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

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 = 0.500073329650332Basically, 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.