Dice With All Pairwise Sums Represented
You want to design a 6-sided die whose faces take values from $\{1, 2, 3, 4, 5, 6\}$ (with repeats allowed). The requirement is that when you roll two identical copies of this die, every integer sum from $2$ through $12$ must be achievable with positive probability.
Two dice are considered the same if they have the same multiset of face values, i.e., the same number of faces showing each value, regardless of physical orientation. Each of the 6 faces is equally likely on each roll.
How many distinct dice satisfy this property?
Hints
- Start with the extreme sums: only one pair of values can produce $2$, and only one can produce $12$. What does that force?
- After locking in values for sums $2, 3, 11, 12$, list all sums you can already make with those values and see what is missing.
- You need at least one of $3$ or $4$ among the remaining two faces. Count the configurations using inclusion-exclusion on that requirement.
Worked Solution
How to Think About It: Constrain from the extremes inward. The sums at the edges of the range ($2, 3, 11, 12$) can each be made in only one way, so they force specific values onto the die. Once those are fixed, check which middle sums are still missing and count by complement.
Quick Estimate: After the forced faces $\{1,2,5,6\}$, there are $2$ free faces, i.e. a multiset of size $2$ from $6$ values: $\binom{7}{2} = 21$ options. The ones that fail use only values from $\{1,2,5,6\}$: $\binom{5}{2} = 10$ options. So $21 - 10 = 11$.
Approach: Determine which face values are forced, find the sums those values miss, then count the ways to fill the remaining two faces so the missing sums are covered.
Formal Solution:
Because the two dice are identical copies, a sum like $4 = 2+2$ only needs one face showing $2$ (that face can come up on both dice).
- Extreme sums force $1$ and $6$: The only way to get $2$ is $1+1$, so $1$ must appear. The only way to get $12$ is $6+6$, so $6$ must appear.
- Next extremes force $2$ and $5$: The only way to get $3$ is $1+2$, so $2$ must appear. The only way to get $11$ is $5+6$, so $5$ must appear.
- Check which sums are already covered with $\{1,2,5,6\}$:
$$2 = 1+1,\; 3 = 1+2,\; 4 = 2+2,\; 6 = 1+5,\; 7 = 2+5$$ $$8 = 2+6,\; 10 = 5+5,\; 11 = 5+6,\; 12 = 6+6$$
The missing sums are $5$ and $9$.
- Covering 5 and 9: Using only $\{1,2,5,6\}$ neither is reachable, and $5 = 1+4 = 2+3$, $9 = 3+6 = 4+5$. A single $3$ covers both ($5 = 2+3$, $9 = 3+6$), and a single $4$ covers both ($5 = 1+4$, $9 = 4+5$). So the die works if and only if at least one of $3, 4$ appears.
5. Counting: The $4$ forced values use $4$ faces, leaving $2$ free faces, which form an unordered pair (multiset of size $2$) from $\{1,\dots,6\}$. Count pairs containing a $3$ or a $4$ by inclusion-exclusion: - Pairs containing a $3$: $\{3,1\},\{3,2\},\{3,3\},\{3,4\},\{3,5\},\{3,6\}$, that is $6$. - Pairs containing a $4$: likewise $6$. - Pairs containing both: only $\{3,4\}$, that is $1$.
$$6 + 6 - 1 = 11$$
Check by complement: $\binom{7}{2} - \binom{5}{2} = 21 - 10 = 11$.
Answer: There are $\boxed{11}$ distinct dice satisfying the property.
Intuition
This problem is a clean exercise in working from the extremes inward. The sums near the boundaries ($2, 3, 11, 12$) are the most constrained because there are very few ways to produce them, so they force specific values onto the die. Once those forced values are in place, most of the middle sums come for free. The only remaining gap requires a simple inclusion-exclusion count. This "constrain from the extremes" technique shows up frequently in combinatorics and in trading contexts. For example, when pricing exotic payoffs, the extreme scenarios (deep ITM, deep OTM) often pin down most of the structure, and the interesting design choices live in the middle.