Is it possible to get all possible sums with the same probability if I throw two unfair dice together?

Question

I throw 2 unfair dice, suppose that $p_i$ is the probability that the first die can give an $i$ if I throw it, for $i =1,2,3,..6$ and $q_i$ the probability that the second die can give an $i$. If I throw the dice together, is it possible to get all possible sums $2,3,4,...12$ with the same probability?

Here's what I've tried so far, the probability that I get a $2$ if I throw both dice is $p_1q_1$, the probability that I get $3$ is $p_1q_2+p_2q_1$, and generally the probability that I get $n$ is $$\sum_{i+j=n} p_iq_j$$ where $i=1,2,...6$, $j=1,2,...6$.

So now in order for all possible sums to appear with the same probability, it must be true that $$p_1q_1=p_1q_2+p_2q_1$$ $$p_1q_2+p_2q_1=p_1q_3+p_2q_2+p_3q_1$$ $$........$$ has a solution, this is where I am stuck I can't find a way to prove that the system above has a solution, can you help?

Answer 1

This is a classical problem. Without changing the problem, we can let the digits on the dice be $0, \ldots, 5$ instead of $1, \ldots, 6$ to make our notation easier. Now we make two polynomials: $$ P(x) = \sum_{i=0}^5 p_ix^i,\qquad Q(x) = \sum_{i=0}^5q_ix^i. $$ Now we can succinctly phrase your condition on $p_i, q_i$: it is satisfied if and only if $$ P(x)Q(x) = \frac1{11}\sum_{i=0}^{10} x^i. $$ Let's multiply both sides by $11 \times (x-1)$, and you get $$ 11(x-1)P(x)Q(x) = x^{11} - 1. $$ The 11 zeroes of the polynomial on the right are the 11th roots of unity, which means those are also the zeroes of the polynomial on the left. The term $(x-1)$ takes care of one of the zeroes, and since $P, Q$ are both of degree 5, that means that they each have to have 5 of the other 10 zeroes.

But now note: besides $1$, all of the 11th roots of unity are complex numbers, while $P, Q$ are real polynomials. If a complex number is the root of a real polynomial, then so is its complex conjugate. That means that $P, Q$ must each have an even number of complex zeroes, but we just showed that they also have to have 5 each.

We have reached a contradiction: such $P, Q$, and thus such distributions $p_i, q_i$, do not exist.

Answer 2

Let take a simpler system. There are 2 outcomes for 2 coins $\{C_1,C_2\}$ - $\{1,2\}$ with probabilities $p_1,p_2$ for $C_1$ and $q_1,q_2$ for $C_2$.

Given condition: probability after 2 coin throws of the sum being $\{2,3,4\}$ is the same.

Question: Can you find some $p_1,p_2,q_1,q_2$ that satisfies the condition?

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In this the given condition is $p_1q_1 = p_1q_2 + p_2q_1 = p_2q_2 -(G)$. But we also have the implicit conditions: $p_1 + p_2 = 1-(eq1)$, $q_1+q_2 = 1-(eq2)$. Consider the first and third expressions of the given condition (G). \begin{align*} p_1 q_1 &= p_2q_2\\ p_1 q_1 &= (1-p_1)(1-q_1)\\ p_1 q_1 &= 1-p_1-q_1+p_1q_1\\ p_1+q_1 &= 1\tag{eq3}\\ p_2+q_2 &= 1\tag{eq4}\\ \end{align*} Comparing the equations (eq1) and (eq3) and (eq1) and (eq4), we get $p_2 = q_1$ and $p_1 = q_2$.

Finally from the first two equalities of the given condition, we have: \begin{align*} p_1q_1 &= p_1q_2 + p_2q_1\\ p_1p_2 &= p_1p_1 + p_2p_2\\ -p_1p_2 &= p_1^2 + p_2^2 - 2p_1p_2\\ -p_1p_2 &= (p_1 - p_2)^2\\ \end{align*} Now we have a contradiction as the left side is $-ve$ and the right side is $+ve$

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Since we can't find a solution for this simpler system, with just 2 equalities, it is quite unlikely that a solution for the more complicated die system with $\binom{6}{2}$ equalities exists.

Note: This is of course not a proof that no solution exists for the die system.

Answer 3

The probability generating function of the dice are $P(x) = \sum_{i=1}^6 p_i x^i$ and $Q(x) = \sum_{i=1}^6 q_i x^i$. The probability generating function for their sum is $R(x) = P(x)Q(x)$. You want all possible sums $2, \ldots, 12$ to have the same probability, or equivalently you want $R(x) = \frac{1}{11} x^2 (1+x+\cdots+x^{10})$.

Hence an equivalent way to state your problem is: can we factor $1+x+\cdots+x^{10} = p(x)q(x)$ where $p, q \in \mathbb{R}_{\geq 0}[x]$ have degree 5?

We can factor $1+x+\cdots+x^{10} = \frac{1-x^{11}}{1-x}$ over the complex numbers as $\prod_{k=1}^{10} (x-\exp(2\pi i k/11))$. Grouping together complex conjugate pairs using \begin{align*} (x-\exp(2\pi i k/m))(x-\exp(-2\pi i k/m)) &= x^2 - (\exp(2\pi i k/m) + \exp(-2\pi i k/m))x + 1 \\ &= x^2 - 2\cos(2\pi k/m) x + 1 \end{align*} gives \begin{align*} 1 + x + \cdots + x^{10} = \prod_{k=1}^5 (x^2 - 2\cos(2\pi k/11)x + 1). \end{align*}

But that says $1 + x + \cdots + x^{10}$ has no real factors of degree $5$ whatsoever, so there is no solution.

Suppose you weaken the requirements and don't insist the dice both have values in $\{1, \ldots, 6\}$. The subsets of factors with non-negative coefficients are $\{\}$,$\{3\}$,$\{4\}$,$\{5\}$,$\{2,4\}$,$\{2,5\}$,$\{3,4\}$,$\{3,5\}$,$\{4,5\}$,$\{2,3,4\}$,$\{2,3,5\}$,$\{2,4,5\}$,$\{3,4,5\}$,$\{2,3,4,5\}$,$\{1,2,3,4,5\}$. Since $1$ is only in one of these, namely $\{1, 2, 3, 4, 5\}$, the only complementary pair is the trivial one, $\{\}, \{1,2,3,4,5\}$ and there are no interesting solutions to the weaker version either.

Answer 4

Let's assume that this is possible. We can derive a contradiction from this assumption.

The probability of rolling a total of $2$ must be $1/11$, and the probability of rolling a total of $12$ must also be $1/11$, so

$$ \begin{align} p_1 q_1 &= 1/11,\ \text{and}\\ p_6 q_6 &= 1/11. \end{align} $$

The probability of rolling a total of $7$ must also be $1/11$. At the same time, the probability of rolling a total of $7$ is greater than or equal to $p_1 q_6 + p_6 q_1$. So,

$$p_1 q_6 + p_6 q_1 \le 1/11.$$

The numbers $p_1$, $p_6$, $q_1$, and $q_6$ are all probabilities, so they can't be negative. Also, if any one of them were $0$, then either $p_1 q_1$ or $p_6 q_6$ would be $0$, which we already know is not the case. So, all four of these numbers are positive. This means that $p_1 q_6$ and $p_6 q_1$ are both positive, and so

$$ \begin{align} p_1 q_6 &< 1/11,\ \text{and}\\ p_6 q_1 &< 1/11. \end{align} $$

If we combine these inequalities with the equations above, we find that

$$ \begin{align} p_1 q_6 &< p_1 q_1,\\ p_6 q_1 &< p_1 q_1,\\ p_1 q_6 &< p_6 q_6,\ \text{and}\\ p_6 q_1 &< p_6 q_6. \end{align} $$

Since the numbers $p_1$, $p_6$, $q_1$ and $q_6$ are all positive, we may cancel them when they appear on both sides of one of these inequalities. Doing that, we conclude that

$$ \begin{align} q_6 &< q_1,\\ p_6 &< p_1,\\ p_1 &< p_6,\ \text{and}\\ q_1 &< q_6. \end{align} $$

But this is impossible.

Answer 5

If you want to generate numbers 1 through 12 with uniform probability, it is possible by re-labeling the faces of fair dice. One die has the faces labeled 1 through 6. The other has the faces labeled -2, 0, 2, 4, 6, 8. If the total is not in the 1 through 12 range, roll again. This is an acceptance-rejection technique.

Also, there is a theorem in probability theory than any desired probability can be constructed by a sequence of Bernoulli trials (coin flips). Simulate coin flips by having the faces of a die be 0, 0, 0, 1, 1, 1, another die with faces 0, 0, 0, 2, 2, 2, a third die with faces 0 and 4, and a fourth die with faces 0 and 8. You will roll 0 through 15 with equal probability, and roll again if your number is not in the 1 through 12 range. Another acceptance-rejection method, with dice simulating coins.

Forgive my "engineering" approach; I can't help it.