Sums of 100 Rolls of Two Six Sided Dice

Question

Suppose you have two six-sided dice. Roll the pair 100 times, calculating the sum of each pair. Print out the number of times each sum occurred. If a sum was never rolled, you must include a zero or some way to identify that that particular sum was never rolled.

Example Output: [3, 3, 9, 11, 15, 15, 11, 15, 7, 8, 3]

The number of times a sum was rolled is represented in the sums index - 2

In this example, a sum of two was rolled 3 times ([2-2]), a sum of three 3 times ([3-2]), a sum of four 9 times ([4-2]), and so on. It does not matter the individual dice rolls to arrive at a sum (5 and 2 would be counted as the same sum as 6 and 1)

"Ugly" outputs are fine (loads of trailing zeros, extra output, strange ways of representing data, etc.) as long as you explain how the data should be read.

Answer 1

Jelly, 13 12 bytes

³Ḥ6ẋX€+2/ṢŒr

A niladic link. Output format is a list of lists of [value, count].

(Zero rolls means no such entry is present in the output - e.g. an output of[[6, 12], [7, 74], [8, 14]] would identify that only sums of six, seven and eight were rolled.)

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How?

³Ḥ6ẋX€+2/ṢŒr - Main link: no arguments
³            - 100
 Ḥ           - double = 200
  6          - 6
   ẋ         - repeat -> [6,6,6...,6], length 200
    X€       - random integer from [1,z] for €ach (where z=6 every time)
       2/    - pairwise reduce with:
      +      -   addition (i.e. add up each two)
         Ṣ   - sort
          Œr - run-length encode (list of [value, length] for each run of equal values)

Answer 2

Python 2, 84 77 76 bytes

-7 bytes thanks to @JonathanAllan
-1 byte thanks to @FelipeNardiBatista

from random import*
a=[0]*13
exec'a[%s]+=1;'%('+randint(1,6)'*2)*100
print a

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The output has two leading zeros

Answer 3

Perl 6, 30 bytes

bag [Z+] (^6).pick xx 100 xx 2

(^6).pick is a random number from zero through five. xx 100 makes a hundred-element list of such numbers. xx 2 produces two such lists. [Z+] zips those two lists with addition, producing a hundred-element list of two-die rolls. Finally, bag puts that list into a bag, which is a collection with multiplicity. Example REPL output:

bag(1(4), 9(4), 0(4), 4(14), 5(18), 3(9), 10(2), 6(19), 7(13), 2(3), 8(10))

That means 1, 9, and 0 occurred four times each, four occurred fourteen times, etc. Since the "dice" in this code produce a number from 0-5, add two to each of these numbers to get the rolls a pair of standard 1-6 dice would produce.

Answer 4

05AB1E, 21 19 bytes

-2 bytes thanks to @Emigna

TÝÌтF6Lã.RO¸ì}{γ€g<

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TÝÌтF6Lã.RO¸ì}{γ€g<
TÝÌ                   Range from 2 to 12
   тF                 100 times do:
     6L                 Range from 1 to 6
       ã                Cartesian product (creates all possible pairs of 1 and 6)
        .RO             Choose random pair and sum
           ¸ì           Prepend result to initial list
             }        end loop
              {γ€g<   Sort, split on consecutive elements, count and decrement

Answer 5

Mathematica, 50 bytes

r:=RandomInteger@5
Last/@Tally@Sort@Table[r+r,100]

Straightforward implementation. If any sum is never achieved, the 0 is omitted from the list.

Answer 6

MATL, 17 bytes

6H100I$Yrs!11:Q=s

Output is a list of 11 numbers (some of them possibly 0) separated by spaces, indicating the number of times for each pair from 2 to 12.

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For comparison, the theoretical average number of times each pair will appear on average can be computed as 6:gtY+36/100*.

If the number of rolls is increased the obtained values approach the theorerical ones. See for example the obtained and theoretical values with 10000 rolls.

Answer 7

CJam, 18 20 bytes

100{;6mr6mr+))}%$e``

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Answer 8

R, 45 37 35 bytes

`[`=sample;table(6[100,T]+6[100,T])

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-7 bytes thanks to Jarko Dubbledam

Answer 9

PHP, 53 Bytes

prints an associative array. key is result of two dices and value is the count of these results

for(;$i++<100;)$r[rand(1,6)+rand(1,6)]++;print_r($r);

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Answer 10

S.I.L.O.S, 99 bytes

i=100
lbla
x=rand*6+rand*6
a=get x
a+1
set x a
i-1
if i a
lblb
c=get b
printInt c
b+1
d=11-b
if d b

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Rolls the dice, and stores them in the first 11 spots of the heap, then just iterates through the heap printing each counter. This is one of the first recorded uses of the rand keyword combined with an assignment operator.

It is worth noting, that a few modifications can be made to output a histogram of the rolls.

Unfortunately it must be run from the offline interpreter.

i=4000
lbla
x=rand*6+rand*6
a=get x
a+1
set x a
i-1
if i a
canvas 1100 1000 Output
lblb
c=get b
printInt c
d=c*1
y=1000-d
x=b*100
newObj 0 100 d
moveObj b x y
b+1
d=11-b
if d b
wait 10000

Answer 11

Elixir, 157 118 bytes

l=&Enum.random(1..&1)
p=fn(o,s)->y=l.(6)+l.(6)
s=List.update_at(s,y,&(&1+1))
if Enum.sum(s)<100 do s=o.(o,s) end
s end

Tried something harder than Jelly.

Explanation:

1. Define function that returns a random number between 1 and 6 inclusive.
2. Define the function anonymously and let y be the variable with the roll sum.
3. update the appropriate place in the list by adding 1.
4. if we are 100 rolls in, quit. Else call yourself again passing in yourself and the updated list.
5. return the updated array.

Should be called like p.(p,[0,0,0,0,0,0,0,0,0,0,0,0,0]). It will raise a warning, but it will return the desired array with 13 elements, the first 2 should be ignored.

Answer 12

Java 8, 104 bytes

A lambda returning an int[] of frequencies. Assign to Supplier<int[]>.

()->{int o[]=new int[11],i=0;while(i++<100)o[(int)(Math.random()*6)+(int)(Math.random()*6)]++;return o;}

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Ungolfed lambda

() -> {
    int
        o[] = new int[11],
        i = 0
    ;
    while (i++ < 100)
        o[(int) (Math.random() * 6) + (int) (Math.random() * 6)]++;
    return o;
}

Answer 13

q/kdb+, 31 28 25 bytes

Solution:

sum!:[11]=/:sum(2#100)?'6

Example:

q)sum!:[11]=/:sum(2#100)?'6
1 3 5 11 16 21 16 9 8 9 1i

Explanation:

Roll a dice 100?6 , roll a dice again and add the vectors together. Then see where each results matches the range 0..10, then sum up all the trues in each list:

sum til[11]=/:sum(2#100)?'6 / ungolfed solution
                 (2#100)    / 2 take 100, gives list (100;100)
                        ?'6 / performs rand on each left-each right, so 100 & 6, 100 & 6
              sum           / add the lists together
    til[11]                 / the range 0..10
           =/:              / apply 'equals?' to each right on left list
sum                         / sum up the results, e.g. how many 1s, 2s, 3s.. 12s

Notes:

'Golfing' is mostly swapping out q keywords for the k equivalents, namely each and til.

Answer 14

JavaScript, 72 70 bytes

Outputs an array of up to 11 elements for each of the roll counts. Except for the first element, if the count is 0 then the element will be empty or, if it's at the end of the array, omitted completely.

f=(a=[n=0])=>n>99?a:f(a,a[x=(g=_=>Math.random()*6|0)()+g(++n)]=-~a[x])

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Answer 15

QBIC, 45 bytes

[100|h=_r1,6|+_r1,6|-2┘g(h)=g(h)+1][0,z|?g(b)

Explanation:

[100|         FOR a = 1 to 100
h=_r1,6|       set h to a random value between 1-6
 +_r1,6|       + another rnd(1,6) (2, 3 ... 11, 12)
 -2            - 2 (index: 0 ... 10
┘             Syntactic linebreak
g(h)          When using array parenthesis on an undefined array,
              it is interpreted as an array with 10 indexes of all zeroes.           
    =         Of array g, set the value of index h (0 ... 11)
      g(h)+1  to one higher (all indices start out as 0)
              Note that we need to track 11 values. Fortunately, QBasic'set
              empty, 10-sized array has 11 indices, because of base 0 / base 1 ambiguity.
]             NEXT set of dice
[0,z|         FOR b = 0 to 10
?g(b)           PRINT the tracker array

Answer 16

APL, 14 bytes

,∘≢⌸+/?100 2⍴6

Presents data as a table with the left column representing the sum and the right representing the number of occurrences.

Explained

        100 2⍴6  ⍝ create an 2×100 array of 6
       ?         ⍝ roll for each cell from 1 to 6
     +/          ⍝ sum every row
   ⌸            ⍝ for every unique sum
,∘≢              ⍝ get the sum and the number of indexes

---

Previous post:

APL, 36 31 bytes

5 bytes saved thanks to @Adám

(11⍴⍉⌽f)[⍋11⍴⍉f←,∘≢⌸+/?100 2⍴6]

Explanation

f←,∘≢⌸+/?100 2⍴6
          100 2⍴6    ⍝ create an 2×100 array of 6
         ?           ⍝ roll for each cell from 1 to 6
       +/            ⍝ sum every row
     ⌸              ⍝ for every unique sum
  ,∘≢                ⍝ get the sum and the number of indexes

(11⍴⍉⌽f)[⍋11⍴⍉f]  ⍝ ⍋x returns the indexes of the sorted x in the current x  
                     ⍝ x[y] find the yth elements of x
                     ⍝ x[⍋y] reorders x the same way that would be required to sort y

            11⍴⍉f   ⍝ the column of sums - see below
 11⍴⍉⌽f            ⍝ the column of counts - see below

How does 11⍴⍉⌽f works?

⍝ ⌽ - Reverses the array
⍝ ⍉ - Transposes the array

  ⍝   f
 9 14   ⍝ Sum - Occurences
 4  9
 7 17
 8 18
 6 15
 5  7
10  3
11  5
 3  6
 2  2
12  4

  ⍝   ⍉f
 9 4  7  8  6 5 10 11 3 2 12  ⍝ Sum
14 9 17 18 15 7  3  5 6 2  4  ⍝ Occurences

  ⍝   ⍉⌽f
14 9 17 18 15 7  3  5 6 2  4  ⍝ Occurences
 9 4  7  8  6 5 10 11 3 2 12  ⍝ Sum

Answer 17

><>, 93 bytes

00[0[v
v 1\v/4
v 2xxx5
v 3/^\6
>l2(?^+]laa*=?v0[
  /&1+&\ v1&0]<
=?/ :?!\}>:@@:@
oa&0n&}< ^+1

Try it online, or watch it at the fish playground!

The ugly output format is a sequence of numbers separated by newlines, where the nth number says how many times the sum was n — it's ugly because it prints forever, for all positive integers n, although most of the lines will be 0. (The TIO link is modified to stop after n=12, at the cost of 5 bytes.)

The fish playground is fairly slow — it takes about three and a half minutes to print up to n=12 at top speed — so you may want to modify it to roll 10 pairs of dice instead of 100 by changing the aa* in the 5th line to a   (that is, a followed by two spaces).

The random dice rolls are done by this bit:

1\v/4
2xxx5
3/^\6

The xs change the fish's direction randomly. Assuming that's implemented with equal probabilities, it's clear that the die roll result is a uniform distribution by symmetry.

Once the fish has rolled 100 pairs of dice, it counts how many times the sum was n with this bit (unwrapped for clarity, and starting in the top left):

v       /&1+&\
>:@@:@=?/ :?!\}
^   +1oa&0n&}<

We keep n at the front of the stack, and use the register to count the number of times n appears.

Answer 18

Javascript 85 75 characters

Thanks Shaggy!

a=[]
for(i=100;i--;)a[o=(f=_=>Math.random()*6|0)()+f()]=(a[o‌​]|0)+1
alert(a)

History

85

a={}
f=_=>Math.random()*6
for(i=0;i++<100;)a[o=-~f()-~f()]=(a[o]||0)+1
console.log(a)

Answer 19

Perl 5, 64 bytes

map$s{2+int(rand 6)+int rand 6}++,1..100;say"$_ $s{$_}"for 2..12

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Output format:

<sum> <# rolls>

For sums with zero rolls, the rolls column is blank.

Answer 20

PHP, 65 bytes

while($i++<100)${rand(1,6)+rand(1,6)}++;for(;++$k<13;)echo+$$k,_;

prints a leading 0_ and then the occurences of 2 to 12, followed by an underscore each.
Run with -nr or try it online.

Answer 21

K (oK), 24 22 bytes

Solution:

+/(!11)=/:+/(2#100)?'6

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Explanation:

k 'port' of my q solution. Evaluation occurs right-to-left, hence brackets around the til (!)

+/(!11)=/:+/(2#100)?'6 / the solution
            (2#100)    / the list (100;100)
                   ?'6 / take 6 from each left/each right (roll the dice twice)
           +/          / sum rolls together
  (!11)                / til, performs range of 0..n-1, thus 0..10
       =/:             / equals each right (bucket the sum of the rolls)
+/                     / sum up to get counts per result

Edits:

• -2 bytes switching the each-left for an each-both, and the each-left + flip for each-right

Answer 22

Pyth, 21 bytes

V100aY,O6O6)VTlfqsTNY

Outputs each step in the creation of the rolls, then outputs frequency of each sum 0 - 10 on a separate line.

---

V100aY,O6O6)VTlfqsTNY Full program, no input, outputs to stdout
V100                  For N from 0 to 100
    a ,O6O6           Append a pair of random ints below 6
     Y                To a list Y, initialized to the empty list
           )          Then
            VT        For N from 0 to 10
              f     Y Print Y filtered to only include pairs
                q  N  For which N is equal to
                 sT   The sum of the pair

Answer 23

Java (OpenJDK 8), 95 bytes

a->{int r[]=new int[11],i=0,d=0;for(;i++<200;)r[d+=Math.random()*6]+=i%2<1?1-(d=0):0;return r;}

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Explanations

a->{
  int r[] = new int[11],     // Rolls or result
      i   = 0,               // Iteration
      d   = 0;               // Dice accumulator
  for (;i++<200;)
    r[d+=Math.random()*6] += // Accumulate a new die and start an addition
     i % 2 < 1               // Accumulate up to two dice
       ? 1 - (d = 0)         // If we're at 2 dice, reset the accumulator and add 1
       : 0;                  // If we only have one die, add 0
  return r;
}

Answer 24

Vyxal, 9 bytes

₁ƛ6℅6℅+;Ċ

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Returns a list of [roll, count]. If a roll total didn't happen, then it isn't included in the list.

Explained

₁ƛ6℅6℅+;Ċ
₁ƛ        # for each item in range(1, 100 + 1):
  6℅6℅    #   push a random number from range(1, 6 + 1) and then another random number from that range
      +   #   and add them together
       ;  # end map - this leaves 100 random dice rolls in a list.
        Ċ # get the count of each item

Answer 25

Burlesque, 18 bytes

1J6rn200.+2CO)++f:

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1J6rn # Infinite random d6 rolls
200.+ # Take first 200
2CO   # Chunks of 2
)++   # MapSum
f:    # Frequency list