this video is a slightly modified version of a talk I gave for the Greater Boston chapter of the ACM about a dice problem that blew my mind the problem statement goes like this when you roll two standard six-sided Dice and add them together there's a probability distribution on the sum that you get for example the most probable sum is seven and the probability of getting Snake Eyes is 1 and 36 does there exists another non-standard way to label the faces of the two dice with positive integers that gives the same Distribution on the sum the two dice don't have to be labeled the same way and a die is allowed to have multiple faces with the same label a very sensible way to go about solving this problem would be to First narrow down the search as much as possible by reasoning about it on pen and paper and then if you're comfortable with computer programming write a program that exhaustively searches the smaller space the solution I'm about to show you is not that it's a kind of miraculous pen and paper only solution that a friend showed me some years ago it goes like this start by supposing you have a die with custom labels and consider this polinomial where the coefficient of x to the K is the number of faces that have label k for example for a standard die it would be x + x^2 + x cubed + x 4th plus X 5th plus X 6th where all the coefficients are 1 for a non-standard example here's a die that has a five on two of the faces and nines on the rest of the faces its polinomial is 2x 5th + 4x 9th okay so why are we looking at these polinomial believe it or not the dice problem becomes simpler when we view it this way because it turns out taking the probability distribution on the sum of two dice corresponds to just multiplying the polinomial together by way of a concrete example let's ask how many ways are there to get a seven as the sum of two standard dice choosing outcomes for the two dice that add up to seven is the same thing as choosing a term from the first polinomial and a term from the second polinomial whose exponents add up to seven and the number of ways to do that is the coefficient of x to the 7th in the product similarly we can imagine rolling two of the non-standard dice I showed earlier and this corresponds to multiplying the weird polinomial by itself in this case it's not possible to get a seven as the sum and by the same token the coefficient of x the 7th in the product is zero but it is possible to get a 14 in fact it's very possible there are eight ways to choose one of the fives from the first die and one of the nines from the second die and an additional eight ways to choose one of the nines from the first die and one of the fives from the second die so in total There are 16 ways for the dice to have a combined value of 14 as a last example we can roll one standard die and one weird die again tallying up the number of ways to get each possible total is the same thing as Distributing out the product and collect cting like terms the annotations here show the number of ways to get a combined value of 11 which is the coefficient of x the 11 in the product finding non-standard dice that give the same Distribution on their sum now corresponds to finding two different polinomial that have the same product as the standard one shown here the one- sentence summary of the rest of the talk is that this equivalence we just established combined with the fact that polinomial over the integers have unique factorization means that we no longer have to do a big search on a computer we only have to hand check a few possibilities all we're doing now is Flushing out that argument and throwing in a few more tricks and pictures here's one detail we have to fill in we're trying to replace a question about dice labels with a question about polinomial so we need to show their equivalent we saw how to turn a die into a polinomial but we still need to know how to reverse the process and turn a polinomial into a die on the face of it it's straightforward how to convert a polinomial to a die we just say for each K there are AK faces having label K this does reverse the process when it works but the reality is that not every polinomial can be turned into a die they have to satisfy certain constraints to be valid dice polinomial one of those constraints which may seem pedantic to even mention is that the coefficients AK have to be non- negative integers there can't be a negative coefficient because for example there can't be a negative number of faces that have label three the second constraint is that the labels on the faces have to be positive so Z is not a valid label what this translates to for the polom is that a0 has to be zero the third constraint is that the die has six faces so the sum of all the coefficients has to be six notice that we can rewrite the second and third constraints in terms of the values the polinomial takes at specific settings of X the second constraint says that evaluating the polinomial at x equals 0 gives 0 and the third constraint says that evaluating the polinomial at X equal 1 gives six these versions of the second and third constraints will come in handy later now we have the full picture of the equivalence between dice labels and pols we can always convert a die to a polinomial and we can convert a polinomial to a die if and only if the polinomial meets the three constraints so now we really can replace the search for dice with a search for pols we're looking for two different polinomial say B and C whose product is the same as the one for standard dice shown here but again we don't need to do a big search all we need to do is decompose everything into irreducible factors the polinomial for a standard die factorizes as x * 1 + x * 1 - x + x^ 2 * 1 + x + x squ so substituting that here's the pile of irreducible factors that need to be allotted to B and C in some way I'm going to represent B and C as baskets Each of which can hold sum of the factors and bear in mind that at the end of the day a the product of what we put in each basket has to satisfy the three constraints we established before let's first look at the constraint that each polinomial has to evaluate to zero when x equals 0 only two of the irreducible factors do that namely the two x's so each basket has to get 1 x next we have the constraint that each polinomial evaluates to 6 when x equal 1 if we substitute xal 1 into each irreducible factor we get the values annotated on top here to get a product of each basket has to contain one of the factors annotated with a two and one of the factors annotated with a three that leaves us with just the two factors of 1 - x + x^2 to divvy up if we put one in each basket then the contents of the baskets are the same and we get the standard dice polinomial so let's try putting both factors into the same basket and see what we get after expanding all the products we end up with these pols x + 2x^2 + 2x Cub + x 4th and x + x Cub + x 4th + x 5th plus X 6th plus X 8th these polinomial work we did get lucky here we didn't do anything special to make sure B and C satisfy the constraint of having non- negative coefficients but they do I'm not aware of any deep reason this had to be the case as far as I know that's a coincidence and the fruits of that coincidence are this pair of non-standard dice whose sum has the same distribution as the sum of two standard dice these are called the sierman dice named after George sierman who invented them in 1978 what blows my mind about this pen and paper solution that my friend showed me after the interview is not so much that it finds an alternate set of labels but that it proves that that's the only alternate set of labels and the heavy lifting in the proof is done by unique factorization of polinomial as far as I can tell this solution isn't just dressing up a different solution in fancy algebraic machinery the use of polinomial really is essential and it's very not obvious to me upon looking at the original question about dice that polinomial would be relevant so that's the dice problem whose solution blew my mind thank you