Occupancy Probability Page 3
Our next calculation will be for 5 spins of a threenumbered roulette wheel.
Using our previous method, this would require listing 243 five digit numbers and then choosing how many of those contain 1, 2 and 3.
At this point, you can see that using this "bruteforce" method is very laborious and the possibility of making a mistake is extremely high.
Now, the calulations will be more mathemaicallyoriented.
STEP 1
First, we calculate the value of "n" (numbers on the roulette wheel) raised to the power of "r" (number of spins).
3^{5} = 243
This represents all possible outcomes when spinning a threenumbered roulette wheel five times.
STEP 2
Next, we calculate each value of "n" raised to the power of "r".
In this case "n" = 3 numbers on the roulette wheel and r = "5" spins.
(And, yes we've already calculated 3^{5}, so now we will calculate 2^{5} and 1^{5}.)
3^{5} = 243
2^{5} = 32
1^{5} = 1
STEP 3
Next, we calculate how many combinations can be made from "n" objects for each value of "n".
3 C 3 = 1
2 C 3 = 3
1 C 3 = 3
Basically, this is saying that
3 objects can be chosen from a set of 3 in 1 way
2 objects can be chosen from a set of 3 in 3 ways
1 object can be chosen from a set of 3 in 3 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 3 numbers.
3^{5} • 1 = 243
2^{5} • 3 = 96
1^{5} • 3 = 3
STEP 5
Then, alternating from plus to minus, we sum the 3 numbers we just calculated.
+ 243
 96
+ 3
TOTAL 150
which equals the total number of ways you can spin a three digit roulette wheel 5 times and have all 3 numbers appear.
STEP 6
So, if we take the number 150 and divide it by
243 (all posiible results of 5 spins of a 3 numbered roulette wheel)
we get the probability of having all 3 numbers appearing after 5 spins.
150 ÷ 243 = 0.617283950617284
So, if we had a roulette wheel with 3 numbers on it, we would need to spin it 5 times in order to have a 50% (or better) chance to get all 3 numbers to appear.
* * * * * * * * * * * * * * * * *
In the next 5 pages, we have calculated the occupancy probabilities of rolling dice that have 4, 6, 8, 12 and 20 sides.
(If you are wondering, these would be dice that are in the shape of the 5 Platonic Solids.)
(Tetrahedron, Hexahedron, Octahedron, Dodecahedron, Icosahedron)
Click here to see the probabilities of a:
4 Sided Die Probability of all 4 numbers in 7 Rolls
6 sided die Probability of all 6 numbers in 13 Rolls
8 Sided Die Probability of all 8 numbers in 20 Rolls
12 Sided Die Probability of all 12 numbers in 35 Rolls
20 Sided Die Probability of all 20 numbers in 67 Rolls
