How you can solve dice puzzles with polynomials 

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

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

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 😏

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.

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...

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.

This is a slick way to look at it.

got an extra "2" in last die

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

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)

Time to delete the channel i guess... can't recover from that

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!

Explain sir

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)

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.

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.

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@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)

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