Still not as hard as Dark Souls

I will define $P(B|A)$ to be the probability that a player eventually passes checkpoint $B$ , given that they are starting from checkpoint $A$ . All other conditional probabilities that I list in this solution will be defined in a similar way.

When player starts at checkpoint $A$ , he will either pass checkpoint $B$ with $\frac{1}{3}$ probability, get another chance with $\frac{1}{3}$ probability, or quit. Thus, $P(B|A)$ can be defined recursively as:

$P(B|A)=\frac{1}{3}+\frac{1}{3}P(B|A)$

Solving this equation for $P(B|A)$ yields $P(B|A)=\frac{1}{2}$ .

Now that the player is at checkpoint $B$ , he will attempt to pass checkpoint $C$ . The player will either pass checkpoint $C$ with $\frac{1}{3}$ probability, get sent back to a previous checkpoint with $\frac{1}{3}$ probability, or quit. If he gets sent back to a previous checkpoint, then he will either get sent back to checkpoint $A$ with $\frac{1}{2}$ probability or checkpoint $B$ with $\frac{1}{2}$ probability. Once again, we will use recursion to define $P(C|B)$ :

$P(C|B)=\frac{1}{3}+\frac{1}{3}\big[\frac{1}{2}P(C|A)+\frac{1}{2}P(C|B)\big]$

Using Bayes' Theorem, we can define $P(C|A)$ :

$P(C|A)=P(B|A)\times P(C|B)=\frac{1}{2}P(C|B)$

We can then substitute and solve for $P(C|B)$ . This yields $P(C|B)=\frac{4}{9}$ and $P(C|A)=\frac{2}{9}$

Once a player passes checkpoint $C$ , he will attempt to pass checkpoint $D$ . This equation is set up in much the same way before. The only exception is that if the player is sent back, he has an equal $\frac{1}{3}$ chance to be sent back to $A$ , $B$ , or $C$ :

$P(D|C)=\frac{1}{3}+\frac{1}{3}\big[\frac{1}{3}P(D|A)+\frac{1}{3}P(D|B)+\frac{1}{3}P(D|C)\big]$

We can use Bayes' Theorem to get $P(D|A)$ and $P(D|B)$ in terms of $P(D|C)$ :

$P(D|A)=\frac{2}{9}P(D|C)$

$P(D|B)=\frac{4}{9}P(D|C)$

We can then solve for $P(D|C)$ . This yields $P(D|C)=\frac{9}{22}$ . Then we can obtain $P(D|A)=\frac{2}{9}\times\frac{9}{22}=\frac{1}{11}$ . Thus the answer to the problem is $1+11=12$ .