Rolling three dice - sum of two results equals third one Is my answer correct?

Three six-sided dice are thrown, what is the probability that the sum of two of them equals the third one?

If order does not matter, there are 9 such possibilities:


*

*(1, 1, 2)

*(1, 2, 3)

*(1, 3, 4)

*(1, 4, 5)

*(1, 5, 6)

*(2, 2, 4)

*(2, 3, 5)

*(2, 4, 6)

*(3, 3, 6)


The total number of possibilities is $\frac{6^3}{3!}=6^2=36$. Therefore, the probability is $\frac{9}{36}$.
 A: If you imagine there are three distinguishable dices (either different colors, or you throw them one after the other), then there are $6^3=216$ possible cases.
Now your enumeration counts down to :
$(1,1,2)$ : $3$ cases
$(1, 2, 3)$ : $6$ cases
$(1, 3, 4)$ : $6$ cases
$(1, 4, 5)$ : $6$ cases
$(1, 5, 6)$ : $6$ cases
$(2, 2, 4)$ : $3$ cases
$(2, 3, 5)$ : $6$ cases
$(2, 4, 6)$ : $6$ cases
$(3, 3, 6)$ : $3$ cases
for a total of $45$ cases, and a probability of $\frac{45}{216}=\frac{5}{24}$.
Hope I did it the right way :-(
A: You seem to have made a mistake by stating that each of these possibility is equiprobable.
Indeed, if you take into account the order of the dice, for (1,2,3) you have 6 possibilites : $(1,2,3)$, $(1,3,2)$, $(2,1,3)$, $(2,3,1)$, $(3,1,2)$ and $(3,2,1)$
However for $(1,1,2)$, you only have three possibilities which are $(1,1,2)$, $(1,2,1)$, $(2,1,1)$.
Therefore you need to consider the possibility of getting a roll with three different values (which is according to what I wrote ${6\over 6^3} = {1\over36}$) and one where two are the same (${3\over 6^3} = {1\over72}$).
Using what you stated, you get that the probability of getting a roll where two dice add up to the third is $3 {1\over 72} + 6 {1\over 36} = {5\over24}$
