I wasn't taking what you said literally; I was merely trying to point out a flaw in your logic. That's all
Anyways...
Original Question:
5) Suppose a dealer for a casino rolls two standard dice. You saw one of the dice; it was a 2. What is the chance that the sum of the face value of the two dice adds up to 6?
I'd prefer to calculate things in my head, but as they always say, "When in doubt, list them all out." Of course, this problem has a small sample space so listing them is reasonable.
So the question says that two dice are rolled together and "you saw one of the dice; it was a 2." Since there 2 dice, and there 6 possible values on each die, then there are 36 possible outcomes. I don't think anyone will disagree on that, for it's just basic math. So here are sample space:
{1 1} {2 1} {3 1} {4 1} {5 1} {6 1}
{1 2} {2 2} {3 2} {4 2} {5 2} {6 2}
{1 3} {2 3} {3 3} {4 3} {5 3} {6 3}
{1 4} {2 4} {3 4} {4 4} {5 4} {6 4}
{1 5} {2 5} {3 5} {4 5} {5 5} {6 5}
{1 6} {2 6} {3 6} {4 6} {5 6} {6 6}
Since you saw a die with a "2", then one of these 11 outcomes must have occured:
{1 1}
{2 1} {3 1} {4 1} {5 1} {6 1}
{1 2} {2 2} {3 2} {4 2} {5 2} {6 2} {1 3}
{2 3} {3 3} {4 3} {5 3} {6 3}
{1 4}
{2 4} {3 4} {4 4} {5 4} {6 4}
{1 5}
{2 5} {3 5} {4 5} {5 5} {6 5}
{1 6}
{2 6} {3 6} {4 6} {5 6} {6 6}
And from those 11 possible outcomes, only 2 of them add up to 6.
{1 1}
{2 1} {3 1} {4 1} {5 1} {6 1}
{1 2} {2 2} {3 2} {4 2} {5 2} {6 2} {1 3}
{2 3} {3 3} {4 3} {5 3} {6 3}
{1 4}
{2 4} {3 4} {4 4} {5 4} {6 4}
{1 5}
{2 5} {3 5} {4 5} {5 5} {6 5}
{1 6}
{2 6} {3 6} {4 6} {5 6} {6 6}
That is why your answer of 1/6 is incorrect. The problem is very counterintuitive, just like the Monty Hall problem. But I have to give it to you, you have good debate skills.