Why does this not seem to be random? I was running a procedure to be like one of those games were people try to guess a number between 1 and 100 where there are 100 people guessing.I then averaged how many different guesses there are.
from random import randint

def averager(times):
    tests = list()
    for _ in range(times):
        l = [randint(0,100) for _ in range(100)]
        tests.append(len(set(l)))
    return sum(tests)/len(tests)

print(averager(100))

For some reason, the number of different guesses averages out to 63.6
Why is this?Is it due to a flaw in the python random library?
In a scenario where people were guessing a number between 1 and 10
The first person has a 100% chance to guess a previously unguessed number
The second person has a 90% chance to guess a previously unguessed number
The third person has a 80% chance to guess a previously unguessed number
and so on...
The average chance of guessing a new number(by my reasoning) is 55%.
 But the data doesn't reflect this.
 A: Suppose that  $n$ guesses are made. For $i=1$ to $100$, let $X_i=1$ if $i$ is not guessed, and  let $X_i=0$ otherwise. If 
$$Y=X_1+X_2+\cdots +X_{100},$$
then $Y$ is the number of numbers not guessed.  
By the linearity of expectation, we have
$$E(Y)=E(X_1)+E(X_2)+\cdots+E(X_{100}).$$
The probability that $i$ is not chosen in a particular trial is $\frac{99}{100}$, and therefore the probability it is not chosen $n$ times in a row is $\left(\frac{99}{100}\right)^n$. Thus
$$E(Y)=100  \left(\frac{99}{100}\right)^n.$$
In particular, let $n=100$. Note that $\left(1-\frac{1}{100}\right)^{100}\approx \frac{1}{e}$, so the expected number not guessed is approximately $\frac{100}{e}$. Thus the expected number guessed is approximately $63.2$, a result very much in line with your simulation. 
In general, if $N$ people choose independently and uniformly from a set of $N$ numbers, then the expected number of distinct numbers not chosen is 
$$N\left(1-\frac{1}{N}\right)^N.$$
Unless $N$ is very tiny, this is approximately $\frac{N}{e}$, and therefore the expected number of distinct numbers chosen is approximately $N-\frac{N}{e}$. Note that the expected proportion of the numbers chosen is almost independent of $N$. 
A: This is an example of the Birthday Paradox / Birthday Problem.
Birthday problem - Wikipedia, the free encyclopedia
I was just looking at my Online Cryptography class video lecture today on this very problem.
Coursera.org: crypto-009
There is an apparent paradox that there is more duplication of numbers than expected when the random numbers are supposedly independent. 
But the Birthday Paradox is just one example of when our intuitive statistical sense is dead wrong.
A: In your 1 to 10 example, it's not true in general that the third chooser has an 80% chance on choosing a new number. It's only the case if the second one has guessed a different number than the first one.
