I understand that the probability of the first die roll being larger then the second is 5/12. I'm having a harder time understanding why, according to a small problem I wrote, the probability of the first die being larger on the first roll then the next two rolls is still 5/12.
It would seem to me that these would be two events. p(a) = 5/12 and p(b) = 5/12. p(a and b) would then be p(a) * p(b) I would think, considering these are independent events?
On
The events A beats B and A beats C are not independent. For intuition, if the first roll turned out to be bigger than the second, it is likely to be biggish, giving it a better than $5/12$ probability of beating the third. So we need to calculate.
If A is $6$, the probability it beats B and C is $\frac{5^2}{6^2}$, for both B and C must be $\le 5$. If A is $5$, the probability it beats B and $C$ is $\frac{4^2}{6^2}$. And so on. So the probability A beats B and C is $$\frac{1}{6^3}\left(5^2+4^2+3^2+2^2+1^2\right).$$
The events are not independent: The higher the first roll, the higher the probability that it is larger than both subsequent rolls.
But your result is not correct anyway. If the first die shows $k$, then the probaility that both next two rolls are smaller is $\frac{(k-1)^2}{36}$. Hence in total this is $$ \frac16\cdot \frac{0^2+1^2+2^2+3^2+4^2+5^2}{36}=\frac{55}{216}$$