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This question was first posted by standupmaths on youtube: https://www.youtube.com/watch?v=xHh0ui5mi_E He already got some answers but I think this community can probably find a better answer.

Imagine you have 3 indistinguishable 6-sided dice inside one big clear plastic die. You can now roll the big die if you need to roll three smaller ones. Or you can roll the big die if you need one die roll - just add up the results of the three dice inside and calculate: $sum $% $6+1$.

But here is the question: Can you find an easy way to calculate from the 3 dice inside one roll of the big die a fair result of a 2 dice roll?

Clarification: We want (3,6) and (4,5) to be different results, i.e. it's not enough to get a fair sum of two dice. But (3,6) and (6,3) can be the same result.

You are not allowed to re-roll the dice or paint numbers on the outside of the big die. You may not discard the die which is closest to the top left corner of the table or anything like that, you must determine the result from only the 3 values of the small dice.

The order of the three little dice should not matter. That is, whatever result you assign to $(1,4,5)$ should also be the same result as the one for $(4,1,5)$, and $(5,4,1)$, and all other permutations.

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    $\begingroup$ @Fimpellizieri In reality of course you could, but then the puzzle is boring, so I specified that you must find the solution from 1 roll of the big die. $\endgroup$ – Sumyrda - Reinstate Monica Apr 16 '16 at 23:26
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    $\begingroup$ It's definitely possible, because every combination of the small dice has probability $\frac{1}{36}$ (20 combinations), $\frac{1}{72}$ (30 combinations), or $\frac{1}{216}$ (6 combinations). The only problem is mapping the values in an "easy" way. $\endgroup$ – f'' Apr 17 '16 at 0:09
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    $\begingroup$ Do we want the 2 values from the roll we are simulating, or just the sum? I.e. do (3,6) and (4,5) count as different results? $\endgroup$ – frodoskywalker Apr 17 '16 at 8:24
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    $\begingroup$ Are (3,6) and (6,3) different results? $\endgroup$ – f'' Apr 17 '16 at 16:57
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    $\begingroup$ "Make 2 dice out of 3 dice"? Can't you just set aside the third one and not roll it? $\endgroup$ – Ian MacDonald Apr 18 '16 at 15:18

10 Answers 10

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All addition is modulo 6 (e.g. 4+3=1, 3+3 = 6, 5+3=2, 6+1=1). $$ \begin{array}{|c|c|}\hline \text{3 Dice Roll} & \text{Resulting 2 Dice Roll}\\\hline\hline \text{Two same, one different: }AAB & AB\\\hline 135 & 14\\\hline 246 & 25\\\hline \text{All same: } AAA & 36\\\hline \text{Three in a row: }\,\,A,A+1,A+2 & AA\\\hline \begin{array}{c}\text{All different, two of which differ by 3: }\\A,B,A+3\\\text{where }B = A+1\text{ or }B = A+2\end{array} & AB\\ \hline \end{array} $$ Credits to Scott M for a suggestion which simplified this answer.

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  • $\begingroup$ I like this one. $\endgroup$ – Sumyrda - Reinstate Monica Apr 19 '16 at 20:55
  • $\begingroup$ How does one roll $5, 5$? $\endgroup$ – Matt Apr 19 '16 at 21:29
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    $\begingroup$ This is really good, and with a little bit of practice I don't think it would even be too bad to actually use. One thing that I think would make it a little cleaner is to remove the "$A$ is low" condition on line five (so $A, A+1, A+2$ always maps to $AA$), and then have lines 2-4 map to $14$, $25$ and $36$ respectively. This would preserve the probabilities and make one less condition the players need to remember. $\endgroup$ – Scott M Apr 20 '16 at 15:41
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    $\begingroup$ This readily supports ordered distributions as well if in the first rule you return the highest value first and in all others the lowest value first. $\endgroup$ – Matt Apr 20 '16 at 16:23
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    $\begingroup$ Nice! This is fairly elegant and I think it's about as good as we're going to get. $\endgroup$ – Deusovi Apr 20 '16 at 22:03
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Here's my take on this.

If all 3 dice are different, we want to map each of the 20 combinations to a single result. Since 20 is the number of ways to roll two different numbers, at most 5, we motivate this bijection.

Let the numbers be $a < b < c$.

If $c = 6$, map it to $(a, b)$

else if $a = 1$, map it to $(c, b)$

else, map it to $(a+b+c-7, 1)$


If two dice are the same, we split into these cases:

Let $x$ be the other number we did not mention.

If one of them is a $6$, we map it to $(x, x)$.

Otherwise, it is one of these cases:

If the smaller number is $1$, map it to $(6, x)$

If the smaller number is $4$, map it to $(6, x-4)$ (or $(6, 1)$)

If the smaller number is $2$, map it to $(x, 6)$

If the smaller number is $3$, map it to $(x-3, 6)$


If all 3 dice are the same, map it to (6, 6).

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NOTE: I originally read the question before it contained the clarification that, e.g., (4,5) and (3,6) should be considered separate results. This answer does not consider those to be separate results. It only provides a single number from 2 to 12 that mimics the sum of a 2-die roll.

No-math solution

This answer doesn't require the die-roller to use any math (aside from adding up the values on the dice), so it might be more accessible to the average person who doesn't understand modular arithmetic or mathematical notation.

I will provide my solution first, then discuss how I arrived at it.

The solution is just a simple lookup table that maps the sum of the 3 dice to a number between 2 and 12 (which are the values available from 2 dice). It does this with identical probabilities, so that you're just as likely to end up with a 2 (or a 7, or a 10) as if you were rolling 2 dice.

The table

Follow the instructions below in order. For example, a roll of 1-1-2 matches rule #2 and rule #3, but since rule #2 comes first, that's the one you should use.

1) If all three dice are identical, you have rolled a $2$
2) If the three dice sum to 4, you have rolled a $4$
3) If two of the three dice are 1s, you have rolled a $3$
4) Otherwise, map the sum of the three dice using this table:

$$\begin{align} 5&\to6\\ 6&\to12\\ 7&\to11\\ 8&\to10\\ 9&\to9\\ 10&\to7\\ 11&\to6\\ 12&\to5\\ 13&\to8\\ 14&\to4\\ 15&\to8\\ 16&\to7\\ 17&\to7\\ \end{align}$$

My method

The first thing I did was to look at the probabilities of rolling each number with 2 dice, and with 3 dice.

With 2 dice, there are 36 possible outcomes (treating order of dice as significant). They are divided as follows:

 #  Number of ways to roll it
--  -------------------------
 2  1  (1-1)
 3  2  (1-2, 2-1)
 4  3  (1-3, 2-2, 3-1)
 5  4  (1-4, 2-3, 3-2, 4-1)
 6  5  (1-5, 2-4, 3-3, 4-2, 5-1)
 7  6  (1-6, 2-5, 3-4, 4-3, 5-2, 6-1)
 8  5  (2-6, 3-5, 4-4, 5-3, 6-2)
 9  4  (3-6, 4-5, 5-4, 6-3)
10  3  (4-6, 5-5, 6-4)
11  2  (5-6, 6-5)
12  1  (6-6)

With 3 dice, there are 216 possible outcomes (treating order of dice as significant). They are divided as follows (I won't list every possibility as I did above, since the number of possibilities is much greater):

 #   Number of ways to roll it
--  -------------------------
 3   1
 4   3
 5   6
 6  10
 7  15
 8  21
 9  25
10  27
11  27
12  25
13  21
14  15
15  10
16   6
17   3
18   1

Since $216=6\times 36$, my first thought was to make 6 of each result with the 3 dice map to a single result with 2 dice. i.e. I wanted to recombine the values in the 3-dice table above to fit the following set, which is 6 times the probabilities for 2 dice: $(6_{(2)}, 12_{(3)}, 18_{(4)}, 24_{(5)}, 30_{(6)}, 36_{(7)}, 30_{(8)}, 24_{(9)}, 18_{(10)}, 12_{(11)}, 6_{(12)})$.

After a couple of attempts, it became clear that the numbers as they were could not be recombined in that way. I began exploring other ways to divide the 3-die rolls, rather than by sum.

One of the easiest things was to treat triplets (three of the same number) as its own case. When I did that, the remaining probabilities all became multiples of 3, which looked much more promising for recombination: $(6_{(\text{triples})}, 3_{(4)}, 6_{(5)}, 9_{(6)}, 15_{(7)}, 21_{(8)}, 24_{(9)}, 27_{(10)}, 27_{(11)}, 24_{(12)}, 21_{(13)}, 15_{(14)}, 9_{(15)}, 6_{(16)}, 3_{(17)})$

In the end, it turned out that that was not enough either. I had to make one more split, where I separated out the double-ones. This left me with the set of probabilities $(6_{(\text{triples})}, 3_{(5)}, 6_{(6)}, 12_{(7)}, 18_{(8)}, 24_{(9)}, 27_{(10)}, 27_{(11)}, 24_{(12)}, 21_{(13)}, 15_{(14)}, 9_{(15)}, 6_{(16)}, 3_{(17)})$ plus 3 double-ones that summed to 4, and 12 others that didn't. So the whole set of probabilities (without the zeroes, which are mostly irrelevant) ended up being $(3_{(4)}, 3_{(5)}, 3_{(17)}, 6_{(6)}, 6_{(16)}, 6_{(\text{triple})}, 12_{(7)}, 12_{(\text{double-1})}, 15_{(14)}, 18_{(8)}, 21_{(13)}, 24_{(9)}, 24_{(12)}, 27_{(10)}, 27_{(11)})$

This finally gave me enough of a fine-grained division that I could recombine the probabilities into the table I listed above.


It's a little cumbersome, but I imagine that anyone needing to use this method for getting a fair 2-die roll using 3 dice would be able to memorize the table and the rules after a short time, and it would then prove to be a quick and efficient method. (In the end, probably quicker than any mathematical solution, since once the table is memorized, results can be determined virtually instantaneously, whereas a mathematical solution requires a calculation every time the dice are rolled.

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    $\begingroup$ maybe I don't get it, but what you show is a mapping from 3 dice to the sum of 2 dice, but it does not give a method to tell, if what the separate values of those 2 dice were. for example you cannot distinguish (2,5) and (3,4), can you? i think this is also part of the problem. $\endgroup$ – elias Apr 19 '16 at 18:31
  • $\begingroup$ @elias You're completely right. I think I originally read the question before that clarification was made, and made an (incorrect) assumption. I'll add a note to my answer to that effect. $\endgroup$ – GentlePurpleRain Apr 19 '16 at 18:38
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The mapping below may fall into the "Something a human can easily do after a short practice" category :)

XXY  -> X,Y              # two same, third diff>erent       
1XY  -> X-1,Y-1          # all different, has 1
346  -> 1,1           
XY6  -> X+Y-4,6          # has 6 (but not 3,4,6)
XY5  -> X+Y-3,X+Y-3      # has 5
234  -> 5,5
XXX  -> 6,6              # all same
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    $\begingroup$ Can you edit and expand this answer a bit? I'm having trouble understanding your notation. $\endgroup$ – Rand al'Thor Oct 19 '16 at 21:41
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    $\begingroup$ move the XXX rule to the top of the list and this is entirely correct, providing the same even distribution as the accepted answer (and as rolling two dice). good job. while I understood the notation, rand al'thor is correct that you may want to make it a little more clear what your rules mean. $\endgroup$ – Rubio Oct 20 '16 at 2:01
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Here's possibly a simpler version.

I call the new dice the "derived" dice, for convenience.

Let the 3 numbers be $a\leq b\leq c$.

The first derived die is fixed at $a+b+c\pmod6$. (Remainders taken from 1 to 6.)

The second derived die is obtained from the value $ab+bc+ca$.

If the first derived die is 3 or 6,

If any 2 of the 3 original dice are the same, the second derived die is a 1.

Else, take the $ab+bc+ca\pmod7$ (remainders taken from 1 to 7), this gives you the second derived die.


Otherwise, propose $ab+bc+ca\pmod6$ as the derived second die.

If there is a pair of dice (among the original 3) with the same value, and the parity of the first derived die and the second derived die are different, add 1 to the second derived die (wrap around 6 to 1).

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Another solution:

Addition, subtraction, neighbouring values, and lowest of consecutive numbers are all ment in a modulo $6$ way, where $6$ represents its congruence class instead of the usual choice of $0$.
This solution uses one less case than @Mike Earnest's approach, and produces ordered pairs by default. Also the first derived die is always the very natural choice of sum of original dice. The method to get the value of the second derived die depends on the first one, or to be more precise, its divisibility by $3$.

1. First case: if the first derived die is not divisible by $3$.
Then among the original dice, there are either two matching values, or two which are neighbouring. Take the value of the third original die to be the second derived die.
Examples: $346 \rightarrow 16$; $122 \rightarrow 51$

2. Second case: if the first derived die is divisible by $3$.
This is a little bit more tricky and less natural, having four different cases:

2.1. If the original dice's values were three consecutive numbers.
Take the value of the lowest one to be the second derived die.
Examples: $123 \rightarrow 61$; $126 \rightarrow 36$

2.2. If any two of the original dice's values had a difference of two.
Take the value of the first derived die to be the second derived die as well.
Examples: $246 \rightarrow 66$; $135 \rightarrow 33$

2.3. If the original dice's values were all same, or all divisible by $3$.
Add two to the value of the first derived die, and use this as the second derived die.
Examples: $222 \rightarrow 62$; $366 \rightarrow 35$

2.4. Two matching values in the original dice, but none of them divisible by $3$
Subtract two to the value of the first derived die, and use this as the second derived die.
Examples: $225 \rightarrow 31$; $114 \rightarrow 64$

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Ummm, maybe I'm misunderstanding something, but here is my try.

Just paint 6 numbers outside the big die and you can roll the 3 small and take the sum as sum % 6 + 1 as the result of the "first" die, and take the result of the other big die as the second one. The only problem is that the 2 dice are not equal, but I think this could work?

Edit: You are not allowed to re-roll the dice or paint numbers on the outside of the big die.

Just read that; my bad. Just so this answer doesn't go to waste, Here is something that may be more in line with the expected results:

Die 1 = mid number
Die 2 = (high number + low number) % 6 +1

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    $\begingroup$ You have to use 3 indistinguishable dice, you are using 4, with one of them distinguishable. $\endgroup$ – Kruga Apr 18 '16 at 14:22
  • $\begingroup$ ya...thought that could be a problem, but hey, it would solve the problem :D nvm $\endgroup$ – Mario Garcia Apr 18 '16 at 14:25
  • $\begingroup$ your second idea also does not provide a correct solution, as it won't produce the desired distribution $\endgroup$ – elias Apr 18 '16 at 15:18
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    $\begingroup$ Your second idea has a bias towards the middle numbers. $\endgroup$ – Sumyrda - Reinstate Monica Apr 18 '16 at 16:33
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Combining ideas behind @Mike Earnest's and my other solution:
All the arithmetic is done modulo $6$.

First, sum up the three original dice.
1. If the sum is not divisible by $3$, let the sum be the first derived die.
In this case, exactly two of the dice will be matching or having neighbouring values. Take the value of the third one to be the second derived die.
Examples: $136 \rightarrow 43$; $223 \rightarrow 13$

The remaining cases are to be used, if the sum is divisible by $3$.
2. If the values of the three original dice are consecutive.
Let the first derived die be $3$ (mnemonic: three-in-a-row), and the second derived die the 'lowest' (still in a modeular meaning) of the original dice.
Examples: $126 \rightarrow 36$; $456 \rightarrow 34$

3. If any two of the original dice's values had a difference of two.
Let the first derived die be $6$ (this will be an umbrella for all the special cases remaining), and the second derived die be $1$ or $2$, whichever matches the parity of the original dice.
Examples: $135 \rightarrow 61$; $246 \rightarrow 62$

4. Triples.
Let the first derived die be $6$, and the second derived die be $3$ (mnemonic: three-of-a-kind).
Examples: $222 \rightarrow 63$; $333 \rightarrow 63$

5. Doubles ($AAB$).
Let the first derived die be $6$, and the second derived die be $4$, $5$ or $6$, whichever can be found among the original dice. (Either $A$ or $B$ will match exactly one of them.)
Examples: $366 \rightarrow 66$; $225 \rightarrow 65$

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Thoughts about uniformity of any mapping:
If all permutations of particular trow of the 3 dices should be treated as one and the same triple then the total number of different 3-dice rolls are 56 which we should either map to 36 different pairs (1,2 is different from 2,1) or 21 ordered pairs (1,2 and 2,1 is one and the same). Both can not be done uniformly.

I prefer to avoid complex mappings with several rules to remember, so I'll try to provide a single expression to obtain the two dice values from the tree numbers and with adequate fear.

We want to end up with two numbers A and B. Let the three thrown numbers are a <= b <= c.

Calculate:

Sum = 3*(a) + 2*(b) + (c).

We can treat this sum as a base 6 number which is between 10(6) and 100(36). So take the first two digits from this number incremented by one for A and B respectively.

For those not familiar with number base systems different from 10 the calculation can be done by:

A = ((Sum div 6) mod 6) + 1 /this is integer division, i.e. take the integer part of the division, and if it is 6 treat it as 0/

B = (Sum mod 6) + 1

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    $\begingroup$ I don't think this results in a uniform distribution (e.g. you can only get 1,1 one way, when the input roll is 6,6,6). $\endgroup$ – 2012rcampion Apr 17 '16 at 18:47
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    $\begingroup$ It is possible; from rolling 3 dice, ignoring order, there are 6 results with probability 1/216 (all 3 dice are the same); 30 results with probability 1/72 (two dice are the same, one is different); and 20 results with probability 1/36 (all 3 dice are different). We can map the first 6 to one output of probability 1/36; the second 30 to 15 outputs of probability 1/36; and the last 20 to 20 outputs of probability 1/36; giving a total of 36 outputs of probability 1/36. $\endgroup$ – 2012rcampion Apr 17 '16 at 19:33
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    $\begingroup$ I don't get what you mean; we can ignore order (and discard that "additional random input") by sorting the dice before passing them to our algorithm (like you do). $\endgroup$ – 2012rcampion Apr 17 '16 at 20:02
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    $\begingroup$ But those 56 outcomes don't have the same probability. $\endgroup$ – elias Apr 18 '16 at 16:53
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    $\begingroup$ @elias Exactly! In fact the probabilities are such that you can group the inputs into exactly the required output probabilities. $\endgroup$ – 2012rcampion Apr 18 '16 at 16:55
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Here is a solution with no look-up tables, and it uses the same method as the solution to a single dice value.

The first dice value will be a+b+c mod 6.

Now we derive a new set of "little" dice.

a' = a+b mod 6

b' = b+c mod 6

c' = a+c mod 6

The second dice value will be a'+ b'+ c' mod 6.

For example (4,4,6) -> (2,4)

First dice = 14 mod 6 = 2

a' = 8 mod 6 = 2; b' = 10 mod 6 = 4; c' = 10 mod 6 = 4;

Second dice = 10 mod 6 = 4

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  • $\begingroup$ First off, if you roll all 6s, then your first dice becomes $6+6+6 \text{ mod } 6 = 0$ which is not a valid die value. The values must be in the range of 1-6 and doesn't include 0. $\endgroup$ – Trenin May 3 '16 at 18:02
  • $\begingroup$ Second, we know that your second dice is $((a+b \text{ mod } 6) +(b +c \text{ mod } 6) + (c + a \text{ mod } 6)) \text{ mod } 6$. The result of the modulo 6 can be done at the end, so this is equivalent to $(a+b+b+c+c+a) \text{ mod } 6 = 2(a+b+c) \text{ mod }6$. $\endgroup$ – Trenin May 3 '16 at 18:06
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    $\begingroup$ This means that you will only ever get 6 values, not all 36 possible pairs where the second dice is simply double the first modulo 6. The possible pairs are then (0,0), (1,2), (2,4), (3,0), (4,2), (5,4). There are many pairs that are not possible with this scheme. $\endgroup$ – Trenin May 3 '16 at 18:09

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