} and B={0,6}. Then A+B={0,1,2,3,4,5,6,7,8,9,10,11} and N=12. Wait! If we let the result of 0+0=0==12, then this result is the exact same you used in the video!! Entertainingly, we can also use A={0, 1, 2, 6, 7, 8} and B={0, 3}, or a personal fav of A={0, 1, 6, 7} and B={0, 2, 4}! I've reached out through Email.
Difference of two n sided dice is most likely 1. Wow. What a great maths fact and one I will use in class tomorrow, very well explained ! Keep these awesome videos coming !! 😎
Great video as always. One thing that I can't help but notice though is the audio becoming weirdly intense during some sounds. Like for just a split second the microphone is resonating along with your voice or something. It's hard to describe what's going on with the sound, but its like some extra low frequency is getting picked up along with your normal speech emphasis and making it kinda painful. 4:00 "Which is mORe likely than seven itSElf..." 5:20 "Well ONE shows up six times now cause it tooK ALL..." Is this just me?
I have tried asking that robot math questions on some recent livestreams on the Combo Class Bonus channel here ( @domotro ) and although the robot has potential, it’s currently comically bad at math haha
I propose that the modular reduction from multiple dice actually does serve a purpose: it reduces the possibility of influencing the outcome. As the system is now a more complicated system with multiple inputs that combine to form the output, manipulating it to produce an intended result would require successfully controlling the result of two (or more!) thrown dice. Assuming that those who seek to manipulate the outcome of dice rolls have a sub-100% chance of succeeding at it, each additional die added to the modular reduction method would exponentially reduce that chance of succeeding at rigging the throw. Also I like the natural progression from modular arithmetic to min/max functions. This all applies so directly to the mathematics objects in the Pure Data visual programming language.
Could you pleeease cover the three sided die (the thick coin which has an equal chance to land on either face *or* it's thich edge) that Matt parker went into a bit but never concluded?!? 😁
This is such a pure series, just fun in a backyard. Complicated yet simple things, such a home hearted feeling, like meeting a friend after a long time. The intro feels special, it feels like a raw intro, no flashy things, just a unique thing captured straight on camera, combo class. Thank you, I am your first die hard fan. Remember me when ya get to the top.
Imagine when you hit 1 million subscriptions you find a weird math pattern and make a video about it on how it grew over time and how it fit that weird pattern.
alright for the simplest strategy for having equal probability algorithms with only a coin: say I want to make a 7 sided die with a coin. H,H,H =>1 H,H,T=>2 H,T,H=>3, T,H,H=>4 H,T,T=>5 T,H,T=>6 T,T,H=> 7 and T,T,T as roll again. because of roll again, we will only have 7 possible outcomes with equal chances.(though you will have a 1/8 chance every roll to roll again)
Now I want to make a D12 composed by a small D6 with red and blue numbers overlapping in each face inside a transparent D6 with half the faces blue and the other helf, red
Sounds like you just need what I like to call the all-in-one dice (other names I have seen for it are the dUltimate and the d-total). You can get the results of a d2, d4, d6, d8, d10, d12, and d20 on a single dice with so many faces that it winds up looking like a ball!
Is there an elegant way to invert the probabilities of the sum of 2 dice? I would call it polar dice, as in this case 2 and 12 would be equally most common and 7 would be least common. So I basically made a virtual 41 sided dice that has six sides with 2 and also for 12, five sides with 3 and also for 11, ... , and one side with 7. I was hoping there would be a more elegant solution. It's just forcing weights on each of the sides.
No, there is no more elegant way to completely invert them. As you found, doing so requires 41 results. 2 d6 have a maximum of 36 combinations. So, missing 5 necessary results shows this is impossible. If your only criteria are that 2 and 12 are the most likely (equally so) and 7 is the least likely, this can be done with 2 d6, but it won't be elegant as it will come down to a lot of arbitrary assignments. In fact, the only way to break down the 36 possible results and maintain symmetry (within the aforementioned constraints) is the following: 2: 5 3: 3 4: 3 5: 3 6: 3 7: 2 8: 3 9: 3 10: 3 11: 3 12: 5 As you can see, this means that every number except 2, 7, and 12 will all be equally likely, which pretty well ruins the intended inversion.
You can make a 3 sided die by basically making a thick coin, where the thickness must be calibrated exactly such that its landing on the side exactly one third of the time. Matt Parker once made a video about that.
Take two slightly offset spheres, e.g. two spheres with radius 10mm and the centers 1mm apart. A die made in the common space would have an earth-like shape and would always land on one of the poles, so it would actually be a 2-sided die. Two such dice would not have 1 as the most frequent difference as 0 and 1 would have an equal change. And a perfect sphere would technically be a 1-sided die, for which the difference will always be 0. They are the only n-sided dice for which 1 wouldn't be the most frequent difference, but they are still exceptions to the statement that 1 is the single most common difference between two equal dice. This of course based on the standard that an n-sided die is numbered from 1 to n, as if you number two 6-sides dice e.g. 1-2-3-4-6-8 each, which would still give the chance of any difference, 1 would not be the most common difference either, it would in fact be 2.
not for the theme of video but important enough I cant fall asleep (even for having 3 hours of sleep in last ~70 hours) and should write this idea I remember you used pre-threeven and post-threeven, but I think that was pretty weird and soulless. I think for two types of throdds we should use words like throddell and throddig, etymologically throdd + small/big. Maybe there is still exist better and cooler variant, but its good too. Those two have some scandinavian vibe. Throddell and Throddig. yeah. Finally that's the advantage to choose the best words we have before that semantic will be widespread represented with that word. Almost like with the symbolic notations in math.
I remember that there was a video with someone trying to make a thick coin as a 3 sided die. The answer in the end contained a square root in theory... But I don't remember if they reached a end result.
god i love this shit. makes me want to snort horse tranquilizers and solve the riemann hypothesis. This could help with my theory on quantum gravity. PS: i shidded my pants
I've played a dice (well knukle bone game) where you roll two dice & if they are both the same even with an even or odd With an odd it is positive meaning you take back & if its an odd & an even it is given forward. you have ten counters & it was used in a battle between a god of Nekromancy (its hard to put them into ephipets) & a man who wanted revenge on his old commanders. He wins even though the god created it & he is humble & doesn't take the gods wives but frees one of his mistress. If he won he'd be taught nekromancy & power to destroy & use souls like the gods of death. If he lost his soul would be the god of nekromancies. Note: it is 20 counters when it goes above 2 as otherwise it goes too fast. Variant includes saving roll when no counters left. but if you don't gain your out.
This feels like taking a class but the professor is a crackhead and the classroom is the dumpster corral behind goodwill and you pay tuition in cigarettes and malt liquor
Have you ever heard of schizophrenic numbers? I've recently learned about them and there's something so oddly alluring about them. If you ever make a video about them I'd definitely love to see it!
13:23 Your 2d6 to 1d12 is wrong. If you could designate dice as "first" and "second", e.g. by saying red/outer dice number and if blue/inner is 4 or greater add 6, then it works, but it's easy to show why you cannot simulate 1d12 with 2d6. You need to assign 3 2d6 results to each 1d12 value, but you cannot distingush results like 6 and 3 from 3 and 6. Anyway, love the energy and I watch your arsonist channel quite a lot.
You don't have to roll them at the same time. In fact, you could use the 2 d6 -> 1 d12 method with just a single d6 (take the value of the first roll then use the second roll to determine +0 or +6, essentially using it as a d2). But even using two identical dice at the same time can work. In separate hands, roll one in one direction and the other in a slightly different direction so you can keep track of which is which. Or roll them together in the same hand and say that whichever dice lands closer to you is the one that will function as the d2. It isn't difficult to come up with ways to make this work.
Did you know Fibonacci is hidden inside the pythagoras theorem? 1,1,2,3,5,8,13,... ==> fibonacci 1 1 3 2 first column: 1x3 = 3 second column: 2 x (2x1) =4 a mix: 1x2+1x3 = 5 3-4-5 ==> pythagoras 3²+4².5² if you keep shifting it will still work. 1 2 5 3 first column: 1x5 = 5 second column: 2x (2x3) = 12 a mix: 1x3 + 2x5 = 13 5-12-13 ==> pythagoras 5²+12²=13² continue 2 3 8 5 first column 2x8=16 second column 2x(3x5)=30 a mix: 2x5+8x3=34 16-30-34 => pythagoras 16²+30²=34² and so on...
Yeah I've seen that pattern before, super cool! It might come up in a future episode about pythagorean triples. Funny enough, the livestream I have scheduled on my bonus channel (for about 3 hours from now) is about using a different method to find pythagorean triples
@@ComboClass Yeah, I saw that. So I thought I'd mention this method. (It generates all of them btw.) Unfortunately, I will be sound asleep a few hours from now. But I'll catch up later.
Thanks for watching! Consider checking out the Patreon I made a couple weeks ago (the names of current supporters are in this video description) www.patreon.com/comboclass
Dude, you're a gem on youtube.
5:50 I already knew that: Matt Parker (standupmaths) has a video about 2d6 to 1d6- this is exactly what he uses to do that transformation- as well as the transformation from 3d6 to 2d6.) In Matt's video, he even uses a 3d6 inside a cube.
6:55 sounds a lot like a clear DND reference: "What would the chart look like if you had an advantage on a saving throw that just uses 1d6?" is essentially what you're saying, but DND typically uses 1d20 for these "saving throws." DND uses all these: d4, d6, d8, d10, d12, d20, d100 (Also called "percentile dice," the d100 is just in the form of 2d10)
technically true, coins are 2 or double-faced and one sided ...😃
Like numberphile but more chaotic
Been in a pretty bad place in life for a while now, and it's the little things like this that keep me going. Thanks for making these! They're silly and weird but education should be more fun in general, and your boundless enthusiasm would surely be enough to overcome the judgement of even the most ardent math-haters. The content and consistent quality is great, and I just wanted to express my appreciation. So thanks again, and to anyone who reads this, I hope you're doing well :)
Thanks for the nice comment, I hope that things go ok for you too :)
hope you're doing well and or that things start easing up in some way if possible.
🫂
Based Anarcho-Communist‽
@@georgemacaroni what do you mean, could you try to explain it if you have the time?
@@uirwi9142 I was asking if they were Anarcho-Communist due to the flag on their PFP, Anarcho- meaning Anarchist, Communist means... well... communist.
All Hail Domotro!!!!
Dude, you're a gem on youtube.
2
10:25 keeping the sound in the video fit's the theme of the channel so well, love it :)
i’ve disliked math for years, but recently things have started making sense and after watching math youtubers like yourself, i’m considering adding a math major to my degree plan. thanks for finally making it interesting and understandable
There are no people who dislike math. There are only people who dislike their bad math teachers.
Or at least that's the understanding I gained from surviving through the education system.
@@btCharlie_ I think that's not really true. I agree there's lot of mediocre or bad math teachers, but it doesn't mean that nobody dislikes it. People do have preferences for certain things (particularly when it comes to just the broad duality of STEM vs non-STEM)
@@MsHojat Fair enough, let me correct it: most people who dislike math actually dislike their bad math teachers.
I'm sure you're aware of this, but in tabletop RPGs, rolling multiple dice and picking the highest result is called rolling with advantage. For 2 n-sided dice, your average result will approach 2n/3 since the equation for the average value is (n+1)(4n-1)/6n^2. If you do the limit as n approaches infinity to, you'll get 1/6 * 1 * 4 = 4/6 = 2/3.
I'm a huge fan of TTRPGs like dungeons and dragons. I actually have probably a few hundred dice from playing!
I'm not very good at the game but it's a lot of fun to play with friends
Love when Math Harlow drops a new educational vid, keep up the good work man
Holy shit bro we're keeping this one he's math harlow now
Math Harlow ahaha
I've always sucked at math, but at the same time found it interesting as hell. Your channel is right up my alley. Thanks for the great content. The only thing stopping your channel from blowing up is lack of exposure. Keep up the good work, you'll be at 100k subs in no time.
Began watching this channel a while ago and subbed pretty quick but never thought to check the subscriber count, because I always assumed it was a multi million subscriber channel; It just caught my eye though. Wdym 20k subscribers?? This is really quality content
so sick to see you getting more popular, hope you keep it up!
go to a doctor.
As a big fan of dice it hurt a bit inside to see so many on the ground.
When it rains and they get stepped on, they are becoming a cool dice carpet
@@ComboClass Only a MANIAC would allow d4s on the ground!! Those things are deadly!!
Hell yeah!!! I was just rewatching your stream on your other channel, so glad I was online to see this
If you treat two dice as a two-digit seximal (base six) number, you can create a number from 0 to decimal 35 if you count 6 as 0. From there, using modulo can get you equal chances for 2, 3, 4, 6, 9, 12, or 18, plus just the raw number for 36
The 1 being the most frequent difference was an interesting fact. Look forward to your future videos on probability, such a fun (and difficult) topic.
My favorite math channel on youtube, great video. I'm always excited for when you drop a new video.
Love your vids!
Highly recommend the game Dicey Dungeons haha. It's not easy, and you'll run into so many tricky sticky probabilistic decisions to make. Seriously, top tier game!
my all time favorite dice rule is from an older DnD book. you roll a number of d6, and sum them up. then you take any dice that rolled a 6 exactly, and re roll those and add that to the sum. you repeat this process until no more 6's were rolled. the math to map n number of dice to s total is actually quite complicated. there was also a relationship where you could roll to hit. if you rolled too low you did zero damage. you increase your chance to hit by reducing the number of d6. if you reduced all but 1 of you dice, you had a way higher chance to hit, but did way less damage. if you took away no d6 then you chances to hit were lower but you did more damage if you did hit. then the question becomes, what is the relationship between how many d6 you roll and your chance to hit. if you roll a d20 to hit and you need to roll a 15, what is the optimal amount of d6 to remove to achieve the highest average damage?
i have solved this using monte carlo, but the covariance solution is exceedingly complicated lol. i still don't have a good model for it that matches the monte carlo results
And what are those results, may I ask?
Could you explain in more detail how the "take away dice" rule works?
After thinking about it for a bit, it doesn't seem too complicated to calculate average sum of any number of d6 with 6 rerolls. For a single dice you get 5/6 * 3 (the average between 1 and 5) + 1/6 * 5/6 * 9 (single reroll) + 1/6 * 1/6 * 5/6 * 15 (two rerolls), which in the end gives you an average of 4.2. "sum 5/6 * 1/6^n * (3 + 6 * n), n=0 to infinity" And because dice are independent, for m dice you get 4.2 * m on average, 8.4 for 2 dice, 12.6 for three, etc.
It is certainly more complicated than the simple charts he shows in the vid, since you have to take weight into account rather than just counting the results (for instance, using 3 d6, even though there are 117 ways to get 20 as a result and only 6 ways to get 5, 5 is the more likely result between the two [2.78% vs 2.08%]) not to mention the fact that the results literally go to infinity.
If you figure out how subtracting dice adds to hit chance, then I'm sure it wouldn't be overly difficult to figure out the mathematically determined optimal approach (you say you need at least 15 on a d20, but how many d6 is that for? and how does that number change as d6 are taken away?). Of course, the pure math can only go so far. When it comes down to actually being used, situation and play style will be the better determiners of whether one should risk missing just to land a hard hit or go for the safer weaker shot.
This is just how I like my math lessons. Unhinged.
my favorite channel on youtube this year, no contest.
Very nice presentation of all the combinations, in a regular grid so you can see the histogram as well as the combinations themselves. Glad to be an early subscriber; extrapolation suggests very early! Keep up the amazing work!
Always struggled with math my whole life butI've been loving the videos. I just might improve!
While it's not strictly rolling dice, there's a tabletop game called Gloomhaven where multiples of numbered cards are drawn (functionally like a roll of the dice) and with some buffs or debuffs, the greater or lesser of the cards are discounted. So it is like rolling two dice and only counting the higher or lower of the two. It's an interesting mechanic.
I really like the fact that you did something different for other math youtuber on the subject of dice, I learned something new today :)
Alright, those dice within dice are stupid cool, might have to get some
Yo! Your conversion of 2D6 into 1D12 or 1D3 etc is literally my PhD. It's a combinatorial structure called Sum Systems, which asks;
For two sets A and B, by taking all sums of each a in A and b in B, can you get
A+B={0,1,2,...,N-1}
for some N in the naturals, and each element occurring exactly once each?
So, if we consider this structure as "modulo N", then you count a result of 0 as N when using these systems in dice (like how 1d100 use 2D10 and a result of 00 and 0 is 100 (almost like the d100 is a Sum System!!! (Spoiler, it is)))
Both sets must have 0, as to make the resultant sum of 0. Otherwise each other element is unique. Your choice of N affects your sets too. The rule is that N = |A| x |B|, i.e. the size of your consecutive integer set is the product of the cardinality of your two "sum system component sets"
Taking each element in a set to be a face on a dice gives you the desired property
But it goes further! Can you have 3 dice? 4 dice? More? What's the minimum sum of these dice? What about if you include subtraction?! (Known as a sum and distance system)
I lloovvee these structures hahaha
Example:
Let A={0,1,2,3,4,5} and B={0,6}. Then A+B={0,1,2,3,4,5,6,7,8,9,10,11} and N=12. Wait! If we let the result of 0+0=0==12, then this result is the exact same you used in the video!! Entertainingly, we can also use A={0, 1, 2, 6, 7, 8} and B={0, 3}, or a personal fav of A={0, 1, 6, 7} and B={0, 2, 4}!
I've reached out through Email.
Difference of two n sided dice is most likely 1. Wow. What a great maths fact and one I will use in class tomorrow, very well explained ! Keep these awesome videos coming !! 😎
"Pardon me, neighbor. Do you have a moment to discuss our completely indifferent overlords, the Dice Gods?"
Your content is a real treat
thanks a bunch, teach
Love this channel!
10:34 what about negative differences?
Great video as always. One thing that I can't help but notice though is the audio becoming weirdly intense during some sounds. Like for just a split second the microphone is resonating along with your voice or something.
It's hard to describe what's going on with the sound, but its like some extra low frequency is getting picked up along with your normal speech emphasis and making it kinda painful.
4:00
"Which is mORe likely than seven itSElf..."
5:20
"Well ONE shows up six times now cause it tooK ALL..."
Is this just me?
This is insane, never thought that about what you can do with dice in that way.
You Domotro are a true gem.
These videos looks like fever dream of my math class
Love this Channel
OFF TOPIC: My divine and burlesque professor, what do you think of gpt 3.5 chat-bot,
about creativity/originality in math research ? Thank you.
I have tried asking that robot math questions on some recent livestreams on the Combo Class Bonus channel here ( @domotro ) and although the robot has potential, it’s currently comically bad at math haha
I propose that the modular reduction from multiple dice actually does serve a purpose: it reduces the possibility of influencing the outcome. As the system is now a more complicated system with multiple inputs that combine to form the output, manipulating it to produce an intended result would require successfully controlling the result of two (or more!) thrown dice. Assuming that those who seek to manipulate the outcome of dice rolls have a sub-100% chance of succeeding at it, each additional die added to the modular reduction method would exponentially reduce that chance of succeeding at rigging the throw.
Also I like the natural progression from modular arithmetic to min/max functions. This all applies so directly to the mathematics objects in the Pure Data visual programming language.
You should be arrested for leaving all the dice on the floor..
Great video as always ;)
Could you pleeease cover the three sided die (the thick coin which has an equal chance to land on either face *or* it's thich edge) that Matt parker went into a bit but never concluded?!? 😁
This is such a pure series, just fun in a backyard. Complicated yet simple things, such a home hearted feeling, like meeting a friend after a long time.
The intro feels special, it feels like a raw intro, no flashy things, just a unique thing captured straight on camera, combo class.
Thank you, I am your first die hard fan. Remember me when ya get to the top.
Remember me as well. I love your channel. I believe you are doing a great job.
Imagine when you hit 1 million subscriptions you find a weird math pattern and make a video about it on how it grew over time and how it fit that weird pattern.
alright for the simplest strategy for having equal probability algorithms with only a coin: say I want to make a 7 sided die with a coin. H,H,H =>1 H,H,T=>2 H,T,H=>3, T,H,H=>4 H,T,T=>5 T,H,T=>6 T,T,H=> 7 and T,T,T as roll again. because of roll again, we will only have 7 possible outcomes with equal chances.(though you will have a 1/8 chance every roll to roll again)
That works, although I was listing ones that two 6-sided dice could duplicate without any “roll again” options
What's the (normal) intro music, used in this video at 11:15
And what's the outro music?
The soundtracks are beats I made myself (I’ll release them someday)
@@ComboClass Oh cool I really like them
Now I want to make a D12 composed by a small D6 with red and blue numbers overlapping in each face inside a transparent D6 with half the faces blue and the other helf, red
Sounds like you just need what I like to call the all-in-one dice (other names I have seen for it are the dUltimate and the d-total). You can get the results of a d2, d4, d6, d8, d10, d12, and d20 on a single dice with so many faces that it winds up looking like a ball!
Good stuff as always!
Is there an elegant way to invert the probabilities of the sum of 2 dice? I would call it polar dice, as in this case 2 and 12 would be equally most common and 7 would be least common. So I basically made a virtual 41 sided dice that has six sides with 2 and also for 12, five sides with 3 and also for 11, ... , and one side with 7. I was hoping there would be a more elegant solution. It's just forcing weights on each of the sides.
No, there is no more elegant way to completely invert them. As you found, doing so requires 41 results. 2 d6 have a maximum of 36 combinations. So, missing 5 necessary results shows this is impossible.
If your only criteria are that 2 and 12 are the most likely (equally so) and 7 is the least likely, this can be done with 2 d6, but it won't be elegant as it will come down to a lot of arbitrary assignments. In fact, the only way to break down the 36 possible results and maintain symmetry (within the aforementioned constraints) is the following:
2: 5
3: 3
4: 3
5: 3
6: 3
7: 2
8: 3
9: 3
10: 3
11: 3
12: 5
As you can see, this means that every number except 2, 7, and 12 will all be equally likely, which pretty well ruins the intended inversion.
You can make a 3 sided die by basically making a thick coin, where the thickness must be calibrated exactly such that its landing on the side exactly one third of the time. Matt Parker once made a video about that.
Sure, but it's easier to just have a triangular prism with rounded ends.
10:30 ambulance coming for me after fainting due to how perplexed i was to hear that the most common difference between two rolled dice is 1🚑
Take two slightly offset spheres, e.g. two spheres with radius 10mm and the centers 1mm apart. A die made in the common space would have an earth-like shape and would always land on one of the poles, so it would actually be a 2-sided die. Two such dice would not have 1 as the most frequent difference as 0 and 1 would have an equal change. And a perfect sphere would technically be a 1-sided die, for which the difference will always be 0. They are the only n-sided dice for which 1 wouldn't be the most frequent difference, but they are still exceptions to the statement that 1 is the single most common difference between two equal dice. This of course based on the standard that an n-sided die is numbered from 1 to n, as if you number two 6-sides dice e.g. 1-2-3-4-6-8 each, which would still give the chance of any difference, 1 would not be the most common difference either, it would in fact be 2.
not for the theme of video but important enough I cant fall asleep (even for having 3 hours of sleep in last ~70 hours) and should write this idea
I remember you used pre-threeven and post-threeven, but I think that was pretty weird and soulless. I think for two types of throdds we should use words like throddell and throddig, etymologically throdd + small/big. Maybe there is still exist better and cooler variant, but its good too. Those two have some scandinavian vibe.
Throddell and Throddig. yeah.
Finally that's the advantage to choose the best words we have before that semantic will be widespread represented with that word. Almost like with the symbolic notations in math.
I remember that there was a video with someone trying to make a thick coin as a 3 sided die. The answer in the end contained a square root in theory... But I don't remember if they reached a end result.
Of course the easy way to do 3-sided is just have a triangular rod with rounded ends, and then mark it like you'd mark a 4-sided die.
This man is wild!
Thanks!
you need more subscribers
I wonder how complex the rules for combining 2d6 would have to be to draw a smily face with the probability table.
Great math channel
Waiting for the new dice innovations)
Non transitive dice are pretty cool.
god i love this shit. makes me want to snort horse tranquilizers and solve the riemann hypothesis. This could help with my theory on quantum gravity.
PS: i shidded my pants
I've played a dice (well knukle bone game) where you roll two dice & if they are both the same even with an even or odd With an odd it is positive meaning you take back & if its an odd & an even it is given forward. you have ten counters & it was used in a battle between a god of Nekromancy (its hard to put them into ephipets) & a man who wanted revenge on his old commanders. He wins even though the god created it & he is humble & doesn't take the gods wives but frees one of his mistress. If he won he'd be taught nekromancy & power to destroy & use souls like the gods of death. If he lost his soul would be the god of nekromancies.
Note: it is 20 counters when it goes above 2 as otherwise it goes too fast. Variant includes saving roll when no counters left. but if you don't gain your out.
This feels like taking a class but the professor is a crackhead and the classroom is the dumpster corral behind goodwill and you pay tuition in cigarettes and malt liquor
interestingly, the average result of the difference of 2 dice is barely under 2 for 6 sided dice, and barely under 4 for 12 sided dice.
So... we're getting an episode on complex numbers on dice... right? Surely that's a yes.
Rolling two (or more) dice is much more satisfying than rolling one.
So that's probably too why games would not use the double die
I love this guy, for some reason i want him to teach me how to read and write in japanese
One word: Factorial. 😊
Dices!
Bros living in an "i spy" picture
Have you ever heard of schizophrenic numbers? I've recently learned about them and there's something so oddly alluring about them. If you ever make a video about them I'd definitely love to see it!
Yes I will definitely cover those sometime!
that was pretty random, but i can roll with it
13:23 Your 2d6 to 1d12 is wrong. If you could designate dice as "first" and "second", e.g. by saying red/outer dice number and if blue/inner is 4 or greater add 6, then it works, but it's easy to show why you cannot simulate 1d12 with 2d6. You need to assign 3 2d6 results to each 1d12 value, but you cannot distingush results like 6 and 3 from 3 and 6. Anyway, love the energy and I watch your arsonist channel quite a lot.
You don't have to roll them at the same time. In fact, you could use the 2 d6 -> 1 d12 method with just a single d6 (take the value of the first roll then use the second roll to determine +0 or +6, essentially using it as a d2). But even using two identical dice at the same time can work. In separate hands, roll one in one direction and the other in a slightly different direction so you can keep track of which is which. Or roll them together in the same hand and say that whichever dice lands closer to you is the one that will function as the d2. It isn't difficult to come up with ways to make this work.
✅
It' all fun and games until we start talking about continuous infinetly-sided dice.
Comment for engagement
Did you know Fibonacci is hidden inside the pythagoras theorem?
1,1,2,3,5,8,13,... ==> fibonacci
1 1
3 2
first column: 1x3 = 3
second column: 2 x (2x1) =4
a mix: 1x2+1x3 = 5
3-4-5 ==> pythagoras 3²+4².5²
if you keep shifting it will still work.
1 2
5 3
first column: 1x5 = 5
second column: 2x (2x3) = 12
a mix: 1x3 + 2x5 = 13
5-12-13 ==> pythagoras 5²+12²=13²
continue
2 3
8 5
first column 2x8=16
second column 2x(3x5)=30
a mix: 2x5+8x3=34
16-30-34 => pythagoras 16²+30²=34²
and so on...
Yeah I've seen that pattern before, super cool! It might come up in a future episode about pythagorean triples. Funny enough, the livestream I have scheduled on my bonus channel (for about 3 hours from now) is about using a different method to find pythagorean triples
@@ComboClass Yeah, I saw that. So I thought I'd mention this method. (It generates all of them btw.) Unfortunately, I will be sound asleep a few hours from now.
But I'll catch up later.
10:10
"If you were to flip four coins" - or just vaguely throw them like an NFL referee...
I think my three-sided dice is broken.
I said 3 1 and 7+or-1
Incoming transitive dice video.
I want to see a video where he is in a bad mood, like explian somethimg that makes him furious
looks like normal distribution, prove it
what probability distribution is sum of gaussians centered on samples from gaussian/normal distribution
assume the sum of gaussians is not "spiky" but kinda "smooth"
no dice, pre-deterministic by God
rats in the garden, dice in the office
Dude, you're a gem on youtube.
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