Secret Santa

In general, assume that there are $n$ people numbered $1,2,\ldots, n$ . Let $I_k$ denote the indicator variable for the event that the $k$ th friend chooses his own name. Clearly, $\mathbb{P}(I_k=1)=\frac{1}{n}, \hspace{5pt} \forall k$ Let the random variable $N$ denote the total number people choosing his own name. Clearly, $N=\sum_{k=1}^{n}I_k$ Taking expectation of both sides and using linearity of expectations , we have $\mathbb{E}N=\sum_{k=1}^{n}\mathbb{P}(I_k=1)=1$

Python 3.4:

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import random
from time import time
friends = list(range(6))
def no_gifts():
    hat = friends[:]
    count = 0
    for friend in friends:
        gift = random.choice(hat)
        if gift == friend:
            count += 1
        hat.remove(gift)
    return count
trials = 0
count = 0
end = time() + 10
while time() < end:
    count += no_gifts()
    trials += 1
print ("Answer:", count/trials)