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Is it possible to build two dice such that the sum of the two dice follows an uniform distribution on $\{2,\dots,12\}$ ?

EDIT : Both dice must have sides 1-6, but the probability that the dice falls on a face can be chosen.

I am looking for a solution that gives the answer and a somewhat detailed reasoning, even though it might seem really easy for some people. I have some difficulties understanding how to solve that kind of problems (what method to follow), any comment towards that issue would be welcomed.

Hippalectryon
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1 Answers1

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Let $a_i, b_i\in[0,1]$ be the probabilities that die $A, B$ shows $i+1$. (According to the edit, $i$ can only range over $\{0,1,2,3,4,5\}$ for both dice). Let $f(z)=\sum a_iz^i$ and $g(z)=\sum b_i z^i$. Then the desired result is that $$f(z)g(z)=\sum_{j=0}^{10}\frac1{11}z^j,$$ a polynomial without real roots. Hence both $f$ and $g$ have no real roots, i.e. they must both have even degree. This implies $a_5=b_5=0$ and the the coefficient of $z^{10}$ in their product is also $0$, contradiction.

Remark: Why does $h(z)=\sum_{j=0}^{10}z^j$ not have real roots? We have $h(z)=\frac{1-z^{11}}{1-z}$ and the numerator has only one real root at $z=1$. But that is no root of $h$.

Hagen von Eitzen
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  • Each dice can have values 1-6, not 0-5 :) (That doesn't affect your reasoning, but I thought you might want to edit it) – Hippalectryon Dec 01 '14 at 19:13
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    @Hippalectryon That's why $a_i$ is the probability that the die shows $i\color{red}{+1}$. This shifting got rid of a factor of $z^2$ and hence justifies the no-real-roots claim – Hagen von Eitzen Dec 01 '14 at 20:33
  • @Hagen von Eitzen In case without shifting ($i \in {0..6})$ you can set $a_0=0, b_0=0$ and the convolution $f(z)g(z) = z^2 \cdot h(z)$, where $0$ is a single root of $f$ and a single root of $g$. The rest of the reasoning follows ($a_6 = b_6 = 0$). Is that correct? – Krzysztof Antoniak Apr 02 '19 at 11:48
  • what is z, f(z) and g(z)? – dksahuji Mar 24 '20 at 10:56