Probability question regarding sequential events 
There are 3 players $A, B, C$ and they throw a die sequentially until somebody wins. $A$ wins if he scores either $5$ or $6$. $B$ wins if the die is an even number and $C$ wins if he scores an odd number.
Find the probability that player $C$ throws the dice at least 3 times. 

Solution:
Let $P_i =$ {Player $P$ wins after throwing the die the i-th time}
$D =$ {Player $C$ throws the die at least 3 times}.
$\bar D =${Player $C$ throws the die 2 times at most}.


*

*$C$ throws the die exactly $0$ times, which is:


$A_1\cup \bar{A}_1B_1$


*

*$C$ throws the die exactly $1$ time, which means that:
Player C wins during the first round or Player A or Player B win the second round, i.e:


$\bar{A}_1 \bar{B}_1 C_1 \cup \bar{A}_1 \bar{B}_1 \bar{C}_1 A_1 \cup\bar{A}_1 \bar{B}_1 \bar{C}_1 \bar{A}_2B_2$
The same argument is for the third case, where Player C throws the die exactly 2 times. 
However, the answer of the textbook is that for the event $\bar{D}$ when we have:
$
\bar{D} = A_1 + \bar{A}_1 B_1 +\bar{A}_1\bar{B}_1C_1 +\bar{A}_1 \bar{B_1}\bar{C}_1\bar{A}_2 \bar{B_2}C_2 + \bar{A}_1 \bar{B_1}\bar{C}_1\bar{A}_2 \bar{B_2}\bar{C}_2 A_3 + \bar{A}_1 \bar{B_1}\bar{C}_1\bar{A}_2 \bar{B_2}\bar{C}_2 \bar{A}_3B_3
$
Why is that the case?
 A: Lulu is definitely correct about the best way to answer this. The book's method is needlessly complicated. But let's examine it anyway:
$$\bar{D} = A_1 + \bar{A}_1 B_1 +\bar{A}_1\bar{B}_1C_1 +\bar{A}_1 \bar{B_1}\bar{C}_1\bar{A}_2 \bar{B_2}C_2 + \bar{A}_1 \bar{B_1}\bar{C}_1\bar{A}_2 \bar{B_2}\bar{C}_2 A_3 + \bar{A}_1 \bar{B_1}\bar{C}_1\bar{A}_2 \bar{B_2}\bar{C}_2 \bar{A}_3B_3$$
We are calculating all the ways that $C$ doesn't throw $3$ times - i.e., all the ways that the game can end before his third throw.


*

*$A_1 = A$ wins on the first throw.

*$\bar A_1B_1 = B$ wins his first throw (after $A$ loses his)

*$\bar{A}_1\bar{B}_1C_1 = C$ wins his first throw (after $A, B$ both lose).

*$\bar{A}_1 \bar{B_1}\bar{C}_1\bar{A}_2 \bar{B_2}C_2 = C$ wins his 2nd throw.

*$\bar{A}_1 \bar{B_1}\bar{C}_1\bar{A}_2 \bar{B_2}\bar{C}_2 A_3 = A$ wins on his 3rd throw.

*$\bar{A}_1 \bar{B_1}\bar{C}_1\bar{A}_2 \bar{B_2}\bar{C}_2 \bar{A}_3B_3 = B$ wins on his 3rd throw.


Note that there is an obvious jump in the progression: the contributions from $A$ or $B$ winning on their 2nd throw are missing.
So the answer to your question "why is that the case?" is: Because your book is wrong. 
Probably somebody skipped a couple terms accidently while formatting this horrendous equation.
