Don't be Caught Out with Probability

Let $A$ be the event of rolling at least one $6$ with 3 dice. Let $B$ be the event of NOT rolling at least one $6.$
The probability of NOT rolling at least one 6 is $P(B) = (\frac{5}{6})^3 = \frac{125}{216}.$
Since A and B are mutually exclusive events, $P(A) = 1 - P(B)$ $P(A) = \boxed{\frac{91}{216}}$

It isn't $\frac{1}{6} + \frac{1}{6}+ \frac{1}{6}=\frac{1}{2}$ because then what would happen with 7 dice? The probability would be $\frac{7}{6}$ , which is impossible. So we try a different approach.

We take the probability of NOT getting a six with one throw: $\frac{5}{6}$ . Now, the probability of NOT getting a single six with 3 throws will be $(\frac{5}{6})^3=\frac{125}{216}$ . But we're not done! This is the complement: the probability of the event P(at least one six)= $1-\frac{125}{216}=\frac{91}{216}$ .

Python:

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from itertools import product
from fractions import Fraction
dice = xrange(1,7)
six_throws = 0
trials = 0
for throw in product(dice, repeat=3):
    if 6 in throw:
        six_throws += 1
    trials += 1
print Fraction(six_throws, trials)