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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.

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    $\begingroup$ Do the answers to this older question help? math.stackexchange.com/questions/2240392/… $\endgroup$
    – Especially Lime
    Commented May 16 at 9:13
  • $\begingroup$ 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 $\{1,1\}$. $\endgroup$
    – Henry
    Commented May 16 at 9:23
  • $\begingroup$ Yes, the answers to the older question definitely helped. My apologies for not going over that before asking my question. $\endgroup$
    – Mitansh
    Commented May 16 at 9:27

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