Non-standard fair dice
Task number: 3873
Construct two dice, each has 6 faces and each face has at lest one pip, with the following property: For any \(k\), the probability that \(k\) pips falls together after a roll of these two dice is the same as for a roll of a pair the standard (honest) dice.
These two dice shall not be the standard ones, but the same number of pips may occur on more faces of the same dice.
Hint
First, describe the sequence of probabilities by a suitable generating function. Then split this function into two to match the non-standard dice you are looking for.
Solution
When throwing two honest standard dice, the probability that the sum of \(k\) falls is equal to the coefficient by \(x^k\) in the following polynomial: \[ P(x) = \frac{1}{36}(x + x^2 + x^3 + x^4 + x^5 + x^6)^2. \] This polynomial fould be adjusted as \[\begin{equation*} \begin{split} P(x) &= \frac{1}{36}x^2(1 + x + x^2 + x^3 + x^4 + x^5)(1 + x + x^2 + x^3 + x^4 + x^5) \cr &= \frac1{36}x^2(1+x+x^2)(1 + x^3)(1+x)(1 + x^2 + x^4). \end{split} \end{equation*}\] Observe that in the first factor we have extracted \(1+x+x^2\), while in the second we have extracted \(1+x\) (both are possible). Now we rearrange the terms and get: \[\begin{equation*} \begin{split} P(x) &= \frac1{36}x(1+x+x^2)(1 + x) \cdot x(1+x^3)(1 + x^2 + x^4) \cr &= \frac1{36}(x + 2x^2 + 2x^3 + x^4) \cdot (x + x^3 + x^4 + x^5 + x^6 + x^8). \end{split} \end{equation*}\] From this we see that if we create two fair dice, the first die with 1, 2, 2, 3, 3, and 4 pips and the second die with 1, 3, 4, 5, 6, and 8 pips, then the probability that the roll of these two dice yields \(k\) pips is equal to the coefficient by \(x^k \) in the polynomial \(P(x)\). This yields the required dice.
