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A loaded die looks like a regular six-sided die, but instead of each number 1-6 being equally likely, some numbers may be more or less likely than others.

Is it possible to have two loaded dice so that, when rolled together, every possible sum 2,3,...,12 is equally likely to occur? The dice may be loaded in different ways.

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    $\begingroup$ You could always just throw a D12. $\endgroup$ – Deacon Jul 28 '15 at 16:05
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    $\begingroup$ Both dice are loaded with plastic explosives and, when they hit the table, they explode. All sums occur with probability 0. ;) $\endgroup$ – Doug McClean Jul 28 '15 at 19:32
  • $\begingroup$ @user12365 You'd actually need a D11, which I don't think exists, since there are only 11 distinct values that can be obtained by summing 2 D6s. $\endgroup$ – GentlePurpleRain Jul 30 '15 at 18:35
  • $\begingroup$ @GentlePurpleRain Just like when creating your slightly above-average DnD characters, you have to reroll the 1's. $\endgroup$ – Deacon Jul 30 '15 at 19:43
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No.

To explain, here is a more formal way to state Ian's answer:

Let the probability of die A rolling $i$ be $a_i$ for $1\le i\le6$, and similarly for die B.

If every sum has the same probability, then they must each have probability $\frac{1}{11}$. Because 2 is only possible from double 1s, and 12 is only possible from double 6s, $a_1b_1=a_6b_6=\frac{1}{11}$. Now the probability of rolling 7 is at least $a_1b_6+a_6b_1\ge 2\sqrt{a_1b_6a_6b_1}=\frac{2}{11}$ (using the AM-GM inequality). Therefore, it is not possible.

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  • $\begingroup$ Thanks for confirming, @f''. My probability math is too rusty to have quickly come up with this. :) $\endgroup$ – Ian MacDonald Jul 28 '15 at 15:54
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    $\begingroup$ Slick! To elaborate, the inequality $x+y\ge 2\sqrt{xy}$ comes from rearranging $(\sqrt{x}-\sqrt{y})^2\ge0$. $\endgroup$ – Mike Earnest Jul 28 '15 at 16:35
  • $\begingroup$ Follow up question: Let $n$ denote the number of indistinguishable bias 6 sided dice labeled 1,2,3,4,5,6. What is minimum value of $n$ such that when they are rolled together, every possible sum from $n,n+1,n+2,\ldots,6n-1,6n$ is equally likely to occur? Or prove that it's impossible. $\endgroup$ – GohPiHan Jul 30 '15 at 11:41
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    $\begingroup$ @GohP.iHan impossible for $n>1$. If $n$ and $6n$ each have probability $\frac{1}{5n+1}$, AM-GM makes the probability of $n+5$ at least $n$ times that. $\endgroup$ – f'' Jul 30 '15 at 15:20
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    $\begingroup$ @GohP.iHan ex. for $n=3$ $P(8)\ge a_1b_1c_6+a_1b_6c_1+a_6b_1c_1\ge3\sqrt[3]{a_1b_1c_6a_1b_6c_1a_6b_1c_1}=3\sqrt[3]{P(3)P(3)P(18)}=\frac{3}{16}$. $\endgroup$ – f'' Jul 30 '15 at 15:23
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Once again not an answer to this question because it has been thoroughly answered. I wrote a short program in R and used the optim command to find the values that get closest to having a uniform distribution. The quality measure was the largest probability (of a number 2-12) - the smallest probability (of a number 2-12)

The solution I got yielded:

\begin{align*} P(2) &= .05 \\ P(3) &= .1 \\ P(4) &= .1 \\ P(5) &= .1 \\ P(6) &= .1 \\ P(7) &= .1 \\ P(8) &= .1 \\ P(9) &= .1 \\ P(10) &= .1 \\ P(11) &= .1 \\ P(12) &= .05 \\ \\ a_1 &= .5 \\ a_2 &= 0 \\ a_3 &= 0 \\ a_4 &= 0 \\ a_5 &= 0 \\ a_6 &= .5 \\ \\ b_1 &= .1 \\ b_2 &= .2 \\ b_3 &= .2 \\ b_4 &= .2 \\ b_5 &= .2 \\ b_6 &= .1 \end{align*}

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This is not a formal proof, but maybe someone could build off it.

My gut says

this is not possible.

If we consider three combinations, it becomes clear (to me, anyway) that this is the case:

Consider 1+1, 6+6, and 1+6. You would need to weigh the 1+1 and 6+6 the same, and weigh them so that they have a higher probability than their normal. However, this increased weighting would then also cause 1+6 to become even more likely than it already is; eternally pushing its probability higher than 1+1 or 6+6.

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I suspect no.

Let's look at 2 and 12, as they can only be achieved by a 1, 1 and a 6, 6.

Let our dies be denoted $a$ and $b$.

Let $a_1$ designate the probability of 1 being rolled on $a$.

So:
$P(2) = a_1 \times b_1$
$P(12) = a_6 \times b_6$.

Without considering any other possible roles, $P(7)$ is currently $a_1 \times b_6 + a_6 \times b_1$. If $a_1 > b_1$ then $a_6 > b_6$ otherwise $a_1\times b_6$ would be greater than either $P(2)$ or $P(12)$.

Thus:
$b_6 > a_6$
$a_1 > b_1$.

$a_1 \times b_1=a_6 \times b_6 \geq a_1 \times b_6 + b_1 \times a_6 $

Since $a_6 < b_6$,
$a_6 \times b_6 > a_1 \times a_6+b_1 \times a_6$ or $(a_1+b_1) \times a_6$.

So $a_1 + b_1$ must be less than $b_6$.

Since $a_1 > b_1$,
$a_6 \times b_6 >b_1 \times b_6+b_1 \times a_6$ or $b_1 \times (a_6+b_6)$.

So $a_6+b_6$ must be less than $a_1$.

However $a_1+b_1$ is less than $b_6$. This is a contradiction so it is impossible.

(apologies for sloppy inequality signage)

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I further show that this sum distribution is impossible for any pair of dice, no matter how many faces they have, what numbers are on those faces, and what weights each face has, barring the trivial case where one die always rolls the same result.

Translate the dice weightings into generating functions

$$A(x) = \sum a_i x^i$$

and likewise for $B(x)$. Since dice sum is distribution convolution is generating function product, we're required to have

$$A(x) B(x) = \frac{x^2+x^3+\dots + x^{11} + x^{12}}{11}$$

Express the geometric series on the RHS as

$$A(x) B(x) = \frac{x^2}{11} \cdot \frac{1-x^{11}}{1-x}$$

The roots of $1-x^{11}$ are the powers of the primitive elevent roots of unity $\omega = e^{2\pi i/11}$, which are $1, \omega, \omega^2, \dots, \omega^{11}$. The $1-x$ in the denominator removes the root $x=1$. So, we can factor into linear terms

$$A(x) B(x) = \frac{x^2}{11} \cdot \prod_{j=1}^{10} (x-\omega^j)$$

Now, because of unique factorization, these linear terms must be split across $A$ and $B$. Moreover, since the coefficients of $A$ and $B$ are probabilities, the resulting polynomials must have non-negative real coefficients.

At this point, we're down to a finite number $2^{10}$ of possibilities, so we could try all of them by computer and check that none work. But, we can simplify the search by hand first.

First, a real polynomial must equal its conjugate, so its complex roots must come in conjugate pairs. Therefore, the conjugate roots must "stick" together when being split into $A$ and $B$. So, we can group them into quadratic units

$$(x-\omega^j)(x-\overline{\omega^j}) = x^2 - 2\mathrm{Re}(\omega^j)x + 1$$

for $j=1,2,3,4,5$.

Now, note that for a product of quadratic units, the coefficient of $x$ is the sum of all $-2\mathrm{Re}(\omega^j)$ of the $j$'s included. Since this coefficient must be positive, the sum of the real parts of the included roots of unity must be negative.

These real parts are roughly $0.84, 0.42, -0.14, -0.65, -0.96$. Already we see that $0.84$ must be grouped with $-0.96$ to make a negative sum, since the other negatives aren't enough. Then, $0.42$ must be grouped with $-0.65$, since the $-0.14$ and the $-0.12$ from the previous pairing aren't enough to make it negative.

So, we have three groups $\{\omega^1,\omega^5\},\{\omega^2,\omega^4\},\{\omega^3\}$ split into two sides. Putting them all onto one side would result in a trivial monomial on the other side, giving a die that always rolls the same. So, we must have one group alone and the other two groups together. Checking all three possibilities directly, each one gives a polynomial that contains negative coefficients. So, the desired split is not possible, and no solution exists.

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  • $\begingroup$ This also shows that it won't work with any number of dice unless there is only one non-constant die. $\endgroup$ – f'' Jul 29 '15 at 12:50
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While I was typing mine, Going hamateur and f' had posted very similar approaches, so check theirs out as well.

There are 11 possible sums (2 thru 12), so the probability of each would have to be $\frac{1}{11}$.

Lets call $X$ die 1, and $Y$ die 2, and let $X_1$ denote the probability rolling a 1 with the first die. Thus, we can say that $X_1*Y_1 = X_6*Y_6 = \frac{1}{11}$ as these are the only ways to get 2 and 12.

To build a little more, lets look at a sum of 7.
- $X_1*Y_6 + X_6*Y_1 = \frac{1}{11}$
- $X_1*Y_6 + X_6*Y_1 = X_1*Y_1$
- $\implies X_1*Y_6 = X_1*Y_1 - X_6*Y_1 $
- $\implies X_1*Y_6 = Y_1(X_1 - X_6) $
- $\implies X_1 \ge X_6$ (since $Y_1 \ge 0$ and $Y_1(X_1 - X_6) \ge 0$

Now, from the other side.
- $X_1*Y_6 + X_6*Y_1 = X_6*Y_6$ (since $X_6*Y_6 = X_6*Y_6$)
- $\implies X_6*Y_1 = X_6*Y_6 - X_1*Y_6 $
- $\implies X_6*Y_1 = Y_6(X_6 - X_1) $
- $\implies X_6 \ge X_1$ (since $Y_6 \ge 0$ and $Y_6(X_6 - X_1) \ge 0$

With both $X_1 \ge X_6$ and $X_6 \ge X_1$, the only value that works is for $X_1 = X_6 = 0$.

Since $X_1*Y_1 = \frac{1}{11}, X_1 \ne 0$, therefore it is not possible.

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Starting from 2 and 12, 3 and 11, ... it follows that the chance of rolling $n$ when one die is cast is equal to that of rolling $7-n$.

If they're $a,b,c,c,b,a$, then:

$a^2$ (getting 2) = $2ab$ (getting 3), so $a$ = $2b$; $2b,b,c,c,b,2b$

$4b^2$ (getting 2) = $4bc$ + $b^2$ (getting 4), so $4b$ = $4c$ + $b$, $3b$ = $4c$; $2b, b, 3b/4, 3b/4, b, 2b$

$4b^2$ (getting 2) = 2*($2b*3b/4$ + $b*3b/4$) (getting 5) = 2*($3b^2/2$ + $3b^2/4$) = $9b^2/2$ (contradiction)

It's impossible.

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