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Tossing a pair of dice is equivalent to tossing one die after the other. The two experiments are independent of each other and hence we distinguish the first die from the other die. The reason you don't count (1,1) twice is because this represents the event that you get {1} on the 1st die and {1} on the 2nd die. Getting {1} on the second die and {1} on the first die (i.e. reverse) is exactly the same event. The case of (2,1) and (1,2) are counted separately because the former represents getting {2} in the 1st die and {1} in the 2nd die whilst the latter represents getting {1} in the 1st die and {2} in the 2nd die i.e. they are two completely different events.I still think the answer should be 5/7 though.
I don't see as to why (1,1) and it's reverse aren't counted as two seperate outcomes whilst (1,2) and (2,1) are.
What makes one situation different from the other?
The sample space here is 36 and perhaps the best way to see this other than drawing a table is to look conditional events.
Suppose we get {1} on the first die then we have the possible pair (1, x) where x = {1, 2, 3, 4, 5, 6}. Thus, GIVEN that we have {1} on the first die we have 6 possibilities.
Now repeat this with the case we get {2} on the first die then we have the pair (2, x) where x = {1, 2, 3, 4, 5, 6} so again there are 6 possibilities GIVEN that we {2} on the first die.
Now repeat this with the remaining possibilities we could get on the first die and it should be clear that the sample space has size 36, not 42.
If it helps, consider this (hopefully) more intuitive example. Consider two draws with replacement of the digits 0 to 9 to make a 'two digit number'. Intuition should tell you that you don't count '11' twice otherwise you are implying that making '11' is more likely than making say '21'. Also, by counting it should be clear that the sample space is size 100 and by symmetry it should also be clear that all 'two digit numbers' are equally likely.
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