Sunday 30 August 2026

Part 1: the uniform sum is impossible

Write the generating functions $$A(x)=\sum _{i=1}^6{a}_i{x}^i,B(x)=\sum _{j=1}^6{b}_j{x}^j.$$ By independence, the probability generating function of $S=X+Y$ is the product $A(x)B(x)$. A uniform law on $\{2,\dots ,12\}$ (eleven values) would require $$A(x)B(x)=\frac{1}{11}\sum _{k=2}^{12}{x}^k=\frac{x^2}{11}(1+x+\cdots +x^{10})=\frac{x^2}{11}·\frac{x^{11}-1}{x-1}.$$

Both dice show values $\ge 1$, so $A$ and $B$ have zero constant term; factor out the smallest power: $$A(x)=x\tilde{A}(x),B(x)=x\tilde{B}(x).$$ The product has degree $12$ and each factor has degree at most $6$, so both must have degree exactly $6$ (equivalently $a_6,b_6>0$). Hence $\tilde{A},\tilde{B}$ are real polynomials of degree $5$, and cancelling $x^2$ gives $$\tilde{A}(x)\tilde{B}(x)=\frac{1}{11}(1+x+\cdots +x^{10})=\frac{1}{11}·\frac{x^{11}-1}{x-1}.$$

Here is the step a candidate misses. The roots of ${\displaystyle \frac{x^{11}-1}{x-1}}$ are the eleven $11$th roots of unity except $1$. Because $11$ is odd, the only real solution of $x^{11}=1$ is $x=1$, which we have removed; therefore the right-hand side has no real root at all.

But $\tilde{A}$ has real coefficients and odd degree $5$, so by the intermediate value theorem (a real polynomial of odd degree changes sign at $\pm \infty$) it must have at least one real root $x_0$. Then $x_0$ is a real root of the product, contradicting that the product has no real roots.

Conclusion. No weighting of two six-sided dice yields a sum uniform on $\{2,\dots ,12\}$. $∎$

(Remark: the same odd-degree argument rules out a uniform sum whenever the number of attainable sums is odd. The tempting wrong instinct — trying to solve the linear system in the $a_i,b_j$ numerically — never terminates in a clean impossibility proof.)

Part 2: relabeling via cyclotomic factorization

For a fair die whose faces are positive integers, take counting generating functions: $A(x)={\sum }_i{x}^{f_i}$ over the six faces, so $A$ has nonnegative integer coefficients, $A(1)=6$, and no constant term. Two such dice reproduce the standard sum distribution iff $$A(x)B(x)=(x+x^2+x^3+x^4+x^5+x^6{)}^2.$$

Factor the single standard die over $\mathbb{Q}$ into cyclotomic polynomials. Since $1+x+\cdots +x^5={\displaystyle \frac{x^6-1}{x-1}}={\Phi }_2{\Phi }_3{\Phi }_6$ with ${\Phi }_2=x+1,{\Phi }_3=x^2+x+1,{\Phi }_6=x^2-x+1$, $$x+x^2+\cdots +x^6=x(x+1)(x^2+x+1)(x^2-x+1).$$ Hence $$A(x)B(x)=x^2(x+1{)}^2(x^2+x+1{)}^2(x^2-x+1{)}^2.$$

Because $\mathbb{Z}[x]$ is a UFD and these cyclotomic factors are irreducible, $A$ and $B$ are each products of a sub-multiset of the eight factors $\{x,x,(x+1),(x+1),(x^2+x+1),(x^2+x+1),(x^2-x+1),(x^2-x+1)\}$. Three constraints pin down the split:

• No constant term / genuine die: each die needs positive faces, and $x^2$ must split as one $x$ to each — otherwise one die has a value $0$ or a repeated global shift; giving both $x$'s to one die forces the other to have constant term, impossible. So each die gets exactly one factor $x$.
• Six faces: evaluate at $x=1$, where $x\mapsto 1,(x+1)\mapsto 2,(x^2+x+1)\mapsto 3,(x^2-x+1)\mapsto 1$. Requiring $A(1)=6=2^1{3}^1$ forces each die to take exactly one $(x+1)$ and exactly one $(x^2+x+1)$. There are precisely two of each available, so this distribution is forced.
• Free factor: the two copies of $(x^2-x+1)$ may be split arbitrarily; let one die take $\beta \in \{0,1,2\}$ of them and the other $2-\beta$.

Thus $$A(x)=x(x+1)(x^2+x+1)(x^2-x+1{)}^{\beta },B(x)=x(x+1)(x^2+x+1)(x^2-x+1{)}^{2-\beta }.$$

• $\beta =1$ gives $A=B=x+x^2+\cdots +x^6$: the standard pair.
• $\beta =0$ (equivalently $\beta =2$ swapped) gives the only nonstandard pair.

Compute the nonstandard dice. With $\beta =0$: $$A(x)=x(x+1)(x^2+x+1)=x^4+2x^3+2x^2+x,$$ so the faces are $\{1,2,2,3,3,4\}$. For the partner, expand $B(x)=x(x+1)(x^2+x+1)(x^2-x+1{)}^2$. Since $(x^2-x+1{)}^2=x^4-2x^3+3x^2-2x+1$ and $(x+1)(x^2+x+1)=x^3+2x^2+2x+1$, their product is $$x^7+x^5+x^4+x^3+x^2+1,$$ and multiplying by $x$ gives $$B(x)=x^8+x^6+x^5+x^4+x^3+x,$$ so the faces are $\{1,3,4,5,6,8\}$.

Conclusion. Up to swapping the two dice, exactly two pairs reproduce the standard two-dice sum distribution:

The unique nonstandard labeling is the Sicherman pair $\{1,2,2,3,3,4\}$ and $\{1,3,4,5,6,8\}$. Both dice remain fair, each face equally likely, yet the sums $2,\dots ,12$ occur with the identical frequencies $1,2,3,4,5,6,5,4,3,2,1$ out of $36$.

(Note: the constraint $A(1)=6$ is what forces the $(x+1)$ and $(x^2+x+1)$ factors to be shared symmetrically; only the self-reciprocal factor ${\Phi }_6=x^2-x+1$ can migrate, which is exactly why there is one and only one alternative.)

Part 1: apply the rule

The BH cutoffs are $\frac{k}{m}q=\frac{k}{10}(0.05)=0.005k$. Tabulate $p_{(k)}$ against $0.005k$:

The rule asks for the largest $k$ satisfying the inequality, not the first failure. The inequality holds at $k=7$ (and fails at $k=6$), and fails for all $k>7$. Hence

$$k^{\star }=7.$$

BH rejects $H_{(1)},\dots ,H_{(7)}$: the seven hypotheses with p-values $0.001$ through $0.034$. The largest rejected p-value is $0.034$.

Part 2: why the shortcut is wrong

The scan-from-top shortcut stops at $k=6$ (first failure) and would reject only the top five. It misses $H_{(7)}$.

BH is a step-up procedure: it starts from the largest index and steps down to the first success, so it takes the largest crossing point. The threshold $0.005k$ grows with $k$, so a p-value can sit above its own cutoff yet still be rejected because a later, larger cutoff is met — here $p_{(6)}=0.031>0.030$, but because $p_{(7)}=0.034\le 0.035$, index 6 is rejected too. Stopping at the first failure is the classic mistake; it turns BH into an overly conservative rule.

Part 3: the FDR guarantee

The Benjamini-Hochberg theorem (1995) states that when the p-values are independent, the procedure at level $q$ controls the false discovery rate at

$$𝔼\left[\frac{V}{max(R,1)}\right]\le \frac{m_0}{m}q,$$

where $m_0$ is the number of true nulls, $V$ the number of false rejections, and $R$ the total number of rejections. The factor $m_0/m$ appears because only true nulls can contribute false discoveries, and each true null's p-value is (sub-)uniform; the true alternatives cannot inflate the false-discovery count.

With $m_0=3$, $m=10$, $q=0.05$:

$$\mathrm{F}\mathrm{D}\mathrm{R}\le \frac{3}{10}(0.05)=0.015.$$

Closing note

Two tempting errors are worth flagging. First, confusing BH with a fixed threshold at $q$: rejecting every $p\le 0.05$ would reject the eight smallest (through $0.034$ — actually $p_{(8)}=0.20$ is above, so seven here) with no FDR guarantee at all. Second, reading $q$ itself as the achieved FDR bound; the actual guaranteed bound is $\frac{m_0}{m}q$, which is why BH is conservative when many alternatives are true. The guaranteed bound in part 3 is $\mathbf{0.015}$.