The question is the following: Can one create two nonidentical loaded 6-sided dice such that when one throws with both dice and sums their values the probability of any sum (from 2 to 12) is the same. I said nonidentical because its easy to verify that with identical loaded dice its not possible.

Formally: Let's say that $q_{i}$ is the probability that we throw $i$ on the first die and $p_{i}$ is the same for the second die. $p_{i},q_{i} \in [0,1]$ for all $i \in 1\ldots 6$. The question is that with these constraints are there $q_{i}$s and $p_{i}$s that satisfy the following equations:
$ q_{1} \cdot p_{1} = \frac{1}{11}$
$ q_{1} \cdot p_{2} + q_{2} \cdot p_{1} = \frac{1}{11}$
$ q_{1} \cdot p_{3} + q_{2} \cdot p_{2} + q_{3} \cdot p_{1} = \frac{1}{11}$
$ q_{1} \cdot p_{4} + q_{2} \cdot p_{3} + q_{3} \cdot p_{2} + q_{4} \cdot p_{1} = \frac{1}{11}$
$ q_{1} \cdot p_{5} + q_{2} \cdot p_{4} + q_{3} \cdot p_{3} + q_{4} \cdot p_{2} + q_{5} \cdot p_{1} = \frac{1}{11}$
$ q_{1} \cdot p_{6} + q_{2} \cdot p_{5} + q_{3} \cdot p_{4} + q_{4} \cdot p_{3} + q_{5} \cdot p_{2} + q_{6} \cdot p_{1} = \frac{1}{11}$
$ q_{2} \cdot p_{6} + q_{3} \cdot p_{5} + q_{4} \cdot p_{4} + q_{5} \cdot p_{3} + q_{6} \cdot p_{2} = \frac{1}{11}$
$ q_{3} \cdot p_{6} + q_{4} \cdot p_{5} + q_{5} \cdot p_{4} + q_{6} \cdot p_{3} = \frac{1}{11}$
$ q_{4} \cdot p_{6} + q_{5} \cdot p_{5} + q_{6} \cdot p_{4} = \frac{1}{11}$
$ q_{5} \cdot p_{6} + q_{6} \cdot p_{5} = \frac{1}{11}$
$ q_{6} \cdot p_{6} = \frac{1}{11}$

I don't really now how to start with this. Any suggestions are welcome.

  • $\begingroup$ Just a side note: A famous example of the generating function techniques described in the answers is the derivation of the "Sicherman dice", two unequal dice with the same distribution for the sum as a pair of ordinary six-sided dice. See en.wikipedia.org/wiki/Sicherman_dice. $\endgroup$ Oct 6, 2010 at 20:31
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    $\begingroup$ I just want to note that I am one of the people voting to close. My concern is that this is a very standard exercise on how to use generating functions to work with probability; I was assigned it as an undergrad and I'm sure I will assign it in turn. $\endgroup$ Oct 6, 2010 at 21:25
  • $\begingroup$ I hadn't seen it, which is why I answered it. (I almost voted to close, actually, until I realized this was not the Sicherman dice problem.) $\endgroup$ Oct 6, 2010 at 21:44
  • $\begingroup$ Fair enough, and your answer is well written. This sort of thing is always going to grey areas, but I didn't like that there were three votes to close with no explanation. $\endgroup$ Oct 6, 2010 at 21:59

6 Answers 6


You can write this as a single polynomial equation $$p(x)q(x)=\frac1{11}(x^2+x^3+\cdots+x^{12})$$ where $p(x)=p_1x+p_2x^2+\cdots+p_6x^6$ and similarly for $q(x)$. So this reduces to the question of factorizing $(x^2+\cdots+x^{12})/11$ where the factors satisfy some extra conditions (coefficients positive, $p(1)=1$ etc.).

This is a standard method (generating functions).

  • $\begingroup$ Ah thanks. Generating functions really seem to be the right choice here. $\endgroup$
    – jakab922
    Oct 6, 2010 at 21:19

You can write a polynomial that encodes the probabilities for each die:

$$ P(x) = p_1 x^1 + p_2 x^2 + p_3 x^3 + p_4 x^4 + p_5 x^5 + p_6 x^6 $$

and similarly

$$ Q(x) = q_1 x^1 + q_2 x^2 + q_3 x^3 + q_4 x^4 + q_5 x^5 + q_6 x^6. $$

Then the coefficient of $x^n$ in $P(x) Q(x)$ is exactly the probability that the sum of your two dice is $n$. As Robin Chapman points out, you want to know if it's possible to have

$$ P(x) Q(x) = (x^2 + \cdots + x^{12})/11 $$

where $P$ and $Q$ are both sixth-degree polynomials with positive coefficients and zero constant term.

For simplicity, I'll let $p(x) = P(x)/x, q(x) = Q(x)/x$. Then we want

$$ p(x) q(x) = (1 + \cdots + x^{10})/11 $$

where $p$ and $q$ are now fifth-degree polynomials. We can rewrite the right-hand side to get

$$ p(x) q(x) = {(x^{11}-1) \over 11(x-1)} $$


$$ 11 (x-1) p(x) q(x) = x^{11} - 1. $$

The roots of the right-hand side are the eleventh roots of unity. Therefore the roots of $p$ must be five of the eleventh roots of unity which aren't equal to one, and the roots of $q$ must be the other five.

But the coefficients of $p$ and $q$ are real, which means that their roots must occur in complex conjugate pairs. So $p$ and $q$ must have even degree! Since five is not even, this is impossible.

(This proof would work if you replace six-sided dice with any even-sided dice. I suspect that what you want is impossible for odd-sided dice, as well, but this particular proof doesn't work.)

  • $\begingroup$ You forgot the curly brackets in $x^{11}$ in one place. $\endgroup$ Oct 6, 2010 at 19:18
  • $\begingroup$ Hans, thanks for pointing that out! It's fixed now. $\endgroup$ Oct 6, 2010 at 21:43
  • $\begingroup$ If I am not mistaken, you can make this solution to work if some of the faces of the dice have the same numbers (i.e. 1,1,2,3,4,5). $\endgroup$
    – Nick S
    Oct 6, 2010 at 22:54

Here is an alternate solution, which I ran across while looking through Jim Pitman's undergraduate probability text. (It's problem 3.1.19.)

Let $S$ be the sum of numbers obtained by rolling two dice,, and assume $P(S=2)=P(S=12) = 1/11$. Then

$P(S=7) \ge p_1 q_6 + p_6 q_1 = P(S=2) {q_6 \over q_1} + P(S=12) {q_1 \over q_6}$

and so $P(S=7) \ge 1/11 (q_1/q_6 + q_6/q_1)$. The second factor here is at least two, so $P(S=7) \ge 2/11$.


I believe the following punchline to the generating function argument doesn’t depend on whether the number of sides on each die is even or odd. Plug $z = \zeta = \exp 2 \pi i /11$ into the purported equality $1+z+z^2+\dots+z^{10} = p(z) q(z)$ with $p,q$ quintic polynomials with nonnegative coefficients. The LHS vanishes but the RHS can’t since all terms contributing to $p(\zeta)$ lie in the upper half-plane and similarly for $q$.

  • $\begingroup$ Dan Ullman reminds me that there’s a different punchline for the case where the number of sides is even (as is the case for 6-sided dice): $p(z)$ and $q(z)$ (having odd degree) must each have a real root, but $(z^{11}-1)/(z-1)$ doesn’t. $\endgroup$ Jul 19, 2022 at 18:59

I've heard this brainteaser before, and usually its phrased that 2-12 must come up equally likely (no comment about other sums). With this formulation (or interpretation) it becomes possible. Namely, {0,0,0,6,6,6} and {1,2,3,4,5,6}. In this case, you can also generate the sum 1, but 2-12 are equally likely (1-12 are equally likely). Without allowing for other sums I do suspect its impossible (and looks like proofs have been given). I arrived at this answer by noting we are asking for equal probability for 11 events that come from 36 (6*6) possible outcomes, which immediately seems unlikely. However equal probability for 12 events from 36 outcomes is far more manageable :)



You can't even solve this with two-sided dice. Consider two dice with probabilities p and q of rolling 1, and probabilities (1-p) and (1-q) of rolling 2. The probability of rolling a sum of 2 is pq, and the probability of rolling a sum of 4 is (1-p)(1-q). These are equal only if p=(1-q). Hence they are equal to one third only if p satisfies the quadratic equation p(1-p) = 1/3. Since this has no real roots, it cannot be done. This logic extends to multisided dice.


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