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: The constraint is that every sum $k$ for $2 \le k \le 12$ must be achievable by adding two face values. Start from the hardest sums to produce -- the extremes. The sum $2 = 1 + 1$ forces a $1$ on the die. The sum $12 = 6 + 6$ forces a $6$. Work inward from there.
Quick Estimate: We need at least $1$ and $6$. We probably need a few more mandatory values, but the die has 6 faces, so there is room for 4 more values chosen from $\{1,2,3,4,5,6\}$. The answer should be a modest number -- likely somewhere between 5 and 20.
Approach: Systematically determine which face values are forced, then count the remaining configurations.
Formal Solution:
- 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: We have $5 = 1+4 = 2+3$ and $9 = 3+6 = 4+5$. So at least one of $\{3, 4\}$ must appear on the die (having $3$ covers $5 = 2+3$ and $9 = 3+6$; having $4$ covers $5 = 1+4$ and $9 = 4+5$).
5. Counting: We have 4 mandatory values ($1, 2, 5, 6$) and 2 remaining face slots. At least one of the 2 remaining faces must be $3$ or $4$. Count by cases: - Fix one face as $3$: the other face can be any of $\{1,2,3,4,5,6\}$ -- that is 6 choices. - Fix one face as $4$: the other face can be any of $\{1,2,3,4,5,6\}$ -- that is 6 choices. - Total: $6 + 6 = 12$, but the pair $\{3, 4\}$ is counted twice (once as "$3$ fixed, other is $4$" and once as "$4$ fixed, other is $3$").
Subtracting the double-count:
$$12 - 1 = 11$$
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.