Q:

Probability function for rolling dice with conditions

I stumbled upon a problem when I was doing math homework.
The problem was this:
An unbiased (except for the condition) dice is rolled 3 times.
The probability that a 6 is rolled at least once is 100%.
What is the probability that a 3 is rolled at least once?
For me the answer was obvious:
Since I know that a 6 is rolled at least once, I effectively have only 2 tries to roll a 3. Because the probability to roll a 3 is 0 after 2 rolls, therefore the third roll doesn't concern me.
That means the probability would be $$\frac{1}{6} + \frac{5}{36} = \frac{11}{36} = 0.30\overline5 $$ 
Though my math teacher said this was wrong and I would have to draw a whole tree diagram. When I did that I got a different probability (${75\over216} = 0.347\overline2 $) but I don't really get why I would have to do that in the first place and why I get different results.
Sadly I cannot really draw a tree in here, so maybe this suffices. 
What am I missing?

A:

Temporarily assume the dice are different colors so that we can more easily assign probabilities by utilizing a sample space where outcomes are equally likely to occur so that we can use counting techniques to approach this.
Assuming the question is "What is the probability that at least one three was rolled given that at least one six was rolled" where it does not matter which color die specifically was a three or a six, etc...
Let $A$ be the event that we rolled at least one three.  Let $B$ be the event we rolled at least one six.  We are tasked with calculating $Pr(A\mid B)$.  To do this, we expand via definition: $Pr(A\mid B) = \dfrac{Pr(A\cap B)}{Pr(B)}$
Now, $Pr(B)=1-Pr(B^c)$ and $B^c$ is the event that none of the dice showed a six, giving us $Pr(B)=1-\left(\frac{5}{6}\right)^3=\frac{91}{216}$
Next, $Pr(A\cap B) = 1 - Pr(A^c\cup B^c) = 1 - Pr(A^c)-Pr(B^c)+Pr(A^c\cap B^c)  = 1 - \left(\frac{5}{6}\right)^3-\left(\frac{5}{6}\right)^3+\left(\frac{4}{6}\right)^3 = \frac{5}{36}$
We get then $Pr(A\mid B) = \frac{30}{91}\approx 0.32967$
The error again in your first approach was that you incorrectly assumed that it must have been the red die that was known to be a $6$ when it was possible that the red die could have been a number different than a six while one of the other dice showed a six.  There is seemingly an error in your second approach or you used an interpretation of the problem different than my own leading to yet another answer.  Without more information I cannot tell you exactly where or what you did that led you to a different answer.
The end result is that this is similar to the boy-girl paradox.  I suggest reading through that problem as well and understanding how the different interpretations of the problem lead to different answers.  It highlights the importance of clarity when describing a problem.

