Occupancy Probability 20 Sided Die
(Icosahedron)

For our next problem, let us calculate the probabilty of getting all 20 numbers after rolling a 20 sided die 67 times.

(Incidentally, a 20 sided die is called an icosahedron and they are used in many games and toys. For example, the "Magic 8-Ball" contains an icosahedron with 20 different "answers" written on each of its 20 faces.)

STEP 1
To determine the number of all the results of rolling a 20 sided die 67 times we raise "n" to the power of "r":

2067
which equals
1.48 × 1087

Then in steps 2, 3, 4 and 5, we will determine how many of those 2067 rolls, will contain all 20 numbers.

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

2067 = 1.48 × 1087
1967 = 4.75 × 1085
1867 = 1.27 × 1084
1767 = 2.75 × 1082
1667 = 4.74 × 1080
1567 = 6.28 × 1078
1467 = 6.17 × 1076
1367 = 4.31 × 1074
1267 = 2.02 × 1072
1167 = 5.93 × 1069
1067 = 1.00 × 1067
967 = 8.60 × 1063
867 = 3.21 × 1060
767 = 4.18 × 1056
667 = 1.37 × 1052
567 = 6.78 × 1046
467 = 2.18 × 1040
367 = 9.27 × 1031
267 = 1.48 × 1020
167 = 1

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

20 C 20 = 1
19 C 20 = 20
18 C 20 = 190
17 C 20 = 1,140
16 C 20 = 4,845
15 C 20 = 15,504
14 C 20 = 38,760
13 C 20 = 77,520
12 C 20 = 125,970
11 C 20 = 167,960
10 C 20 = 184,756
9 C 20 = 167,960
8 C 20 = 125,970
7 C 20 = 77,520
6 C 20 = 38,760
5 C 20 = 15,504
4 C 20 = 4,845
3 C 20 = 1,140
2 C 20 = 190
1 C 20 = 20

Basically, this is saying that
20 objects can be chosen from a set of 20 in 1 way
19 objects can be chosen from a set of 20 in 20 ways
18 objects can be chosen from a set of 20 in 190 ways
.......................................................................................

2 objects can be chosen from a set of 20 in 190 ways
1 object can be chosen from a set of 20 in 20 ways

STEP 4
We then calculate the product of the first calculation of STEP 2 times the first calculation of STEP 3 and do so throughout all 20 numbers.

1.48E+087   ×   1       = 1.48 × 1087
4.75E+085   ×   20     = 9.50 × 1086
1.27E+084   ×   190     = 2.41 × 1086
2.75E+082   ×   1,140     = 3.14 × 1085
4.74E+080   ×   4,845     = 2.30 × 1084
6.28E+078   ×   15,504     = 9.74 × 1082
6.17E+076   ×   38,760     = 2.39 × 1081
4.31E+074   ×   77,520     = 3.34 × 1079
2.02E+072   ×   125,970     = 2.54 × 1077
5.93E+069   ×   167,960     = 9.97 × 1074
1.00E+067   ×   184,756     = 1.85 × 1072
8.60E+063   ×   167,960     = 1.44 × 1069
3.21E+060   ×   125,970     = 4.05 × 1065
4.18E+056   ×   77,520     = 3.24 × 1061
1.37E+052   ×   38,760     = 5.30 × 1056
6.78E+046   ×   15,504     = 1.05 × 1051
2.18E+040   ×   4,845     = 1.06 × 1044
9.27E+031   ×   1,140     = 1.06 × 1035
1.48E+020   ×   190       = 2.80 × 1022
1   ×   20           = 20

STEP 5
Then, alternating from plus to minus, we sum the 20 terms we just calculated.

 + 1.48 × 1087 - 9.50 × 1086 + 2.41 × 1086 - 3.14 × 1085 + 2.30 × 1084 - 9.74 × 1082 + 2.39 × 1081 - 3.34 × 1079 + 2.54 × 1077 - 9.97 × 1074 + 1.85 × 1072 - 1.44 × 1069 + 4.05 × 1065 - 3.24 × 1061 + 5.30 × 1056 - 1.05 × 1051 + 1.06 × 1044 -1.06 × 1035 + 2.80 × 1022 - 20 Total 7.38 × 1086

which equals the total number of ways you can roll a 20 sided die 67 times and have all 20 numbers appear.

STEP 6
So, if we take the number 7.38 × 1086 and divide it by
1.48 × 1087 (all posiible results of 67 rolls of a 20 sided die)
we get the probability of rolling all 20 numbers appearing after 67 rolls.

When we divide 7.38 × 1086 by all posiible results of 67 rolls of a 20-sided die (1.48 × 1087), we get the probability of having all 20 numbers appearing after 67 rolls.

Probability = 7.38 × 1086 ÷ 1.48 × 1087 = 0.500073329650332

Basically, you would have to roll an icosahedron at least 67 times in order to have a better than 50 / 50 chance of rolling all 20 numbers.