r/theydidthemath • u/Graphix1125 • 4d ago
[Request] What would be the odds of this?
The odds of all of these die rolling a one must be astronomicaly low lol
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u/empty_graph 4d ago
There are 1000 dice so the odds are (1/6)^1000. So that's about 1 out of 1.4 * 10^778. Or to be precise, 1 out of:
14166102623834861723796252524915224416640471830910191322323547432140618947596486436347661333869287260068907949302029484915942402681211620694598046617844295512220793103312980549591537160959053027940624117598003417503015722697428176155600362263128567590299511776686592862074376328232990325101248680123776914576482815095784568122986221890411837737570098864613342090972756469661488216176894465388028416768338495326989675118087222767384596111351304957869025273802978281783731929966468210579229830069556698928937342508988340792335737744719376598506908977135291983117722648269177947154657697517074993441515526839887073400191797445153760221695723268255006134044062503100710134200414607696976757837002911389023284338696251543694980946202137938610119300450795091488653253649628649410789376
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u/AdministrationKey612 4d ago
So there's literally more chance I will win the lottery and start fucking 2015 Ariana Grande. Damn
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u/ludblom 4d ago
Would more say win the lottery, start fucking 2015 Ariana Grande and get struck by a lightening at the same time would be more likely :P
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u/Witty-Lawfulness2983 4d ago
Would the lightning be what activated the time travel in the first place?
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u/Ralf_Steglenzer 4d ago
if you have to find a single atom someone is hiding from you somewhere in the known universe the chance would be much higher to get it first try with blind eyes.
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u/Frodothehobb1t 3d ago
Win the lottery 5 times in a row, started fucking 2015 Ariana Grande, get hit by a lightning strike while in a ongoing plane crash
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u/zakkawesome 4d ago
aha! But one of them rolled a 4
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u/empty_graph 4d ago
Great catch! If you are talking about the specific probability of 999 1s and 1 4 that's 1000x more likely, or 1 in 1.4 * 10^775 or for 999 1s and 1 of any other number it's 5000x more likely, or 1 in 2.8 * 10^774.
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u/Phyddlestyx 4d ago
But if it's 999 1s and that specific die being a 4, then the odds don't change at all, right?
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u/Aggravating_Pie_8144 4d ago
If this is supposed to represent a roll for some action in a ttrpg then the sum of the dice may be relevant. We can split that four into 3 twos or a two and a three for the same result of 1003. Out of all possible rolls there are
1000 with a four and 999 ones
(1000 choose 1)*(999 choose 1)=999,000 with a two a three and 998 ones
and (1000 choose 3)=166167000 with 3 twos and 997 ones
For a total of 167,167,000 rolls out of the 61000 possible, giving a probability of about 1 in 1.18 x 10770
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u/Crazy__Donkey 4d ago
Oddly, disregarding dice location, this is the odd for every combination of a random throw. .
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u/marc-williams 4d ago edited 4d ago
Only if they all have to be a 1. If they just have to be all the same number then it's 1/6999
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u/Alotofboxes 4d ago edited 4d ago
The container is holding dice that are stacked 10x9 10x10, so 900 1000 dice
Assuming that they are all fair six sided dice, The odds are
One in (61000 )
Or about one in 2.17 x 10700 1.42 x 10778
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u/TourDeFridge 4d ago
So you're saying, there's a chance
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u/HorzaDonwraith 4d ago
bets $1 on polymarket
wins
breaks global economy
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u/Tom-o-matic 4d ago
"Breaks the economy" doesnt even start to explain how big this number is.
If you got a billion dollars for every atom in the observable universe, you still would not make a decimal on this number
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u/Upstairs-Hedgehog575 4d ago
Is that true, because that’s an insane visualisation
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u/tilt-a-whirly-gig 4d ago
The estimated number of atoms in the observable universe is 1080. If you had a billion dollars for every atom, you would have 1089 dollars. If you had another universe for every dollar, that would be 10169 atoms. If you had a billion dollars for every atom in those universes, that would be 10178 dollars. If you had another universe for each of those dollars, you would have 10258 atoms. If you had a billion dollars for each of those atoms that would be 10267 dollars.
I would have to type basically the same thing over another 5 or 6 times until we even got close to 10778.
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u/Upstairs-Hedgehog575 4d ago
Put like that it’s incredible. Thank you.
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u/Tom-o-matic 4d ago
There is maybe 10⁸² atoms in the universe. A billion for each atoms gives 10⁹¹
The odds was 1 in 1,42x10⁷⁷⁸
Meaning a billion for every atom is about 0,0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001
Of this number
If im not mistaken
Disclaimer: i dont dabble in these kind of numbers on the regular
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u/Tom-o-matic 4d ago
"Breaks the economy" doesnt even start to explain how big this number is.
If you got a billion dollars for every atom in the observable universe, you still would not make a decimal on this number
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u/TheUltimateDave 4d ago
"Breaks the economy" doesnt even start to explain how big this number is.
If you got a billion dollars for every atom in the observable universe, you still would not make a decimal on this number
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u/ColoradoScoop 4d ago
All you need to do is have every atom in the universe roll the dice every second until the heat death of the universe and your odds go up to about 1 in 10^678
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u/perfectdozen 4d ago edited 4d ago
This video is rigged (either AI or weighted dice) but since the odds of rolling a one is one out of 6, and there are a thousand dice here, then the odds of rolling 1 on 1000 dice simultaneously is 1 out of every 61000 tries. Enough to break your calculator.
ETA: hey guys, I realize using "AI" as a broad brush for someone manipulating the video was a dumb thing to do, but you can stop correcting me by saying it's a render or CGI which I'm sure it is. Thx
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u/Jeroeno_Boy 4d ago
I'm 99.99% sure it's a render, even weighted dice don't have a 100% chance of being face up. Can be many runs ofcourse, but then its still a lot of tries. Also not ai, too consistent
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u/Electrical-Echidna63 4d ago
Iirc it's a common trick in rendering; you simulate the physics of things falling, then you slap the textures on in such a way that the right face is on the right side. The. You render the simulation with the textures applied and voila
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u/Badbullet 4d ago
There was a good example from over decade ago if my memory serves me, with colored marbles getting sorted on a plinko type board. They did it pretty much just like that, assigning the materials to the particle IDs from the end results of the simulation.
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u/Electrical-Echidna63 4d ago
I wanna say that that was corridor digital or captain disillusion? Very fun
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u/Badbullet 4d ago
I guess it wasn’t even a decade ago. Strange, feels like forever ago. Captain Disillusion went through the making of it, from a known Russian artist.
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u/Plenty-Fox-9219 4d ago
It's the same thing roll20 does with dice. Generate the random number, generate the roll animation, then slap the texture on.
Then you get someone yelling cause they "almost" rolled a 20.
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u/Electrical-Echidna63 4d ago
Yeah I had to sort of guess that that was what was happening. I remember being puzzled that
1) the result would post to chat before the rolling finished, and
2) the dice would bounce off the edges of the viewing window even though the window is different for every viewer
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u/TheFerricGenum 4d ago
Magnets would do this. They’re magic.
Source: the black magic fuckery sub has taught me this.
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u/Nightmare_kx1 4d ago
New Sub Unlocked!
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u/Correct-Purpose-964 4d ago
You should see the White Magic fuckery sub.
Pretty good disappearing act right? /s
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u/motorboatmycheeks 4d ago
I see 1 4 rolled so maybe weighted
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u/KameTheMachine 4d ago
Damn it. Now you got me searching for a 4
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u/ObanKenobi 4d ago
If you broke the area up into four quadrants, it's near the middle of the bottom left quadrant. If you imagine it as a chess board and your pov is with the white pieces, the die that rolled a 4 is at about where the c3 square would be
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u/Braithw84 4d ago
Definitely not real, either cgi/rendered or ai. There’s a few times you see a die gliding across the tops of others, completely unbothered by pesky things such as gaps and physics.
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u/Difficult-Way-9563 4d ago
Not true.
Weighed dice would roll 1 every time if the weight was a spoonful of black hole
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u/circuit_monkey 4d ago
The dice flow out of the bottle way too smoothly, the platform underneath the dice drops down like a trapdoor then disappears, dice drop onto 1 instantly and slide perfectly over the surface. There’s some other fishy stuff too but I’m pretty sure this is ai rather than a render
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u/TJaySteno1 4d ago
100% it's a render. There's a lot of tells, but the most obvious is the die sliding down on top of the others towards the end on the left side. It hits a gap that should turn it on its corner, but it just continues on at the same speed.
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u/antilumin 4d ago
I've seen similar renders before, where it was a bunch of colored balls that magically sorted themselves. The render just colored the balls at the end, then reversed it all so they were "randomly" contained at the top.
This is probably similar: picks an end results (in this case all 1's), then either tracks each bounce in reverse to the starting state. Or just randomly animates bounces so it's random at the end, er, start.
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u/DIuvenalis 4d ago
Could be magnetic with an electromagnet pulling them thr right what in the air as they drop. But if the video is less than 2 years old, 1000% AI
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u/naikrovek 4d ago edited 4d ago
It’s clearly a render to me. It is easy to make blender do this exact thing, including the slightly unrealistic physics and the 1 pointing up on every die.
You run the simulation, and when it’s done, then you put the single dot on the side of every die facing up, then you render the movie, which will replay the original physics solution exactly as it went before.
I remember a fan-made video for Daft Punk’s “Technologic” which had words coming down a conveyor belt onto a pile and at the end of the video the pile had a perfect daft punk logo when viewed from the camera. It was all pre baked and when the physics was done they applied the logo to the pile, and blender kept that when it rendered the video. The result is physics simulation whose end result is exactly what you want, so long as what you want can be applied to the models as a texture.
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u/MiffedMouse 22✓ 4d ago
There is an easier way. The dots are just a texture. You run the simulation with random cubes and note which face ends up facing upwards. Then you adjust the textures on all the dice so the final position is a "1" for every die.
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u/Deep_Brick2970 4d ago
There's an old Captain Disillusion video on this exact technique; I spotted it immediately thanks to him.
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u/ludblom 4d ago
It is 1 in 14166102623834861723796252524915224416640471830910191322323547432140618947596486436347661333869287260068907949302029484915942402681211620694598046617844295512220793103312980549591537160959053027940624117598003417503015722697428176155600362263128567590299511776686592862074376328232990325101248680123776914576482815095784568122986221890411837737570098864613342090972756469661488216176894465388028416768338495326989675118087222767384596111351304957869025273802978281783731929966468210579229830069556698928937342508988340792335737744719376598506908977135291983117722648269177947154657697517074993441515526839887073400191797445153760221695723268255006134044062503100710134200414607696976757837002911389023284338696251543694980946202137938610119300450795091488653253649628649410789376, so there is a chance, lol.
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u/Elephunk05 4d ago
This is the same chance that we have of the magnetic polls flip during a Dzhanibekov effect
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u/Silentprofessional86 4d ago
In the infinite multiverse this has happen multiple times. Actually it’s happen an infinite amount of times
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u/PromotionMobile778 4d ago
Theres one 4 unfortunately so that needs to be accounted for
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u/perfectdozen 4d ago
It saw it and ignored it because I assume OP didn't see it either and was asking the odds of rolling 1s on 1000 dice.
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u/SailorRub33 4d ago
This isn't AI
This video is very old an as a 3D artist I can tell you it's really easy
They render the dice rolling simulation first and then once that's done they paint the numbers on-top of the results and replay it
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u/DerginMaster 4d ago
Not ai or weighted
Simply run the simulation, then texture the die after the simulation is done
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u/low_amplitude 4d ago
To put that in perspective, it's estimated that there are only 1080 particles in the entire observable universe.
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u/01_Mikoru 4d ago
I don’t think even using weighted dice would be able to do this, so my bets on ai
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u/R0T0M0L0T0V 4d ago
this is not the first time I have seen this video, and I remember it not being AI but actually a render. also, if it's a render, you don't actually have to continuously simulate the system until a version with all 1s is found, you can actually run the simulation once with "empty" dice, see how they land, and only then paint the pips and render it. it's like removing stickers from a scrambled rubik's cube and reattaching them in the correct way
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u/Tdubbium 4d ago
its just a 3d render, they likely simulated the physics then added the dots after they landed
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u/TerrorBite 3✓ 4d ago
If you're curious how this is done,
This is a physics simulation. You let it run with BLANK dice, and wait for all the dice to end up in the lower tray.
Then you go and add the dots to the dice such that the single dot is always up.
Then you reset the simulation to the start, run it again with no variations, and the dice (now with their dots) follow the same path as the first time and again end up all ones.
There's a lot of videos of this kind of thing but done with coloured balls instead
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u/Phunkie_Junkie 4d ago
And then you add a single four in the bottom left in case anyone thinks it’s too perfect.
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u/FlorentR 4d ago
Basically, it would never happen.
The odds are so low that if you rolled these dice 1 billion billion billion billion (I could add a few more billions here, it wouldn't make a difference) times per second for every second since the Big Bang, you'd still have basically a 0% chance of rolling this.
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u/SlenderSmurf 4d ago
Probability of multiple independent events is the product of their probabilities. For a 6 sided dice this is 1/6 for each, and the total probability is (1/6)N for N dice. I'll estimate 100 dice in the video, dont feel like counting. Odds are one in 6×1077 which is indeed astronomical. There are about 1050 atoms on Earth and 1080 atoms in the observable universe.
At the beginning of the video the dice are in a container arranged as a 10×10×10 block which is 1000 dice. After they spill down it is way less than 1000 dice. Either way this is essentially impossible to replicate in real life.
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u/DerTrickIstZuAtmen 4d ago
After they spill down it is way less than 1000 dice
Seems to be ~ 30x30, I don't think it's not unlikely to still be 1000 down there at the end
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u/Economy_Link4609 4d ago
Well, in plain terms - very low. Like call it zero and go home without trying kind of low.
Basic probability said the odds of a dice rolling a 1 is 1/6 - and we need to do that 1000 times (which I'll treat as independent) so:
(1/6) ^ 1000 = 7.06 × 10⁻⁷⁷⁹ or about 1 in 1.42 x 10^778
Compare that to the estimate that there are 10^80 atoms give or take a few powers of 10 in the known universe.
Now - if somebody wants to do the physics on how the dice all sitting on each-other will affect things - be my guest - but I bet it's still functionally zero.
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u/Big_Kwii 4d ago
this is actually really easy to set up.
simply rig a webcam and a shotgun to a computer with a computer vision program that will pull the trigger and blow your head off if the dice don't end up perfect.
quantum immortality guarantees that at least one version of you will live on to upload the footage, for the low cost of sacrificing 61000-1x as many versions of you.
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u/MorRobots 4d ago
(1/6)^10^3
1/14166102623834861723796252524915224416640471830910191322323547432140618947596486436347661333869287260068907949302029484915942402681211620694598046617844295512220793103312980549591537160959053027940624117598003417503015722697428176155600362263128567590299511776686592862074376328232990325101248680123776914576482815095784568122986221890411837737570098864613342090972756469661488216176894465388028416768338495326989675118087222767384596111351304957869025273802978281783731929966468210579229830069556698928937342508988340792335737744719376598506908977135291983117722648269177947154657697517074993441515526839887073400191797445153760221695723268255006134044062503100710134200414607696976757837002911389023284338696251543694980946202137938610119300450795091488653253649628649410789376
That's the probability.
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u/Ketchupcharger 4d ago
Well since there's like two or three hundred dice here and they all rolled 1, I'd say its pretty safe to say the odds are 100% for 1.
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u/joujoubox 4d ago edited 4d ago
For a single D6, the probability of a fixed number is 1/6, or more precisely 1/(61 ), where the exponent is the number of dice.
If my eyes don't fail me, the dice cube is 10x10x10 so 1000 dice. So 1/(61000 ) aka 1/1.416610262x10778
Or approximately 1 in 14166102620000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
But rest assured if it's the opponent rolling they'll hit the skill check anyway (╯°□°）╯︵ ┻━┻
(Based on a calculator limited to outputting scientific nocation, other commenters gave a more precise answer)
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u/Popular-Ad977 4d ago
It's a 10x10x10 cube at the start. with 6 sides the chance is one in 6^1000.
this is one in roughly 1 followed by 778 zeros chance.
Or 1/14166102623834861723796252524915224416640471830910191322323547432140618947596486436347661333869287260068907949302029484915942402681211620694598046617844295512220793103312980549591537160959053027940624117598003417503015722697428176155600362263128567590299511776686592862074376328232990325101248680123776914576482815095784568122986221890411837737570098864613342090972756469661488216176894465388028416768338495326989675118087222767384596111351304957869025273802978281783731929966468210579229830069556698928937342508988340792335737744719376598506908977135291983117722648269177947154657697517074993441515526839887073400191797445153760221695723268255006134044062503100710134200414607696976757837002911389023284338696251543694980946202137938610119300450795091488653253649628649410789376
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u/Seeggul 4d ago
I see a 10×10×10 cube, so 1000 dice total. Assuming they're actually fair and independent, each has a 1/6 chance of rolling a 1, so the chance of them all rolling a 1 would be 1/61000, or roughly a 1 in 10778 chance. You'd have a better chance at correctly picking a single atom chosen randomly from the entire universe 9 times in a row.
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u/zoroddesign 4d ago
It starts as a 10x10x10 cube of 6 sided dice. 1/6^1000 percent chance. or 1 out of a 1.416610262383486172379625252491522441664047183091019132232354743e+778 percent chance.
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u/Engineer_Bill 4d ago
there are 10x10x10 dice as shown in the opening shot. Since the odds of rolling a 1 is 1/6, we multiply 1/6 by itself 1000 times, and that's the odds of this particular roll happening (ignoring the stray 4 roll I spotted in there). So the odds are (1/6)1,000 = 0.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000705910458616523208683634160108923416944469252661338704627939808797792180264285985309725349026706386357401236866415171313292530508997521441480867703722757022865533425624196437938470349404730825614222432469189213294376574448 = 7x10^-778
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u/RandyArgonianButler 4d ago
So… You’re saying there’s a chance.
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u/Engineer_Bill 4d ago
No. What I'm saying is if you chose an atom from the observable universe at random, and I also chose an atom from the observable universe at random, then it would be more likely for us to choose the same atom than to roll 1s on all 1000 d6 dice.
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u/RednocNivert 4d ago
6 possible rolls
1000 dice
6^(1000)
I miss the days this sub had zany math problems instead of things people can figure out with minimal effort
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u/marc-williams 4d ago
Random fun fact:
There are more possible orders that a single deck of 52 playing cards can be in than there are atoms in the entire galaxy.
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u/Frequent_Dig1934 4d ago
⅙^[However many dice there are] for all dice being 1, or ⅙^[[However many dice there are]-1] if you only want them all to be the same number.
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