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.

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.
