A cricketer bats until he is out. For any given shot he takes, he either plays an attacking shot with probability $\frac{2}{3}$ or a defensive one with probability $\frac{1}{3}$. The type of each shot is independent of the last. Given that he plays an attacking shot, he scores a point with probability $\frac{1}{2}$ and gets out with probability $\frac{1}{20}$. Given that he plays an defensive shot, he scores a point with probability $\frac{1}{4}$ and gets out with probability $\frac{1}{40}$. If he doesn't score a point or get out, he stays put i.e. scores no points and continues batting.
I am looking to find the expected amount of points scored. My guess is to tweak with the problem so that it doesn't have 3 outcomes at each stage (making it so that there are 2), and so I could use the geometric distribution. For example, I could set the success probability being $P($point scored | doesn't stay put$)$ = $\frac{50}{55}$ and the failure probability being $P($out | doesn't stay put$)$ = $\frac{5}{55}$. This would mean that the expected amount of points is $\frac{50}{5} = 10$.
Is this answer correct, and is this a sensible approach (making 3 scenarios into 2 to use geometric distribution) to solving these types of problems? Secondly, I am told that this problem can be solved by using that $E(X) = \sum E(X|Y_i)P(Y_i)$, where $Y_i$ the events in a partition. How am I supposed to solve the problem this way, when the choice of an attacking or defensive shot (which I assume are the $Y_i$ here) can change at each different shot?
$$E=\frac23(.5(1+E)+.45E)+\frac13(.25(1+E)+.725E)$$
This is because $2/3$ of the time he makes an offensive swing. When does, half the time he scores a run, and then he he scores his expected number of runs. $45\%$ of the time he doesn't score a run or make an out, so he just scores his expected number of runs.
The second term comes from analyzing the case when he makes a defensive swing.
You have only to solve for $E$.