Uniform Sum Distribution with Non-Fair Dice
You have two 6-sided dice, each with sides labeled 1 through 6. The dice need not be fair -- each die can have any probability distribution over its six faces, as long as the probabilities are non-negative and sum to 1.
Can you choose two such dice, both non-fair, so that the sum of the two dice is uniformly distributed over $\{2, 3, \ldots, 12\}$? That is, can we have $P(\text{sum} = k) = \frac{1}{11}$ for every $k \in \{2, \ldots, 12\}$?
Prove your answer.
Hints
- Start with the events that can only happen one way: the sum is 2 exactly when both dice show 1, and the sum is 12 exactly when both show 6. What does the uniform assumption tell you about $p_1 q_1$ and $p_6 q_6$?
- Use the sum-7 constraint. Since $(1,6)$ and $(6,1)$ both contribute to sum = 7, you get $p_1 q_6 + p_6 q_1 \leq \frac{1}{11}$. Substitute the expressions for $q_1$ and $q_6$ from the extreme constraints.
- After substituting, you get $\frac{p_1}{p_6} + \frac{p_6}{p_1} \leq 1$. Setting $x = p_1 / p_6 > 0$, apply AM-GM: $x + \frac{1}{x} \geq 2$ for all positive $x$, giving the contradiction.
Worked Solution
How to Think About It: Before writing anything, ask: what constraints does "uniform sum" actually impose? The sum equals 2 only if both dice show 1, and the sum equals 12 only if both dice show 6. So those two extreme events each have probability exactly $\frac{1}{11}$. That pins down $p_1 q_1$ and $p_6 q_6$. Now ask: what about the sum equaling 7? That can happen many ways -- it is the most flexible outcome. But here is the trap: two of those ways, rolling (1,6) and (6,1), already have their individual pieces constrained by the above. You can write an inequality between them and the uniform probability, and that inequality forces a contradiction. The punchline is an AM-GM style argument: for any positive real $x$, we have $x + 1/x \geq 2$, and the uniform sum requirement needs $x + 1/x \leq 1$, which is impossible.
Approach: Assume for contradiction that such distributions exist with $P(\text{sum} = k) = \frac{1}{11}$ for all $k \in \{2, \ldots, 12\}$. We derive the contradiction by combining the constraints from the extreme sums with the constraint from the sum equal to 7.
Formal Solution:
Let $p_i = P(\text{die 1 shows } i)$ and $q_j = P(\text{die 2 shows } j)$ for $i, j \in \{1, \ldots, 6\}$.
Assume the sum is uniform, so $P(\text{sum} = k) = \frac{1}{11}$ for each $k \in \{2, \ldots, 12\}$.
Step 1: Pin down the extreme products.
The sum equals 2 if and only if die 1 shows 1 and die 2 shows 1 (the only way to make 2 with faces in $\{1, \ldots, 6\}$). Similarly, the sum equals 12 if and only if both dice show 6. So:
$$p_1 q_1 = \frac{1}{11} \qquad \text{and} \qquad p_6 q_6 = \frac{1}{11}$$
In particular, all four quantities $p_1, q_1, p_6, q_6$ are strictly positive.
Step 2: Bound the sum-7 probability.
The event $\{\text{sum} = 7\}$ can be decomposed as the union of mutually exclusive events $\{(i, j) : i + j = 7\}$. The pairs $(1,6)$ and $(6,1)$ are among them, so:
$$P(\text{sum} = 7) \geq p_1 q_6 + p_6 q_1$$
By the uniform assumption, $P(\text{sum} = 7) = \frac{1}{11}$, giving:
$$p_1 q_6 + p_6 q_1 \leq \frac{1}{11}$$
Step 3: Substitute using Step 1.
From Step 1: $q_1 = \frac{1}{11 p_1}$ and $q_6 = \frac{1}{11 p_6}$. Substitute:
$$p_1 \cdot \frac{1}{11 p_6} + p_6 \cdot \frac{1}{11 p_1} \leq \frac{1}{11}$$
Multiply both sides by 11:
$$\frac{p_1}{p_6} + \frac{p_6}{p_1} \leq 1$$
Step 4: Apply AM-GM to get a contradiction.
Let $x = \frac{p_1}{p_6} > 0$. The inequality becomes $x + \frac{1}{x} \leq 1$. But by AM-GM (or because $(\sqrt{x} - 1/\sqrt{x})^2 \geq 0$ implies $x + 1/x \geq 2$), we have $x + \frac{1}{x} \geq 2$ for all $x > 0$. This contradicts $x + \frac{1}{x} \leq 1$.
Answer: No such dice exist. The assumption of a uniform sum forces $\frac{p_1}{p_6} + \frac{p_6}{p_1} \leq 1$, which contradicts $x + \frac{1}{x} \geq 2$ for all positive $x$.
Intuition
The core idea here is that extreme events -- the minimum and maximum sums -- are always achieved in exactly one way, so their probabilities pin down products of face probabilities at the boundary. Once you know $p_1 q_1$ and $p_6 q_6$, you know $q_1$ and $q_6$ as functions of $p_1$ and $p_6$. The sum-7 constraint then creates a tension: it says the contribution from the pairs $(1,6)$ and $(6,1)$ alone cannot exceed $\frac{1}{11}$, but the pinned values force that contribution to violate AM-GM. This kind of "extremes pin the boundary, middle creates contradiction" argument appears frequently in combinatorics and discrete probability impossibility proofs.
In a trading context, this is related to why certain joint distributions are infeasible given marginals -- you cannot always achieve a target joint distribution just by adjusting marginal weights. The uniform sum constraint here is extremely rigid: it forces a very specific correlation structure between the two dice, and that structure turns out to be self-contradictory. A related practical lesson: when someone claims a financial instrument has "perfectly flat" payout probabilities, check whether the marginal constraints on the extreme outcomes are consistent with the interior ones. Often they are not.