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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.

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  • $\begingroup$ Must the two dice be identical to each other? (P.S. 'dice', not 'dices', is plural of die). $\endgroup$
    – lemon
    Dec 1, 2014 at 11:44
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    $\begingroup$ You can use the dice ${0,0,0,6,6,6}$ and ${1,2,3,4,5,6}$. I can't really give a good explanation how I got it though, it just popped up in my head. $\endgroup$
    – Arthur
    Dec 1, 2014 at 11:45
  • $\begingroup$ Sure. Each die is in the shape of a regular dodecahedron. One die has sides numberd from $1$ to $12$, the other has $0$ on every side. $\endgroup$
    – bof
    Dec 1, 2014 at 11:46
  • $\begingroup$ Sum $1$ is impossible by definition. $\endgroup$ Dec 1, 2014 at 19:05
  • $\begingroup$ @barakmanos Sorry that was a typo $\endgroup$ Dec 1, 2014 at 19:13

1 Answer 1

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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 $h$.

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  • $\begingroup$ 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) $\endgroup$ Dec 1, 2014 at 19:13
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    $\begingroup$ @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 $\endgroup$ Dec 1, 2014 at 20:33
  • $\begingroup$ @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? $\endgroup$ Apr 2, 2019 at 11:48
  • $\begingroup$ what is z, f(z) and g(z)? $\endgroup$
    – dksahuji
    Mar 24, 2020 at 10:56

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