Occupancy Probability 12 Sided Die
(Dodecahedron)



For this problem, we'll calculate the probabilty of getting all 12 numbers after rolling a 12-sided die 35 times.

(Incidentally, the 12 sided die shape is called a dodecahedron. You have probably seen this shape in the form of a plastic desk calendar with one month printed on each of the 12 faces.)


STEP 1
The set of numbers on the die (1 through 12), would be called called "n" and the number of trials or attempts is called "r" which in this case is 35 rolls.
Calculating the value of "n" raised to the power of "r":

1235
which equals
5.91 × 1037

This number represents all the possible results from rolling a 12 sided die

In steps 2, 3, 4 and 5, we will determine how many of those 1235 rolls, will contain all 12 numbers.

STEP 2
We first must calculate each value of "n" raised to the power of "r".
Rather than explain, this is much easier to show:

1235 = 5.91 × 1037
1135 = 2.81 × 1036
1035 = 1.00 × 1035
935 = 2.50 × 1033
835 = 4.06 × 1031
735 = 3.79 × 1029
635 = 1.72 × 1027
535 = 2.91 × 1024
435 = 1.18 × 1021
335 = 5.00 × 1016
235 = 3.44 × 1010
135   =   1  


STEP 3
Next, we calculate how many combinations can be made from "n" objects for each value of "n".
This is much easier to show than explain:

12 C 12 = 1
11 C 12 = 12
10 C 12 = 66
9 C 12 = 220
8 C 12 = 495
7 C 12 = 792
6 C 12 = 934
5 C 12 = 792
4 C 12 = 495
3 C 12 = 220
2 C 12 = 66
1 C 12 = 12

          Basically, this is saying that
12 objects can be chosen from a set of 12 in 1 way
11 objects can be chosen from a set of 12 in 12 ways
10 objects can be chosen from a set of 12 in 66 ways
9 objects can be chosen from a set of 12 in 220 ways
8 objects can be chosen from a set of 12 in 495 ways
7 objects can be chosen from a set of 12 in 792 ways
6 objects can be chosen from a set of 12 in 934 ways
5 objects can be chosen from a set of 12 in 792 ways
4 objects can be chosen from a set of 12 in 495 ways
3 objects can be chosen from a set of 12 in 220 ways
2 objects can be chosen from a set of 12 in 66 ways
1 object can be chosen from a set of 12 in 12 ways


STEP 4
We then calculate the product of the first number of STEP 2 times the first number of STEP 3 and do so throughout all 12 numbers.

  5.91 × 1037     ×   1     = 5.91 × 1037
    2.81 × 1036   ×   12     = 3.37 × 1037
      1.00 × 1035     ×   66     = 6.60 × 1036
    2.50 × 1033   ×   220   = 5.51 × 1035
    4.06 × 1031   × 495     = 2.01 × 1034
    3.79 × 1029   × 792     = 3.00 × 1032
    1.72 × 1027   × 934     = 1.59 × 1030
    2.91 × 1024   × 792     = 2.31 × 1027
    1.18 × 1021   × 495     = 5.84 × 1023
    5.00 × 1016   × 220     = 1.10 × 1019
    3.44 × 1010   × 66     = 2.27 × 1012
  1     ×   12     = 12


STEP 5
Then, alternating from plus to minus, we sum the 12 terms we just calculated.

    + 5.91 × 1037
    - 3.37 × 1037
    + 6.60 × 1036
    - 5.51 × 1035
    + 2.01 × 1034
    - 3.00 × 1032
    + 1.59 × 1030
    - 2.31 × 1027
    + 5.84 × 1023
    - 1.10 × 1019
    + 2.27 × 1012
        - 12
Total 3.14 × 1037

which equals the total number of ways you can roll a 12 sided die 35 times and have all 12 numbers appear.



STEP 6
So, if we take the number 3.14 × 1037 and divide it by
5.91 × 1037 (all posiible results of 35 rolls of a 12 sided die)
we get the probability of having all 12 numbers appearing after 35 rolls.

3.14 × 1037 ÷ 5.91 × 1037 = 0.53182114905432

So, if you had a 12 sided die, you would need to roll it at least 35 times in order to have a better than 50 / 50 chance of rolling all 12 numbers.


Click here to see the probabilities of a:

4 Sided Die
6 Sided Die
8 Sided Die
20 Sided Die


Return To Home Page

Copyright © 1999 -

1728 Software Systems