An $N$ -sided die

This is an example of the famous "Coupon Collector's Problem" .

The expectation value for "collecting" $N$ different objects (or in this case numbers on a die) is given by:

$E(N) = N(1/1 + 1/2 + ... + 1/N)$

This gives $E(\boxed{24}) = 90.62$ . ( $91$ when rounded to the nearest integer)

Here is the code so that you can simulate the "coupon collector problem" using monte-Carlo simulations.

EXPLANATION :
1. This code gets a random objects from the set S == {1, 2, 3 ... . totaldistinctobjects }.
2. And then do the above line numberofsimulations times.
3. For each unique value that we got we add the value to a set s.
4. Remember a set contains only unique keys.
5. If the number of unique elements in the set is equal to totaldistinctobjects then we found a way to reach the state of getting N unique elements.
6. Finally lets say that X is a random variable and X defines the number of ways to get all N unique elements.
7. Now I define this similar to the number of coin tosses required to get the first head problem, like this : Number of simulations needed to find a way to get all N unique elements. So E[X] = 1/(probability of getting all N unique elements) .

Results

CODE STARTS

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14

def simulate(numberofsimulations, totaldistinctobjects):
    total = 0
    found = 0
    s = set()
    for i in range(numberofsimulations):
        output = random.randint(1, totaldistinctobjects)
        s.add(output)
        total += 1
        if len(list(s)) == totaldistinctobjects:
            found += 1
            # clear the set
            s.clear()
   # return the expected value.  
return float(total)/float(found)