# In determining probability using 2 dice rolls why are permutations (x,x) not counted twice?

So I've been working in probability regarding dice rolls. I came across this problem:

If you roll 2 dice, what is the probability the first die is a 6 given that you rolled an 8?

This is clearly a conditional probability where E = {the event you roll an 8} and S = {event you roll a 6}. So the P(S) = 1/6. Then P(E) = 5/36. P(E) is calculated using the fact that there 36 possible permutations of rolls and 5 ways to rolls an 8 ({(2,6), (6,2), (3,5), (5,3), (4,4)}).

I'm just curious why (4,4) isn't counted twice? What is the logic behind this?

I'm assuming it has something to do with it just not being distinguishable? As in if I say dice 1 = 5 and dice 2 = 3 versus dice 1 = 3 and dice 2 = 5, there's a distinction, but saying dice 1 = 4 and dice 2 = 4 is not distinct form dice = 4 and dice 2 = 4, but this isn't satisfactory explanation for me.

• Each of the events, the five ways to get an $8$, have the same probability, namely $\frac 1{36}$. No reason to count any of them twice. – lulu May 25 '19 at 14:29

Are $$(2,6)$$ and $$(6,2)$$ different? Yes, indeed!
Now what about $$(4,4)$$? Nope!
The only way to get $$(4,4)$$ is if both dice show $$4.$$ By contrast, there are two distinct (if not distinguishable) ways for the dice to show a $$2$$ and a $$6,$$ which is why we count both $$(2,6)$$ and $$(6,2).$$