1

Given a roll of $2$ regular fair $6$ sided die, the most likely outcome is $7$.

Is there a way to construct $2, 6$ sided die, so all outcomes are still equally likely? (just like rolling a single die)

These are guidelines, not strict rules:

  • The highest number does not have to be 12, it can be higher if that is needed (eg, binary counting up to $2^{12}$).
  • The dies does not have to be a continuous sequence of numbers (die:"$1, 4, 7,\ldots $"), but the rolled total of the $2$ die should preferably be able to reach any number in the sequence (eg: "$1$ to $2^{12}$" ok. but if it for some reason can't roll $3$ then that's not ok)
  • The die does not have to start at $1$ (or $2$), but it would be preferable if it starts as low as possible.
  • The dies shouldn't have repeating sides (eg. dice: "1,1,1,1,1,1" not ok)
Sebastian Norr
  • 111
  • See my answer to this question for a proof that you can't do this with the ordinary faces ($1$ through $6$). Perhaps the argument can be adapted, I haven't thought about that. – lulu Mar 11 '23 at 12:41
  • 1
    You should add some additional constrains to discard dice like $1,1,1,1,1,1$ which would give always $1+1=2$. – jjagmath Mar 11 '23 at 12:53
  • As stated, I don't think the question is interesting...it's easy to distinguish the outcomes if you have no constraints. Even with unweighted dice. If, say, one die has $6$ distinct small faces and the other has huge faces which are very far from each other, then knowing the sum tells you the faces that came up. I think you need to add constraints to get a good combination. – lulu Mar 11 '23 at 12:54
  • @jjagmath Thanks, I added that constrain too. – Sebastian Norr Mar 11 '23 at 15:21
  • @lulu I imagined the dice should look like "normal" fair dice, not have any strange geometry, or no "unfair weights" inside either. Should i add a constrain for that too? – Sebastian Norr Mar 11 '23 at 15:23
  • You should certainly add some constraints, or else the problem is boringly trivial. – lulu Mar 11 '23 at 15:37

2 Answers2

4

Edit: The underlying assumption here is that the two dices are allowed to have different faces.

If you use one regular dice and one dice where the sides have the following numbers: $$\{0, 6, 12, 18, 24, 30\} $$ Then the sum of the two dice throws will be between 1 and 36 and each outcome will be equally likely.

You can use several different combinations on the dices. Just make sure that each possible outcome can be achieved by the same number of dice combinations.

Mevve
  • 658
0

Suppose the first die has numbers $a_1 < a_2 < a_3 < a_4 < a_5 < a_6$ and the second die has numbers $b_1 < b_2 < b_3 < b_4 < b_5 < b_6$.

Then there is exactly $1/36$ chance to get the result $a_1 + b_1,$ because any other sum of two faces must be larger.

That implies you must have exactly $36$ distinct possible results that these dice can produce. And you can have at most $36$ possible results, because there are only $36$ total ways to choose which side of the first die is used and which side of the second die is used.

You say you don't want any numbers skipped, that is, the dice must be able to produce any integer between the minimum and maximum result (so "$1$ to $2^{12}$ but not $3$" is given as an example of a disallowed set of outcomes). Therefore you must have a sequence of $36$ consecutive integers.

The integers $2$ through $37$ or $1$ through $36$ are possible such sequences, but some things work out a little neater if we consider the integers from $0$ through $35$.

This answer suggests one possible numbering (if you subtract $1$ from all sides of the first die in order to make the results run from $0$ through $35$ instead of $1$ through $36$):

$$ \{0,1,2,3,4,5\}, \{0,6,12,18,24,30\}. $$

But there are many ways to put numbers on the dice that satisfy the conditions in the question, even if we restrict ourselves to dice that produce the consecutive integers from $0$ through $35$. For example, we could have any of the following pairs of dice:

$$ \{0, 1, 2, 6, 7, 8\}, \{0, 3, 12, 15, 24, 27\}; $$ $$ \{0, 1, 4, 5, 8, 9\}, \{0, 2, 12, 14, 24, 26\}; $$ $$ \{0, 1, 2, 9, 10, 11\}, \{0, 3, 6, 18, 21, 24\}; $$ $$ \{0, 1, 2, 18, 19, 20\}, \{0, 3, 6, 9, 12, 15\}. $$

A way to derive these arrangements and several others is to consider writing the numbers $0$ through $35$ in a mixed-radix number system with two "base $2$" digits and two "base $3$" digits, $d_3d_2d_1d_0.$ We then assign two of the digits to one die and two to the other so that the numbers on the dice add up to a number from $0$ through $35$; for example, if $d_0$ and $d_1$ are base $2$ and the others are base $3$, one possible pair of dice adds up like this:

$$ d_3d_2d_1d_0 = d_30d_10 + 0d_20d_0. $$

In order for each die to have $6$ values it must have one base-$2$ digit and one base-$3$ digit (the other two digits being zero always).

There are $6$ ways to choose which digits are "base $2$" (which determines which are "base $3$"), and for each of those ways there are $2$ ways to choose which base-$3$ digit is paired with the less significant base-$2$ digit. So we have $12$ possible pairs of dice.

For any other sequence of $36$ consecutive integers, determine the first number in the sequence, write that number as a sum of integers $a + b,$ and then add $a$ to all sides of one die and add $b$ to all sides of the other die. Note that either $a$ or $b$ could be zero.

We could also add $a$ to one die and $-a$ to the other die in order to obtain the same sequence in a slightly different way.

These possibilities are less restricted if we allow negative numbers to appear on the faces of the dice.

For the sequence $0$ through $35$, assume we "normalize" the two dice by taking the least value on one die, $m,$ and subtracting it from all faces, then adding $m$ to all faces of the other die. The $0$ is the least number on each die.

One of the dice must have a $1$ in order to produce $1.$ The other die cannot have a $1$ since that would let $1$ be produced in two different ways. So we have $\{0, 1, \ldots\}$ and $\{0, n, \ldots\}$ where $n > 1.$ It seems easy enough to rule out $n=4$ and $n=5$ by trial and error.

By such methods I think we can show that the $12$ possible numberings described above are the only ones possible. But a more elegant proof would be nice to have.

David K
  • 98,388