Consider Tom plays a Mario game. On any given try, he passes the first level with a probability of 2/5 (skipping a bonus level), he fails the level but passes with a bonus try with a probability of 1/5 , and he fails both with a probability of 2/5 . If he keeps playing until he passes the level or the bonus, what is the probability that he won the bonus?
As I am pretty new to random walks and Markov chains, I am having difficulty to understand how exactly I should compute it?
Is $P$ simply $(1-\frac{2}{5})\cdot\frac{1}{5} = \frac{3}{25}$ because it is the probability of passing the bonus level in a single try? It seems wrong to me because he has infinite tries, so computing the probability of a single try is absurd.
The logic says it is $3/25 + 3/25 + \cdots + 3/25$ which is obviously wrong. I cant think intuitively on how i should find this probability. Is it related somehow to hitting times or anything else? would like to get some help here, Thank you!
Intuitively, at any given turn, the probability that he passes the level is twice that he passes the bonus. This ratio holds for every attempt, so it will hold overall. That is, if $p$ is the probability that he passes the bonus, then the probability that he passes the level itself will be twice that, at $2p$. Further, since these are mutually exclusive events and he will play until one of them occurs, their sum is $1$. That is $$p + 2p = 1 \implies p = \frac13.$$
(This solution can be made formal by using conditional probability)
More explicitly, we can show that in order to finish with a bonus round, he must first lose both the level and the bonus $i = 0, 1, 2, \ldots$ times (note $i=0$ means he never loses both) with probability $(2/5)^i$ and then win the bonus round with probability $1/5$. Put together, this gives the probability of $$\sum_{i=0}^\infty \left(\frac25\right)^i\cdot \frac15 = \frac{\frac15}{1-\frac25} = \frac{\frac15}{\frac35} = \frac13$$