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?

The second situation is easiest, so let's start there. In the second situation, you throw two six-sided dice. There are 6*6=36 different outcomes (for example [1,6] and [6,1] are two different outcomes), and six of them have a sum of 7. So  6/36 = 1/6 outcomes have a sum of 7, as you said.
In the first situation, the situation is unusual.  Instead of picking out of the 6*6=36 different outcomes, you're supposed to pick from the set of outcomes where you can't tell the difference between the two dice. That is, [1,6] and [6,1] are considered to be the same event.
There are 21 different pairings in this case (not 23?): {1,1} ... {1,6}, {2,2}, ..., {2,6}, {3,3},...{3,6}, ..., ..., {5,6}, {6,6}. In the physically realistic case, the probability of each pairing is not the same. For example, when rolling dice it's twice as likely to get {2,4} than to get {3,3} because there are two of the 36 outcomes yield {2,4} and only one yields {3,3}.  In a physically unrealistic case, you can imagine a model where you choose one of these pairings uniformly at random. There are only three of those {1,6}, {2,5}, {3,4}, so the chances of getting a sum of 7 are 3/21 = 1/7 in this physically unrealistic case.
