Odds of rolling a $6$ in three chances I'm working on a dice game to play with the kids I work with. I'm trying to recall calculating probability but its been a while. Its for a table top DND type game and they are rolling for the properties of their character.
What are the odds of rolling a $6$ with a D6 given $3$ chances? They are trying to get a $6$ and they are given $3$ rolls to do so. 
I want to say its $50\%$ ($\frac16 + \frac16 + \frac16 = \frac36$) but it's been a while since I studied this. Am I on the right track?
Thanks for your time.
 A: You can't simply add the three $1/6$ together because you neglect the probability, in that case, of rolling more than one $6$.  That expression will, however, give you the expected number of $6$'s, though, which is indeed $1/2$.
The way to derive the probability of rolling at least one $6$ is a bit backward, but comes naturally once one gets used to it.  We first calculate the probability of rolling no $6$ at all.  Assuming the die rolls are independent, as we usually do, the probability of rolling a die three times and getting no $6$ is
$$
P(\text{no $6$ in three rolls}) = \left(\frac{5}{6}\right)^3 = \frac{125}{216}
$$
so the probability of getting at least one $6$ is
$$
P(\text{at least one $6$ in three rolls}) = 1-P(\text{no $6$ in three rolls})
    = \frac{91}{216}
$$
A: You can most easily calculate this with the counter probability.
What is the probability of rolling no 6. 
$$P(\text{no } 6) = \frac 56 \frac 56 \frac 56=\frac{125}{216}$$
Then the probability of rolling at least one 6 (which I think is the outcome you're interested) is 
$$P(\text{ at least one 6}) = 1 -\left(\frac 56\right)^{3} = 1- \frac{125}{216} = \frac{91}{216} \approx 42\%. $$
Note that your calculation is incorrect as you would get one 6 for sure if you roll 6 times. But this is obviously incorrect when you played your favorite board game.
A: It's easier to calculate the chances of not obtaining any $6$ in three rolls. The probability of not otaining a $6$ in a roll is $5/6$, so in three rolls you have
$$5/6\cdot 5/6 \cdot 5/6= \sim 0.58$$
So the probablity of not obtaining any $6$ is $\sim 58 \% $, hence the probability of obtaining a $6$ is  $\sim 42 \%$.
