Unknown Die with Uniform Sum
Alan has two fair 6-sided dice -- each of the six faces of each die is equally likely. One is a standard die with faces labeled $1$ through $6$. The other die has unknown face values -- they need not be the integers $1$-$6$ and may be repeated -- but all values are positive integers.
Whenever Alan rolls both dice together, each integer sum from $2$ to $13$ (inclusive) is equally likely. What is the sum of all six face values on the unknown die?
Hints
- You cannot determine the individual face values, but you may not need to -- what single quantity determines the sum of all faces on the mystery die?
- The expected value of a die equals the average of its faces, which equals (sum of faces) / 6. So finding $E[D_2]$ immediately gives the sum.
- Use linearity of expectation: $E[D_1 + D_2] = E[D_1] + E[D_2]$. You know $E[D_1 + D_2]$ from the uniform constraint and $E[D_1]$ from the standard die.
Worked Solution
How to Think About It: We don't know the individual faces of the mystery die, but we can pin down their SUM using only expected values: the sum of a die's faces equals six times its mean. The condition that the total is uniform on $\{2,\dots,13\}$ fixes $E[\text{sum}]$, and linearity of expectation then fixes $E[\text{mystery die}]$.
Quick Estimate: A uniform distribution on $\{2,\dots,13\}$ has mean $\frac{2+13}{2}=7.5$. A standard die has mean $3.5$. So the mystery die must have mean $7.5-3.5=4$, and with $6$ equally-likely faces its face-sum is $6\times4=24$.
Approach: Both dice are fair, so each die's mean is the plain average of its six faces and the face-sum is $6$ times the mean. That turns the question into a one-line application of linearity of expectation -- no need to identify the faces themselves. We then confirm separately that a die meeting the constraint really exists, which is what licenses the expectation argument in the first place.
Formal Solution: Let $D_1$ be the standard die and $D_2$ the mystery die, with $S=D_1+D_2$ uniform on $\{2,\dots,13\}$. Then $$E[S]=\frac{2+13}{2}=\frac{15}{2},\qquad E[D_1]=\frac{1+2+3+4+5+6}{6}=\frac{7}{2},$$ so by linearity $E[D_2]=E[S]-E[D_1]=\frac{15}{2}-\frac{7}{2}=4$. Since $D_2$ has six equally-likely faces, the sum of its face values is $$6\,E[D_2]=6\times4=24.$$
Does such a die exist? Yes -- and the expected-value argument is only valid because it does. Matching generating functions, $(x+\cdots+x^6)\,G_2(x)=3(x^2+\cdots+x^{13})$ forces $G_2(x)=3x+3x^7$: the mystery die has three faces showing $1$ and three showing $7$, i.e. $\{1,1,1,7,7,7\}$. Check: the mystery die is equally likely $1$ or $7$, so sums $2$–$7$ (mystery $=1$) and $8$–$13$ (mystery $=7$) each occur with probability $\tfrac{1}{12}$ -- uniform. Its face-sum is $3(1)+3(7)=24$.
Answer: The sum of all six face values on the mystery die is $\mathbf{24}$.
Intuition
The elegance here is that you do not need to figure out the actual face values on the mystery die -- just its average. The uniform constraint on the sum is a constraint on the distribution of $D_1 + D_2$, and extracting an expected value from that distribution is much easier than recovering the full joint structure. Linearity of expectation does the heavy lifting.
This is a general principle: expected values are additive regardless of the dependence structure (or in this case, regardless of the mystery die's layout). You can solve many underdetermined problems by working with expectations rather than trying to pin down the full distribution. In market making and risk, you often similarly extract moments from observable quantities without needing to know the full distribution of the underlying.