So this is a question with its solution below to which I don't understand 2 things. How is P(B) derived? And, why is $P(D_i)$=55/72 and not $(55/72)^i$. Since, for example, obtaining heads in the n trial, the probability is $(1/2)^n$, since every time we flip, we multiply by 1/2 and we do this n times. Am I right?
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(1) How is $P(B)$ derived?
According to the definition of $B$ these are the possibilities:
$ 2,2,1\\ 3,3,2 \text{ | }1\\ 4,4,3 \text{ | } 2,1\\ 5,5,4 \text{ | } 3,2,1\\ 6,6,5 \text{ | } 4,3,2,1 $
Altogether $15$. However, $(2,2,1)$ can also be $(1,2,2)$ or $(2,1,2)$. Similarly for other triplets. Thus $15$ needs to be multiplied by $3$, hence $45$ cases out of $216$.
(2) Why is $P(D_i)=55/72$ and not $(55/72)^i$?
It's explained in the answer above -- any throwing of three dice is independent of the previous ones, as if the previous trials didn't happen three dice are thrown again, hence the probability is the same in every trial.
$B$ is the event that two from the three die show the same face, and that that facing is higher than that of the remaining die. Now there are $3$ ways we can select two from three die, and for each possible face we count the ways the remaining die can have a lower face.
$$\begin{align}\mathsf P(B)~&=~\dfrac{\dbinom 32\sum\limits_{k\in\{6,5,4,3,2\}}(k-1)}{6^3}\\[2ex]&=~\dfrac{3\times(5+4+3+2+1)}{216}\\[2ex]&=~\dfrac{3\times((5)+(4+1)+(3+2))}{216}\\[2ex]&= \dfrac{3\times 3\times 5}{216}\end{align}$$
$D_i$ is the event that trial $i$ results in a decision, assuming the trial does take place.
$(55/72)^i$ is the probability that this result occurs in all of the first $i$ trials.
On the other hand $(17/72)^{i-1}$ is the probability that the result does not occur in all of the first $i-1$ trials. Thus $(17/72)^{i-1}(55/72)$ is the probability that the first decision occurs in trial $i$.