Rolling two dice: different probabilities Today my professor said that if I roll two identical dice at the same time then there will be $21$ outcomes and the probability of getting sum of $7$ is $1/7$:
\begin{array}{c c c c c c}
\{1,1\}, & \{1,2\}, & \{1,3\}, & \{1,4\}, & \{1,5\}, & \{1,6\},\\
& \{2,2\}, & \{2,3\}, & \{2,4\}, & \{2,5\}, & \{2,6\},\\
& & \{3,3\}, & \{3,4\}, & \{3,5\}, & \{3,6\},\\
& & & \{4,4\}, & \{4,5\}, & \{4,6\},\\
& & & & \{5,5\}, & \{5,6\},\\
& & & & & \{6,6\}
\end{array}
but when I roll two different dice, then the probability of getting sum of 7 is $1/6$
\begin{array}{c c c c c c}
(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6)\\
(2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6)\\
(3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6)\\
(4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6)\\
(5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6)\\
(6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6)
\end{array}
How this is possible?

If you roll two dice then there are $6*6=36$ possible outcomes.
6 of those outcomes add up to 7.
So yes , the probability of rolling the sum of 7 is $\frac{6}{36}=\frac{1}{6}$
But in your first statement you considered $(2,1)$ and $(1,2)$ as one outcome. These are Different possible outcomes. If you consider those as one outcome then there are 21 possible outcomes. In other words, there are 21 different possible number combinations that the dice could show. But in reality you must treat these separately to calculate probabilities.
