Slight error I just want to address here (and small spoiler ->). I should have said that the faces of the dice are required to be positive integers, at least that's what makes this problem have a unique answer. If you allow either die to have 0 on it's face then you get more solutions (for example, one die could be labelled 0,1,2,3,4,5 and the other die would be 2,3,4,5,6,7 and you'd get the exact same probability distribution). If you allow negative numbers on the faces you would have infinitely many solutions.
Finally! Someone talked about Sicherman dice. I swear, every time numberphile does a dice video, I'm always expecting them to talk about Sicherman dice, and they never did.
When I thought of doing this video my first thought was 'ah man but I bet numberphile has already covered this in pretty much the exact same way'. Then I looked through all his dice videos and just kept expecting it to show up and it never did, I was really surprised but happy haha.
I assume you mean the only pair of dice where only numbers >0 are allowed, because you could decrease all sides of one dice by 1 and increase the other dices sides
Came here to say this, as there’s actually infinite possible dice if the faces can be any integers. Subtracting k from all sides of dice 1 and adding k to all sides of dice 2 will give a valid solution for all values of k.
those 2 points we add to get from a 0-10 repartition to the normal 2-12 repartition can be found in factor of the massive polinomial as the x^2 at its beginning this is funny because you can choose to put those 2 x either is the f or g function and get valid results you can as well put negative powers in one of the function and keep getting valid results i think it should be said in the problem that you want >0 faces so you dont bother and put one x in each function yet the +1 -1 solution is probably the most clever one 😏
To avoid the technical bits in the video, you can also do the following. Assume that 1,2,3,4,5,6 are the sums of two other dice: a two-sided die, and a three sided die. We can get these sums with equal probability with the following two pairs of dice: - 1,2 and 0,2,4. - 1,4 and 0,1,2. Hence, rolling all four dice together yields the right distribution of sums. So all we need to do is pair them differently: -1,2 + 0,1,2 yields 1,2,2,3,3,4 -1,4 + 0,2,4 yields 1,2,3,4,5,6,8 Just like the solution in the video.
Your bad at maths, you have no natural skill. How can you claim to be a dr of maths but not have a clue about this simple generating function. Clearly you never participated in olympiads...
thanks Zach for this 'outside the box' idia. I realy enjoy your vidios. In total, I found 5 solutions for this problem: 2 with 1 or more dots in each side (as shown): (1,2,3,4,5,6) , (1,2,3,4,5,6) (1,2,2,3,3,4) , (1,3,4,5,6,8) 3 with 0 or more dots in each side: (0,1,2,3,4,5) , (2,3,4,5,6,7) (0,1,1,2,2,3) , (2,4,5,6,7,9) (0,2,3,4,5,7) , (2,3,3,4,4,5)
We studied generating functions on one of my introductory courses to Computer Science at college and they were the first thing that came to mind to solve the problem.
4:58 here's how you can do such factorization: Let P(x) = x1+x2+x3+x4+x5+x6 P(x) = x(x0+x1+x2+x3+x4+x5) Note that (x-1)P(x) = x(x6 -1) Now x6 -1 is easier to factor as a difference of cubes x6 -1 = (x3+1)(x3 -1) which can be factored as a sum of cubes and a difference of cubes x6 -1 = (x+1)(x2 -x+1)(x-1)(x2+x+1) Multiply by x to get (x-1)P(x) x(x6 -1) = x(x+1)(x2 -x+1)(x-1)(x2+x+1) Now we just divide by (x-1) to get P P(x)=x(x+1)(x2 -x+1)(x2+x+1) Square this and you get f(x) f(x)=x^2*(x+1)^2*(x2 -x+1)^2*(x2+x+1)^2
zach do you think you could check out how programming languages are essentially just fancy maths? such as a for loop is almost / exactly the same as sigma in math depending on how you use it. and how if statements can be represented by piecewise brackets i think it would be amazing if you could cover that topic since all everything computers are, are just math machines
A loop is not the same as sigma. That would entirely depend on what the loop is doing. You could easily have one be Pi. For example: o=1 for i is 1 through n o=o*n Is literally a product, to calculate n!, not a sum as sigma would suggest. I would say it is more a really complex piecewise function.
@@jeffreyblack666 yes i know... thats why its more of a POC since programming isnt exactly the same as math. although when you break down computers down to the fundamentals they are entirely math. think of the ALU, booths multiplier, division, etc etc
This is a very simplified version of the modularity theorem where coefficients from the elliptical curve of a Galois group representation correspond to coefficients of modular forms. The Langlands program hopes to generalize this. In this case it is the 1-dimensional case with class field theory noting Kronecker-Weber’s theorem on abelian extensions, cyclomatic fields, and their cyclomatic polynomials. The generating function is derived from cyclic group. The cyclic group of the roll of dices determines the Cayley table and the generating function is the cyclomatic polynomial.
I juste wish to say that I love the fact that you don’t have background music in your videos, which usually distract me while being most of the time not that pleasing to listen to.
I already knew this technique but instead of number of ways to achieve a result, the coefficients were probabilities to achieve this result. There's btw a nice way to prove that 2 dice can't be cheated so that each outcome (from 2 to 12) has an equal probability to occur (1/11 in this case). It uses factorization of a certain polynomial and is really pretty even though it doesn't provide much insight into the concrete problem.
Yeah -- this clearly works -- he made a clarification in the comments that the faces should be positive integers, so 0 would be excluded as a possible face. Using negative numbers, you could take this further by just repeating your step -- e.g. (-1, 0, ..., 4) and (3, 4, ..., 8)
Well, since you said 'integers' and not 'positive integers', I'm going to assume 0 and negative count! So, you could just subtract 1 from all faces of one dice and add 1 to all faces of other. Each time you do that, you get a new pair of dice that give out the required solution!
How exactly was that pair of polynomials found to be each with a sum of coefficients of 6? Was it just brute force or was there other ideas at play? I’m still in secondary school so I’m not familiar with polynomial factorisation just yet enough to know how it works really well.
Consider a generating polynomial for a dice f(x). The value of f(1) is easily seen to be the number of faces of the dice (because you're summing the coefficients), so you can easily check which factorizations are good without doing the full calculation every time.
2:16 "There is only one way to change the dice in such a way that the probability of getting each number is exactly the same" Me: (0, 1, 2, 3, 4, 5) and (2, 3, 4, 5, 6, 7). Checkmate.
Please make a video on these topics -Architecture -Engineering Physics -Petroleum Engineering -Systems Engineering -Marine Engineering -Nuclear Engineering
Generating functions are awesome. You can find the number of ways to make change for a dollar by finding the x^100 coefficient in 1/(1 - x^1 - x^5 - x^10 - x^25), and explain why 1/9899 = 0.0001010203050813213455 gives the Fibonacci numbers. Also you can prove that the number of ways to write 1,000,000 as a sum of odd positive integers equals the number of ways to write 1,000,000 as a sum of distinct positive integers.
@@legendgames128 I would use the new dice since you can pick the double you want to go for. Doubles 3s has an advantage for the player with the new dice. With normal dice you shouldn't play those bets since they have some of the worst odds on the craps table
You say that its the only pair of dice with the same values but you can subtract any number from each face of 1 die and add the same number to the other die and have the exact same probability distribution, (a 0-5 die and a 2-7 die for example) so technically there are infiinitely many ways to relable dice without changing craps
2:17 I don’t believe you, assuming you stated the problem correctly; I’ve thought of infinite. If you limit yourself to *positive* integers I haven’t thought of any but if you limit yourself to *non-negative* integers I’ve thought of one (or technically two, by switching the dice).
Ah, I see you have a pinned comment about this. That was the solution I’d found when allowing 0, and the general form was (1, 2, 3, 4, 5, 6) ± (n, n, n, n, n, n) (or, in generating function form, multiply one die’s polynomial by x^n and multiply the other’s by x^-n)
My first instinct was add 1 to each face of one die and subtract 1 from the other. Assuming they're positive integers though, there are indeed only two solutions
We could abstract all the polynomial "language" to simply talk about vectors and convolutions, right? With the advantage that convolution makes it clear that we're" sliding" the vectors and with that, taking different combinations of it
Hello sir, please can you make a video on to write a lab report? I have just started with my electrical engineering course and it is giving me a tough time.
I feel like you're spying on my life or something lol Here I am, a fresh mechanical engineering graduate and since I couldn't find a job as an engineer I took a job as a dealer at a casino And what game do I deal the most? That's right... blackjack oh yeah there's a lot of craps involved too
It's not hard to do this yourself, the only extra thing you need to know is that if you plug in 1 in the polynomial, you will get the number of labels on the die, and that the polynomial only represents a die if there are no negative terms in the polynomial. If you want the answer, here it is: you can only have dice of the form (1, 2, 2, 3, 3, 4), (1, 2, 3, 4, 5, 6), and (1, 3, 4, 5, 6, 8), with the same number of (1, 2, 2, 3, 3, 4) and (1, 3, 4, 5, 6, 8) dice.
So while crisis come's up .Bitcoin climbs lol 😆 Now i say crypto is better than stock investing in crypto should be in every wise individual to do list .
The Rice become richer by spending like the poor and investing Non stop,While the poor stay poor by spending like the rich yet not making any investment.
@Axeuweuecar There’s a glitch where an unlisted video that becomes public can have their upload date changed (under certain circumstances that I don’t really know)
Can you break down and analyze the Dream speedrun case. It is a Minecraft speedrunning case where the speedrunning officials used stats to try to decide whether he cheated or not on his speedrun, and they said he did, but I would love to hear the opinion and take on the info from a stats expert like you. Thanks.
See...you do have to watch the video. He defines (or clarifies) the question in the video, and your solution does not work. For example, the probability of rolling an 8 with two standard six-sided dice is 5/36. But if each dice has only one number, then you will either have a 0/36 chance or a 36/36 chance of getting an 8, which is not the same as rolling two standard dice.
@@hebernelski3898 Neigh sorry but the Question is stated "Can you change the numbers on a pair of dice and have the same probablility of getting any given sum." So since Number is not defined here I assume doubles. So lets call the sum d. Since change the *numbers* (plural) on a *pair* (both) of dice means: I can change any number of both dice on every of the six sides, I do so. Set each Number to be d/2. Now the probability to roll on each dice the number d/2 is 100% Which means that the sum will always be 2*(d/2). Now we have to proove that A: 2*(d/2) == d. Which I now will demonstrate via Induction. Observe: d=1: 2*(1/2) = 1 == d now with d+1 (induction step) 2*((d+1)/2) == d+1 2*(d+1)/2 == d+1 2*(d+1)*0.5 == d+1 2*(d/2)+2*(1/2) = d+1 with A: d+2*(1/2) == d+1 d+1 == d+1 qed.
@@Sonnentau1 The issue is the phrase "any given sum". What do we consider to be a "given sum". Arguably, a given sum is a sum that we are given. We are given 11 sums, 2 through 11. Now when we say "the same probability", do we mean the same probability for each sum? This is unlikely -- since we have 36 possible rolls and only 11 sums -- perhaps if have an extra sum included -- then each of our given sums could theoretically be 3/36 (say if we could make 1-12 possible outcomes). More likely what is meant is that each sum will have the same probability after we change the numbers on the dice as they did before we changed the numbers on the dice. Your solution does not preserve any original probabilities, nor can it make each of the original given sums. Your solution solves a different question: "Can you change numbers on two dice such that each new possible sum has the same probability as each other?" Note that we need to refer to "new sums", since your solution does not preserve the "given sums". Also, note that these probabilities will have nothing to do with the original probabilities. I don't understand why you have included a proof by induction, or a proof at all for that matter. Induction is excessive at best in this situation, and does little more than obfuscate any relevant point you might have. Proving 2*(d/2) = d (which is trivial) does not resolve the fact that your solution neither preserves sums, nor probabilities. Your solution and argument depend on not understanding the question that is being posed. I agree that different interpretations of the original questions posed in the title are possible -- thus the reason you needed to have watched the video before giving your solution. Your understanding of the original question was not the intent of the original question.
Slight error I just want to address here (and small spoiler ->). I should have said that the faces of the dice are required to be positive integers, at least that's what makes this problem have a unique answer. If you allow either die to have 0 on it's face then you get more solutions (for example, one die could be labelled 0,1,2,3,4,5 and the other die would be 2,3,4,5,6,7 and you'd get the exact same probability distribution). If you allow negative numbers on the faces you would have infinitely many solutions.
Yes
what?! 6 hours ago?
@@aashsyed1277 videos can be privately uploaded before they’re published
@@briangeer1024 ok but what about a viewer who can't see them?
Started watching, immediately thought about that
Zach has forgotten his engineering roots, he's become a pure mathematician ;(
That's okay. It's an important step in everyone's progress.
Pure mathematician>Applied mathematician>Physicist>Engineer. He levelled up.
;(
@@MrAlRats physicist > applied mathematician. better a fool than a traitor
@@MrAlRats Physicist > Chemist > Mathematician > engineer > Doctor > Psychologist > Sociologist > Economist > Lawyer > Drug Dealer > Politician > sweeper > Homeless > Beggar
Never knew polynomials could be used like that… blew my mind
Finally! Someone talked about Sicherman dice. I swear, every time numberphile does a dice video, I'm always expecting them to talk about Sicherman dice, and they never did.
When I thought of doing this video my first thought was 'ah man but I bet numberphile has already covered this in pretty much the exact same way'. Then I looked through all his dice videos and just kept expecting it to show up and it never did, I was really surprised but happy haha.
I assume you mean the only pair of dice where only numbers >0 are allowed, because you could decrease all sides of one dice by 1 and increase the other dices sides
Came here to say this, as there’s actually infinite possible dice if the faces can be any integers.
Subtracting k from all sides of dice 1 and adding k to all sides of dice 2 will give a valid solution for all values of k.
those 2 points we add to get from a 0-10 repartition to the normal 2-12 repartition can be found in factor of the massive polinomial as the x^2 at its beginning
this is funny because you can choose to put those 2 x either is the f or g function and get valid results you can as well put negative powers in one of the function and keep getting valid results
i think it should be said in the problem that you want >0 faces so you dont bother and put one x in each function
yet the +1 -1 solution is probably the most clever one 😏
To avoid the technical bits in the video, you can also do the following. Assume that 1,2,3,4,5,6 are the sums of two other dice: a two-sided die, and a three sided die. We can get these sums with equal probability with the following two pairs of dice:
- 1,2 and 0,2,4.
- 1,4 and 0,1,2.
Hence, rolling all four dice together yields the right distribution of sums. So all we need to do is pair them differently:
-1,2 + 0,1,2 yields 1,2,2,3,3,4
-1,4 + 0,2,4 yields 1,2,3,4,5,6,8
Just like the solution in the video.
This is a slick way to look at it.
got an extra "2" in last die
Cool puzzle! Maybe my algebra is weak lol but I didn’t anticipate the unique factorization to jump into this
hi!
my brain left the chat
Your bad at maths, you have no natural skill. How can you claim to be a dr of maths but not have a clue about this simple generating function. Clearly you never participated in olympiads...
Love your videos Zack. Very interesting and engaging. Keep up the great work!
I learned all the tools to answer the question and still fascinated by the elegance of it. Great riddle!
I just watched a veritasium video on clickbait and this is the most accurate title I've ever seen
hey i watch veritasium 2 haha
Same.
Same
I saw it and thought it meant each sum having the same probability as each other. i.e. you are just as likely to get a 2 as you are to get a 7.
Always hyped for these vids!
thanks Zach for this 'outside the box' idia.
I realy enjoy your vidios.
In total, I found 5 solutions for this problem:
2 with 1 or more dots in each side (as shown):
(1,2,3,4,5,6) , (1,2,3,4,5,6)
(1,2,2,3,3,4) , (1,3,4,5,6,8)
3 with 0 or more dots in each side:
(0,1,2,3,4,5) , (2,3,4,5,6,7)
(0,1,1,2,2,3) , (2,4,5,6,7,9)
(0,2,3,4,5,7) , (2,3,3,4,4,5)
We studied generating functions on one of my introductory courses to Computer Science at college and they were the first thing that came to mind to solve the problem.
4:58 here's how you can do such factorization:
Let P(x) = x1+x2+x3+x4+x5+x6
P(x) = x(x0+x1+x2+x3+x4+x5)
Note that
(x-1)P(x) = x(x6 -1)
Now x6 -1 is easier to factor as a difference of cubes
x6 -1 = (x3+1)(x3 -1) which can be factored as a sum of cubes and a difference of cubes
x6 -1 = (x+1)(x2 -x+1)(x-1)(x2+x+1)
Multiply by x to get (x-1)P(x)
x(x6 -1) = x(x+1)(x2 -x+1)(x-1)(x2+x+1)
Now we just divide by (x-1) to get P
P(x)=x(x+1)(x2 -x+1)(x2+x+1)
Square this and you get f(x)
f(x)=x^2*(x+1)^2*(x2 -x+1)^2*(x2+x+1)^2
Dude you are so charismatic and likeable. Glad i found your channel.
zach do you think you could check out how programming languages are essentially just fancy maths?
such as a for loop is almost / exactly the same as sigma in math depending on how you use it.
and how if statements can be represented by piecewise brackets
i think it would be amazing if you could cover that topic since all everything computers are, are just math machines
A loop is not the same as sigma.
That would entirely depend on what the loop is doing.
You could easily have one be Pi.
For example:
o=1
for i is 1 through n o=o*n
Is literally a product, to calculate n!, not a sum as sigma would suggest.
I would say it is more a really complex piecewise function.
@@jeffreyblack666 yes i know... thats why its more of a POC since programming isnt exactly the same as math. although when you break down computers down to the fundamentals they are entirely math. think of the ALU, booths multiplier, division, etc etc
i love your videos! these are awesome! always hyped! :)
This is a very simplified version of the modularity theorem where coefficients from the elliptical curve of a Galois group representation correspond to coefficients of modular forms. The Langlands program hopes to generalize this.
In this case it is the 1-dimensional case with class field theory noting Kronecker-Weber’s theorem on abelian extensions, cyclomatic fields, and their cyclomatic polynomials. The generating function is derived from cyclic group. The cyclic group of the roll of dices determines the Cayley table and the generating function is the cyclomatic polynomial.
I have missed you So very much Zach♥️
really really awsome videos zach! thank you so much! subbed again :) please keep it up
I juste wish to say that I love the fact that you don’t have background music in your videos, which usually distract me while being most of the time not that pleasing to listen to.
Interesting generating functions. I didn't think such simple polynomials can describe probabilities in such a way
"Dice dice baby"
~ Vanilla Dice
*An excellent explanation*
That was such an interesting proof!
I was watching the last Veritasium's video and it's funny, it talks more to you than to me.
A die is usually labeled such that any two parallel faces add up to seven. I see some of the CGI dice don't adhere to this standard.
Time to delete the channel i guess... can't recover from that
I learned so much and nothing at the same time ❤️
I already knew this technique but instead of number of ways to achieve a result, the coefficients were probabilities to achieve this result.
There's btw a nice way to prove that 2 dice can't be cheated so that each outcome (from 2 to 12) has an equal probability to occur (1/11 in this case). It uses factorization of a certain polynomial and is really pretty even though it doesn't provide much insight into the concrete problem.
Great video. I recently stumbled across an idea called Cyclotomic Polynomials. Your video reminded me of it.
Just add 1 (or any number) to one dice and subtract 1 to the other. Dice A: 0, 1,..., 5 and dice B: 2, 3,..., 7. 🤔
Yeah -- this clearly works -- he made a clarification in the comments that the faces should be positive integers, so 0 would be excluded as a possible face. Using negative numbers, you could take this further by just repeating your step -- e.g. (-1, 0, ..., 4) and (3, 4, ..., 8)
Brilliant sponsorship strikes back!
Sir u can say about aerospace engineering. And what skill need to improve... Pls explain sir
Explain sir
0,1,2,3,4,5 and 2,3,4,5,6,7 also works. in the examples you allowed zeros.
this is VERY VERY COOL
This is amazing, never thought of this
Well, since you said 'integers' and not 'positive integers', I'm going to assume 0 and negative count! So, you could just subtract 1 from all faces of one dice and add 1 to all faces of other. Each time you do that, you get a new pair of dice that give out the required solution!
Can you please show us how did you turn the equation from that 5:03 to that 5:06
How exactly was that pair of polynomials found to be each with a sum of coefficients of 6? Was it just brute force or was there other ideas at play? I’m still in secondary school so I’m not familiar with polynomial factorisation just yet enough to know how it works really well.
Consider a generating polynomial for a dice f(x). The value of f(1) is easily seen to be the number of faces of the dice (because you're summing the coefficients), so you can easily check which factorizations are good without doing the full calculation every time.
Treat all the x as 1, and you get the sum of coefficient.
Got it. Thanks!
2:16 "There is only one way to change the dice in such a way that the probability of getting each number is exactly the same"
Me: (0, 1, 2, 3, 4, 5) and (2, 3, 4, 5, 6, 7). Checkmate.
wow! when i read the title first it thought we were looking for a pair of dice which would have the same probability of getting a given sum !!!
generating functions!
Cool stuff
What video editing software do you use to make your videos?
Please make a video on these topics
-Architecture
-Engineering Physics
-Petroleum Engineering
-Systems Engineering
-Marine Engineering
-Nuclear Engineering
Generating functions are awesome. You can find the number of ways to make change for a dollar by finding the x^100 coefficient in 1/(1 - x^1 - x^5 - x^10 - x^25), and explain why 1/9899 = 0.0001010203050813213455 gives the Fibonacci numbers. Also you can prove that the number of ways to write 1,000,000 as a sum of odd positive integers equals the number of ways to write 1,000,000 as a sum of distinct positive integers.
Which set of dice would give you the advantage in terms of those bets we didn't care about? The original or the new set?
You can no longer win double 2,5,6
Double 1,4 stays the same
Double 3 is better with the new dice
@@iamanoob9747 That's a lot of disadvantages to advantages for the new dice... maybe the original is better.
@@legendgames128 I would use the new dice since you can pick the double you want to go for. Doubles 3s has an advantage for the player with the new dice. With normal dice you shouldn't play those bets since they have some of the worst odds on the craps table
@@iamanoob9747 Oh, did I misunderstand your first comment? That double 3s may be a game changer for the better for the player with the new dice.
You say that its the only pair of dice with the same values but you can subtract any number from each face of 1 die and add the same number to the other die and have the exact same probability distribution, (a 0-5 die and a 2-7 die for example) so technically there are infiinitely many ways to relable dice without changing craps
2:17 I don’t believe you, assuming you stated the problem correctly; I’ve thought of infinite. If you limit yourself to *positive* integers I haven’t thought of any but if you limit yourself to *non-negative* integers I’ve thought of one (or technically two, by switching the dice).
Ah, I see you have a pinned comment about this. That was the solution I’d found when allowing 0, and the general form was (1, 2, 3, 4, 5, 6) ± (n, n, n, n, n, n) (or, in generating function form, multiply one die’s polynomial by x^n and multiply the other’s by x^-n)
Funniest skit yet !!!
My first instinct was add 1 to each face of one die and subtract 1 from the other.
Assuming they're positive integers though, there are indeed only two solutions
We could abstract all the polynomial "language" to simply talk about vectors and convolutions, right? With the advantage that convolution makes it clear that we're" sliding" the vectors and with that, taking different combinations of it
5:17, small error. I think you meant to say factors, not terms. Great video!
Hey, do you use the "manim" python module for some of the math graphics? it resembles a bit 3blue1brown's graphics
Cyclotomic polynomials!
can you give any tips for a first year mechanical engineer?
Hello sir, please can you make a video on to write a lab report? I have just started with my electrical engineering course and it is giving me a tough time.
....Did ANYONE check the expansion? There is a sign error, and it is in the book
Primes also are irreducible polynomials!
I feel like you're spying on my life or something lol
Here I am, a fresh mechanical engineering graduate and since I couldn't find a job as an engineer I took a job as a dealer at a casino
And what game do I deal the most? That's right... blackjack oh yeah there's a lot of craps involved too
Can you make a video on Engineering Technology
Change the 6 to 1, 5 to 2, 4 to 3, and vice versa. Problem, mathematicians?
what's freaky is that I can actually understand this
me at 2:20: let's see how long he's gonna take to say the answer...
This guy looks really similar to the dude that does those funny D&D skits
What about a game where you have 3, 4 or n dice?
It's not hard to do this yourself, the only extra thing you need to know is that if you plug in 1 in the polynomial, you will get the number of labels on the die, and that the polynomial only represents a die if there are no negative terms in the polynomial.
If you want the answer, here it is: you can only have dice of the form (1, 2, 2, 3, 3, 4), (1, 2, 3, 4, 5, 6), and (1, 3, 4, 5, 6, 8), with the same number of (1, 2, 2, 3, 3, 4) and (1, 3, 4, 5, 6, 8) dice.
I'd love to play monopoly with those dice
So while crisis come's up .Bitcoin climbs lol 😆 Now i say crypto is better than stock investing in crypto should be in every wise individual to do list .
Investments are the stepping stones to success, investing is what creates wealth.
Recently, I invested in both stock and
crypto but currently i believe crypto is doing more better !!
The best thing to do in this period is to buy silver and crypto
The Rice become richer by spending like the poor and investing Non stop,While the poor stay poor by spending like the rich yet not making any investment.
@@@@@@ Most intelligent words I've heard today
How calculation P And B in Baccarat score.
Zach can u do use of triple integration in real life
Who needs two dice? One 36-sided die will do nicely and no polynomials needed.
Wow
wait, can't you just add 1 to each number on one of the die, and subtract one from each number on the other die
Well i got it in rint theory
So, that would work for Catan game
Can you shorten the title?
Can't you just number one die 0-5 and the other 2-7?
First!
@Axeuweuecar There’s a glitch where an unlisted video that becomes public can have their upload date changed (under certain circumstances that I don’t really know)
Can you break down and analyze the Dream speedrun case. It is a Minecraft speedrunning case where the speedrunning officials used stats to try to decide whether he cheated or not on his speedrun, and they said he did, but I would love to hear the opinion and take on the info from a stats expert like you. Thanks.
That's a very long title.
GF's FTW!!!!!!!
3+1=3
Zach could you shorten the title please?
yes easy, dont even have to watch the video: Just make all numbers the same. You can choose any number.
See...you do have to watch the video. He defines (or clarifies) the question in the video, and your solution does not work. For example, the probability of rolling an 8 with two standard six-sided dice is 5/36. But if each dice has only one number, then you will either have a 0/36 chance or a 36/36 chance of getting an 8, which is not the same as rolling two standard dice.
@@hebernelski3898 Neigh sorry but the Question is stated "Can you change the numbers on a pair of dice and have the same probablility of getting any given sum."
So since Number is not defined here I assume doubles. So lets call the sum d.
Since change the *numbers* (plural) on a *pair* (both) of dice means: I can change any number of both dice on every of the six sides, I do so.
Set each Number to be d/2.
Now the probability to roll on each dice the number d/2 is 100%
Which means that the sum will always be 2*(d/2).
Now we have to proove that A: 2*(d/2) == d.
Which I now will demonstrate via Induction.
Observe:
d=1:
2*(1/2) = 1 == d
now with d+1 (induction step)
2*((d+1)/2) == d+1
2*(d+1)/2 == d+1
2*(d+1)*0.5 == d+1
2*(d/2)+2*(1/2) = d+1
with A:
d+2*(1/2) == d+1
d+1 == d+1
qed.
@@Sonnentau1 The issue is the phrase "any given sum". What do we consider to be a "given sum". Arguably, a given sum is a sum that we are given. We are given 11 sums, 2 through 11. Now when we say "the same probability", do we mean the same probability for each sum? This is unlikely -- since we have 36 possible rolls and only 11 sums -- perhaps if have an extra sum included -- then each of our given sums could theoretically be 3/36 (say if we could make 1-12 possible outcomes). More likely what is meant is that each sum will have the same probability after we change the numbers on the dice as they did before we changed the numbers on the dice.
Your solution does not preserve any original probabilities, nor can it make each of the original given sums. Your solution solves a different question: "Can you change numbers on two dice such that each new possible sum has the same probability as each other?" Note that we need to refer to "new sums", since your solution does not preserve the "given sums". Also, note that these probabilities will have nothing to do with the original probabilities.
I don't understand why you have included a proof by induction, or a proof at all for that matter. Induction is excessive at best in this situation, and does little more than obfuscate any relevant point you might have. Proving 2*(d/2) = d (which is trivial) does not resolve the fact that your solution neither preserves sums, nor probabilities.
Your solution and argument depend on not understanding the question that is being posed. I agree that different interpretations of the original questions posed in the title are possible -- thus the reason you needed to have watched the video before giving your solution. Your understanding of the original question was not the intent of the original question.
@@hebernelski3898 That... was a joke. You might have guessed it. As you said: _"Induction is excessive [...] Proving 2*(d/2) = d (which is trivial) "_
@@Sonnentau1 I assumed you were joking and being pedantic solely for amusement -- I was just playing into it for fun too. ;p
I'm guessing yes cus. 12 sides
XD bro you gotta work on your video titles
I hate math