Rolling dice game - who first will roll number 3 Let's consider very easy game with players A and B - they roll a dice starting with player A. If any of players roll a three, then he wins. I want to calculate probability that player B wins.
Intuition
Intuition is that $P(\textrm{player B wins}) < P(\textrm{player A wins})$
because they have even chances on winning, and player A starts, so player A has one more roll, therefore bigger chance to win. In other words player A is one roll ahead of player B so what should hold is that:
$$P(\textrm{player A wins}) = P(\textrm{player B wins}) + \frac 16$$
Out of this we can already calculate desire probability $P(\textrm{player B wins}) = \frac{5}{12}$
Normal approach
I want to calculate this normally (without any tricks) to compare the results. Please see the probability tree that I've created:

Out of this tree we can see that:
$$P(\textrm{B won}) = \frac{5}{6} \cdot \frac 1 6 + (\frac{5}{6})^2 \cdot \frac{5}{6} \cdot \frac 1 6 + (\frac{5}{6})^4 \cdot \frac 1 6 + ... = \sum_{n = 0}^\infty (\frac 5 6)^{2n}\frac{5}{6}\frac{1}{6} = $$
$$= \sum_{n = 0}^\infty(\frac{25}{36})^n\frac{5}{6}\cdot \frac 1 6 = \frac{1}{1 - \frac{25}{36}} \cdot \frac{5}{36} = \frac{36}{11} \cdot \frac{5}{36} = \frac{5}{11}$$
Question
As you can see those two probabilities differ. Second result also matches our intuition that $P(\textrm{player B wins}) < P(\textrm{Player A wins})$ but I want to ask you - which result is correct and where is the mistake with the wrong one?
 A: Denoting the ultimate probability of $A$ winning by $a$,
either $A$ wins immediately, or both fail and we are back to square $1$
So we have $a = \frac16 + \frac{5}6\frac56\cdot{a}$
which yields $a = \frac6{11},\;\; b = \frac5{11}$
You should be able to now spot the error in your first formulation
A: The first approach is not okay.
It would be okay if after a first roll that is not a three a new game starts with equal chances for both. However that is not the case. After such roll they go on with a game with advantage for player $B$.

Let $p$ denotes the probability that the player $A$ will win.
Then:$$p=\frac16+\frac56(1-p)$$leading to: $p=\frac6{11}$ which is in accordance with the second approach.
In the expression on RHS term $\frac16$ is the probability that the first roll is a three. If this does not happen (probability on that is $\frac56$) then in a certain sense a "new" game starts in which player $B$ is the starting player. So in that new game player $A$ has probability $1-p$ to win.
