Weird Dice: An Application of Unique Factorization I EXAMPLE 12 Consider an ordinary pair of dice whose faces are labeled 1 through 6. The probability of rolling a sum of 2 is 1/36, the probabil- ity of rolling a sum of 3 is 2/36, and so on. In a 1978 issue of Scientific American [1], Martin Gardner remarked that if one were to label the six faces of one cube with the integers 1, 2, 2, 3, 3, 4 and the six faces of (x6 + x + x4 +x³ + x² + x)(xó + x³ + x4 + x³ + x² + x). Observe that we pick up the term x in this product in precisely the fol- lowing ways: dence between pairs of labels whose sums are 6 and pairs of terms whose products are x6. This correspondence is one-to-one, and it is valid for all sums and all dice-including the Sicherman dice and any other dice that yield the desired probabilities. So, let a,, a,, az, a4, ag, ɑç and b¡, b,, b3, b, b,, b, be any two lists of positive integer labels for the faces of a pair of cubes with the property that the probability of rolling any particular sum with these dice (let us call them weird dice) is the same as the prob- ability of rolling that sum with ordinary dice labeled 1 through 6. Using our observation about products of polynomials, this means that • x', x* • x² ,x³. x³, x² · xª, x'.x. Notice the correspon- another cube with the integers 1, 3, 4, 5, 6, 8, then the probability of obtaining any particular sum with these dice (called Sicherman dice) would be the same as the probability of rolling that sum with ordinary dice (that is, 1/36 for a 2, 2/36 for a 3, and so on). See Figure 17.1. In this example, we show how the Sicherman labels can be derived, and that they are the only possible such labels besides 1 through 6. To do so, we utilize the fact that Z[x] has the unique factorization property. (x6 + x + x +x³ + x² + x)(x6 + x + x+ x + x² + x) = (x1 + x42 + x*3 + x4 + x"s + x%s) · (xi + x2 + xbz + xbs + xbs + x). • •• • • . 3 4 7 2 3 3 4 4 5 (1) Now all we have to do is solve this equation for the a's and b’s. Here is where unique factorization in Z[x] comes in. The polynomial x + x + x* + x³ + x2 + x factors uniquely into irreducibles as 3 4 5 6 8 4 7 4 56 7 8 9 5 6 7 8 9 10 789 10 11 9 10 11 12 6 77 8 8 8 9 8 8 9 9 10 5 6 • . 6 7 7 x(x + 1)(x + x + 1)(x² – x + 1) 7 so that the left-hand side of Equation (1) has the irreducible factorization 7 8 9. 10 10 11| 11 12 x(x + 1)2(x² + x + 1)²(x² – x + 1)2. So, by Theorem 17.6, this means that these factors are the only possible irreducible factors of P(x) = xªı + x42 + x*3 + x4 + xs + x6. Thus, P(x) has the form Figure 17.1 To begin, let us ask ourselves how we may obtain a sum of 6, say, with an ordinary pair of dice. Well, there are five possibilities for the two faces: (5, 1), (4, 2), (3, 3), (2, 4), and (1, 5). Next we consider the product of the two polynomials created by using the ordinary dice labels as exponents: x*(x + 1)'(x? + x + 1)'(x² – x + 1)“, where 0 < q, r, t, u<2. To restrict further the possibilities for these four parameters, we evalu- ate P(1) in two ways. P(1) = 1ª1 + 1ª2 + · . . + 1ª6 = 6 and P(1) = 192"3'1“. Clearly, this means that r = 1 and t = 1. What about q? Evaluating P(0) in two ways shows that q + 0. On the other hand, if q = 2, the smallest possible sum one could roll with the corresponding labels for dice would be 3. Since this violates our assumption, we have now reduced our list of possibilities for q, r, t, and u to q = 1, r = 1, t = 1, and u = 0, 1, 2. Let's consider each of these possibilities in turn. When u = 0, P(x) = x + x + r³ + x² + x2 + x, so the die labels are 4, 3, 3, 2, 2, 1–a Sicherman die. When u = 1, P(x) = x + x + x + x + x² + x, so the die labels are 6, 5, 4, 3, 2, 1–an ordinary die. When u = 2, P(x) = x + x + x + x* + x + x, so the die labels are 8, 6, 5, 4, 3, 1-the other Sicherman die. This proves that the Sicherman dice do give the same probabilities as ... • I. • . .....