Probability of $n$ times a $\frac1n$ event I never studied probability at school and this problem has been bothering me for a long time:
Let's say I have a perfectly fair die. If I roll it, the odds of it landing on $6$ are $\frac{1}{6}$. If I roll two dice, the odds of at least one of them landing on 6 are $\frac{1}{6}\times 2 =\frac{1}{3}$.
But what about if I roll six dice? What are the odds that one will land on $6$? Based on the previous reasoning, it should be:
$$\frac{1}{6}\times 6 = 1$$
But that can't be true. It's actually possible that I roll six dice and none of them land on 6. What about if I roll $100$ dice? It's still possible that none of the land on 6. So what are the odds that at least one will?
 A: 
If I roll two dice, the odds of at least one of them landing on 6 are $\frac16×2=\frac13$

Not true - these are easily proved by thinking about it the other way - what are the odds that neither die lands on 6?
That is simply $$\frac56 \cdot \frac56 = \frac{25}{36}$$
So the odds that either of them lands on 6 is $$1- \frac{25}{36} = \frac{11}{36}$$ which is slightly less than $\frac13$.
The next example works the same way, the odds that none of the 6 dice land on 6 is
$$\frac56 \cdot \frac56 \cdot \frac56 \cdot \frac56 \cdot \frac56 \cdot \frac56 = \frac{5^6}{6^6} \approx0.33489$$
So the odds that any of the dice lands on 6 is the complement: about 66.5%.
A: The probability of rolling at least one 6 is $1$ - the probability of rolling no sixes.  This is then
$$1-\left (\frac{5}{6}\right)^6 \approx 0.665$$
This works for any number of dice.
A: 
Let's say I have a perfectly fair die. If I roll it, the odds of it landing on 6 are 1/6. If I roll two dice, the odds of at least one of them landing on 6 are 1/6×2=1/3.

The chance of rolling the first 6 is 1/6, including the following cases (1/36 chance each):
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
The chance of rolling the second 6 is also 1/6, which counts the following cases:
(1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (6, 6)
Using this method, you've actually counted the odds of rolling two 6's twice.
This overcounting gets worse with three+ dice: (6, 6, 2) (6, 4, 6) (3, 6, 6), (6, 6, 6), etc.
The easiest way is to find the complement probability (odds of rolling no 6's) and subtracting that from 1 to find your answer with 6 dice:
$$1-(\frac{5}{6})^6 = 1 - \frac{15625}{46656}= \frac{31031}{46656} \approx 66.5\%$$
Or you could count the cases of rolling a 6, two 6's, three 6's, ... six 6's. But that's too much casework.
A: As you have discovered, probability of independent events is not additive. If it was, then by rolling 10 dice, you'd have a probability of getting a 6 as $10\cdot \frac{1}{6} > 1$. Clearly this is false, since you could get $\{1,1,1,1,1,1,1,1,1,1\}$.
There are two ways of looking at this: either you count the number of events where no 6s come up and subtract this from the total number of events, or you work the variables as separate events:
$$P(\textrm{at least one 6}) = P(\textrm{first die is a 6}) + P(\textrm{first die is not a six AND second die is a six}) \\
= \frac{1}{6} + \frac{5}{6}\cdot \frac{1}{6} = \frac{6}{36}+\frac{5}{36} = \frac{11}{36}.$$
A: If you list the $36$ possibilities and count, you will find the chance of getting at least one $6$ in two rolls is $\frac {11}{36}$.  Your calculation of $2 \cdot \frac 16$ gives the expected number of sixes, but you get two of them at once when you roll $12$, so only eleven different rolls have at least one six.  Similarly, when rolling six dice, on average you will get one six.  Sometimes you will get more, sometimes less.
A: Bernoulli's trials is the best solution. For an event occurring $n$ times , if  '$p$' gives the probability of  a success and '$q$' gives the probability of a failure, then the probability of the event occurring '$r$' times is given by
$P(r) =  \binom{n}{r}p^rq^{n-r}$
