This is a genuine question and I don't know the "correct" answer to this. I'm also of two minds if it is really on-topic on the site. Finding the answer is surely "a puzzle" (if possible), but it might actually also be impossible? In worst case, this post can serve as a generalized post for a group of puzzles like f.e. this, this, or this. Please be gentle and just turn a blind eye if you find this puzzle offending. (Ther story is fictional.)

As the chef-designer of a British dice-making company, you have been given the task to create the ideal set of dice for Scrabble. Your boss is not really into nitty-gritty details, he just wants something nice as a product. After some thought, you've decided that the following restrictions and conditions apply:

  • The set can only consist of dice shapes, which are produced by the company. These are the five platonic solids (D4, D6, D8, D12, D20) and a ten-sided die (D10).

  • A "dice roll" should produce an outcome as closely matching the letter frequency table for English, e.g. the likelihood of getting a particular letter should best possible match this distribution.
    (In other words: Doing a lot of "dice rolls" should produce this distribution.) enter image description here

  • The set may consist of as many dice as needed, but the lower the number of dice, the better the set can be sold.
    (Packaging costs for an infinite set of dice are, well, rather high...)

  • Each side of any die may only show letters (A-Z) or a blank. No "wildcards", symbols of special meaning etc. However, a side may show more than a single letter.

  • No complicated "evaluation" rules allowed. At any "dice roll", the outcome set of letters is the sum of all letters shown on the dice faces.

  • A single "dice roll" is defined as rolling all N dice of the set and producing M letters

  • As each "dice roll" produces M letters (follwing the distribution of the letter frequency table), the dice set with the lowest M is preferred.
    (It is hard to build a word with too many letters, but easy to allow multiple rolls for a word...)

What dice set would you propose?

The "winner" here is the most practical set. It should reproduce the frequency table in good approximation and allow "rolling up a few letters" with relative ease.

Bonus, if a method of constructing the set is provided, allowing it to be customized to different languages, i.e. different frequency tables.

However, the answer must contain some simple to follow instructions on how to produce the set, suitable to be passed on to your boss and colleages at production. (Who are all down-to-earth men with few mathematical skills.) So, while mathematical deductions and proofs are welcome, there also needs to be a "final" example set of dice listed explicitly. (i.e. Number of dice, their type and what is on their faces. )


The puzzle was inspired by this die:

D-Scrabble ( 30 sides, 4 "wild"cards )

It is a nice die, but rolling it N times will give you N letters of even distribution (and the wildcards), so I was wondering for something better than that, which could be really used in Scrabble. And then I realized, that different languages would need different sets due to their different letter frequencies... And the puzzle/question was born.

Results so far:

Compiled from the answers below, the following distributions have been achieved:
Link to results table (image)

  • $\begingroup$ In the academic literature, f.e. is usually written as e.g.. $\endgroup$ – Matsmath Dec 8 '16 at 8:57
  • $\begingroup$ Nice try, Mx. British Dice Company Employee. $\endgroup$ – Aza Dec 8 '16 at 9:19
  • $\begingroup$ "a side may show more than a single letter": Does that mean a player has to choose one of the letters, or would it count as two letters? $\endgroup$ – friedemann_bach Dec 8 '16 at 10:14
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    $\begingroup$ @Emrakul Lemming of the BD, Lemming of the BD, Lemming of the BDC! $\endgroup$ – Rand al'Thor Dec 8 '16 at 11:46
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    $\begingroup$ I think that the set should contain exactly 7 dice, since that's the number of tiles a player gets in Scrabble... Presumably, when you play "Scrabble Dice", you roll the dice and then try to construct words based on your roll. 7 thus seems natural. $\endgroup$ – GoldenGremlin Dec 8 '16 at 15:59

I chose to use a standard set of D&D dice, as you might get from a game shop, for ease of manufacturing and rolling. Incidentally, it contains seven dice, which is the number of tiles that form a scrabble player's 'hand'.

The dice I came up with are as follows:

d20 = x,z,qu,k,j,v,m,w,u,f,t,i,h,o,l,s,r,a,p,g,y
d12 = g,p,y,c,a,t,i,i,h,r,r,d
d10 = b,f,u,c,i,e,r,t,n,d
d10 = w,r,t,a,l,n,o,s,n,d
d8 = m,t,a,h,e,o,l,s
d6 = e,t,n,s,h,i
d4 = e,e,a,o

The most common low frequency here is about .7%, although q dips slightly lower, down to .6%.

Here is a terrbily constructed frequency chart, 27 representing 'QU' enter image description here

I don't think that effects any of the odds on the non-d20 by more than a tenth of a percent, but I might be wrong.

Everything seems to be within a couple tenths of a percent of where it 'should' be, except for Z, Q, X, and J, which can't show up infrequently enough.

  • 2
    $\begingroup$ Good idea to match the regular number of scrabble tiles. $\endgroup$ – friedemann_bach Dec 8 '16 at 16:12
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    $\begingroup$ Very nice. You don't have a "q" on its own (#17), and according to the rules rolling a "qu" will count as "Q+U" (two letters) which can but don't have to be used. (So your U distribution will go up from 2.14-->2.85%, which is good.) Great tool, btw! Needs to be added to our puzzling-tools meta post! $\endgroup$ – BmyGuest Dec 9 '16 at 9:06
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    $\begingroup$ Comparing with the solution from friedemann_bach below, you've spent to more dice but get close to the distribution. Form a practical perspective, selling a "full RPG set" and getting "7 Scrabble letters" seems two very good arguements though, so "the boss" is very much inclined to accept this suggestion. (I keep it open a bit more in case another variant is posted.) $\endgroup$ – BmyGuest Dec 9 '16 at 9:37
  • $\begingroup$ @BmyGuest Unless my math is wrong, the benefit to U from 'QU' should only be .0625% (one eighth of the total that shows up one time in twenty) Which would bring it to ~2.77. This also has a very small negative effect on every letter that is on a die other than the d20, but I haven't managed to get a good calculation for that yet. $\endgroup$ – Sconibulus Dec 9 '16 at 14:23

Boss, here is my proposal for our scrabble dice set. I wanted to make it funny, so I thought the set should be composed of different dice. As it should be a set you can carry in your pocket, I limited the number of dice to five. I propose a composition of five dice: D6, D8, D10, D12 and D20 with one letter on each side.

Calculating the frequencies seems easy: The frequency of each letter would be the frequency of each side of that dice divided by the total number of dice, e.g. for the letter q, which shows only on D20, it would be 1 / 20 / 5 = 1% (I trust that our statistics department will confirm this).

To match the frequencies for English as best as possible, I came up with a table ordered by letter frequency.

    D6 D8 D10 D12 D20   f(dice) f(real)  diff
e   2   1   1       1   12,17%  12,70%  -0,54%
t   1   1   1            7,83%   9,06%  -1,22%
a   1   1   1            7,83%   8,17%  -0,33%
o   1   1   1            7,83%   7,51%   0,33%
i   1   1   1            7,83%   6,97%   0,87%
n       1   1   1        6,17%   6,75%  -0,58%
s       1   1   1        6,17%   6,33%  -0,16%
h       1   1   1        6,17%   6,09%   0,07%
r           1   2        5,33%   5,99%  -0,65%
d           1   1   1    4,67%   4,25%   0,41%
l               1   2    3,67%   4,03%  -0,36%
c               1   1    2,67%   2,78%  -0,12%
u               1   1    2,67%   2,76%  -0,09%
m               1   1    2,67%   2,41%   0,26%
w               1   1    2,67%   2,36%   0,31%
f               1   1    2,67%   2,23%   0,44%
g                   2    2,00%   2,02%  -0,02%
y                   1    1,00%   1,97%  -0,97%
p                   1    1,00%   1,93%  -0,93%
b                   1    1,00%   1,49%  -0,49%
v                   1    1,00%   0,98%   0,02%
k                   1    1,00%   0,77%   0,23%
z                   1    1,00%   0,74%   0,26%
j                   1    1,00%   0,15%   0,85%
x                   1    1,00%   0,15%   0,85%
q                   1    1,00%   0,10%   0,91%

Each row shows how one letter is distributed amongst the dice sides. f(dice) shows the frequency for many dice rolls, while f(real) shows the frequency in real language, followed by the difference.

It is obvious that a small dice set will favor the less frequent letters; however, I decided that it would be ok for them to appear more often in the dice rolls than in reality, just to make the set more fun.

To adjust the frequencies, we could add a second letter on each side of D20 (e.g. one of etaoinshrdlcumwfgypb).

For other languages, we could sell a substition D20 which not only adjusts the frequencies, but also adds foreign letters. A German D20 could be: eennuggypbvkzjxpäöüß (which still favors the o a lot, but it might be acceptable for a start).

What do you say? When will we start production? :)

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    $\begingroup$ +1 for style and the "perfect pitch" to the boss. With only 5-dice you beat the 7-dice solution of Sconibulus at the cost of about doubling the deviation from the distribution both st.dev and max (at t). At that point, I think it is a bit a "matter of taste" to decide which solution is superiour. The "7-characters for Scrabble" arguement, and the the "Standard RPG set" arguement for selling (plus bigger (sensibel) set = more money) will likely convince the boss, though. see $\endgroup$ – BmyGuest Dec 9 '16 at 9:32

I know this isn't the intended solution but I'll try something a little different...

Use a huge D20 dice, and a single regular six sided dice. The D20 dice should be large enough to hold five letters per face. The second die has the numbers 1-5, the six face forces a reroll (not ideal, but we dont have a five sided dice...). The combination of both dice rolls allow a possibility of 100 possible cases, which fits a regular set of scrabble (unfortunately, which is 98 letters+2 wildcards. Also a little off the intended question). A roll of the D20 will result in five letters that would be sorted alphabetically, while the second die determines the letter to be selected.

  • $\begingroup$ You can do all sorts of "lookup-tables" - and there have been (easier) solutions suggested using simply two D10 (=1D%) - but that does not fit the requirements of the puzzle. Also: Your suggestions would give an even distribution of all letters - and is much more compllicated than the actula D30 shown at the end of the post. $\endgroup$ – BmyGuest Dec 9 '16 at 11:34
  • $\begingroup$ I did not know that a "Scrabble set" has 100 pieces, though. So your solution would indeed "match to the Scrabble game", but not to the letter-distribution of the language (or only as good, as the Scrabble set matches that one.) I maybe should reword my question... $\endgroup$ – BmyGuest Dec 9 '16 at 11:38
  • $\begingroup$ @BmyGuest this isn't a lookup-table per se though, its literally having all 100 cases of letters on the dice. $\endgroup$ – skyeriding Dec 11 '16 at 7:22
  • $\begingroup$ Sorry haven't finished typing and couldn't edit my comment. I misread your question and yes you're right, it would not match the letter-distribution of the language. $\endgroup$ – skyeriding Dec 11 '16 at 7:26
  • $\begingroup$ BTW, in your solution I would simply make the letters in five colours and the second die a color die.(Not that that would make the solution more valid though, but it would be nicer) $\endgroup$ – BmyGuest Dec 11 '16 at 7:54

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