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I looked into all the probabilities of two dice. Usually it's just 6 x 6; when I listed out all of the combinations, I am confused about why you don't count when the two die have the same number twice.

List of all of the dice combinations:

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

There are a total of 21 combinations; I assumed that you double 21 to get 42.

For example if I call the first dice "dice_1" and the second dice "dice_2" and I take a scenario when 3 and 5 are rolled, dice_1 can yield a 3 and dice_2 would yield a 5, but dice_1 can also yield a 5 which would make dice_2 yield a 3. This would suggest that I double all of the combinations in the above list.

Why don't I double the scenario when two of the same numbers (e.g. 1,1 or 2,2) are rolled?

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    $\begingroup$ Consider cases when dice are of different vs same color $\endgroup$ – Alex Jun 13 '17 at 17:17
  • $\begingroup$ Look up the difference between combinations and permutations, that might help. $\endgroup$ – Bobson Dugnutt Jun 13 '17 at 17:19
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You're correct: They do double—let's re-write the first few lines of your list considering two different die. Let's say I have $\text{Die A}$ and $\text{Die B}$. For the first few combinations, we have:

$(1_A, 1_B), (1_A, 2_B), (1_A, 3_B), (1_A, 4_B), (1_A, 5_B), (1_A, 6_B) $

$(1_A, 1_B), (2_A, 1_B), (3_A, 1_B), (4_A, 1_B), (5_A, 1_B), (6_A, 1_B)$

$(2_A, 2_B), (2_A, 3_B), (2_A, 4_B), (2_A, 5_B), (2_A, 6_B)$

$(2_A, 2_B), (3_A, 2_B), (4_A, 2_B), (5_A, 2_B), (6_A, 2_B)$

... Notice the first combination in each line, though. They're duplicates of each other. There's only one way for both dice to have the same numbers. Usually, you can switch the numbers. $(2_A, 3_B)$ and $(3_A, 2_B)$ exist as a pair where the numbers are switched. If you try to switch the numbers in $(1_A, 1_B)$, though, you just get... $(1_A, 1_B)$—so there's only a single way for both dice to get the same number.

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Well, in your presentation, 1,2 can be obtained by throwing a 1 and then 2 or a 2 and then a 1 so two ways but 1,1 can only be obtained by throwing a 1 and then another 1.

Write your table as

1,1; 1,2; 1,3; 1,4; 1,5; 1,6

2,1; 2,2; 2,3;

...

...

and no doubling.

It might be a bit clearer why there are 36 cases.

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Let’s color the dice red and green instead. The green die showing a 5 and the red a 3 is clearly different from having a green 3 and red 5, so these are two different events. On the other hand, there’s only one way for both dice to show a 3.

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