When calculating probability for two standard die, 4,2 and 2,4 are treated as two separate outcomes. By that logic, why aren’t doubles treated the same way? One outcome for die A and B rolling 3,3 and another for die B and A rolling 3,3.
In: 5
Put colors on the dice. Say you have a green and a red one. A green 2 and a red 4 are different than a green 4 and red 2. But there is only one way to have a green 3 and a red 3.
Well simple because with two dice you can get 2 and 4 two ways. When the first one is 2 and the second is 4 or when the first one is 4 and the second is true. Now for doubles you can only get them 1 way when both are the same.
So we simply account for this by treating the dice independently as we should. So it’s not that 2,4 and 4,2 are different it’s just that you have more ways to get 2,4 than 3,3. Same with sums certain sums can be achieved by multiple throws like a 7 but 12 can only be achieved by double 6s.
They are treated separately when computing probabilities.
The probability of rolling a given face of a die is equal to the probability of rolling any other face if the die is fair. If you had one unfair die, you could account for this by weighting each of the enumerated combinations by the product of the probability of rolling each pair of faces and summing this to get the probability of rolling a given value (e.g. rolling 6, in this case).
Imagine the dice are actually laying out on the table: die A shows four pips and die B shows two. You could physically pick up A and turn it into a two, and physically pick up die B and turn it into a four.
Now do the same with a pair of threes: You cannot change either of them from a three to a different three. There is only one three on each die, and so there is only one actual result out of 36 potential ones where both roll a three.
**Edit:** If it helps to have a visual, [here](https://www.edcollins.com/backgammon/diceposs.gif) is an image showing every possible result to rolling two six-sided dice.
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