I'm having trouble deciding when to take the material identical and when distinct. [duplicate]

Here is a question. Two fair dice are thrown, find the probability sum of outcomes is 11.

case 1: considering the dice to be distinct, (6,5) and (5,6) are the favourable outcomes out of total of 6x6=36 outcomes. P(E) = 2/36 = 1/18

Case 2: considering the dice to be identical, only 1 favourable outcome is (5,6). For total number of cases, 1 can link with 6 elements (1 to 6), however, 2 can now link with only 5 elements (2 to 6) as (1,2) has already been counted and as dice are identical (1,2) is same as (2,1). Hence, total number of cases are 6x7/2 = 21 Therefore, P(E) = 1/21 Can someone help me understand what is really going on here.

• Do the answers to this older question help? math.stackexchange.com/questions/2240392/… Commented May 16 at 9:13
• As the first comment by @lulu there says, the problem is that the $21$ cases are not equiprobable. $\{1,2\}$ is twice as likely as $\{1,1\}$. Commented May 16 at 9:23
• Yes, the answers to the older question definitely helped. My apologies for not going over that before asking my question. Commented May 16 at 9:27