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?
Best Answer
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