Maximum Expected Fishing Day (Recurrence Relation)
Although the subproblems are defined correctly in the approach quoted in the question, there are some typos and glitches when it comes to the base cases and recurrence relations. 

Please see the corrected base cases and recurrence relations below.





Base case: for any $(m,d,v)$ such that $m$, $d$ or $v$ equals to 0,
$$\text{MEFD}(m,d,v) =0\tag{R0}$$

Recurrence Relations:
$$\begin{align*}
\text{MEFD}(m,d,v) &amp;= \max(\text{MEFD}_{1}(m,d,v),  \text{MEFD}_{2}(m,d,v))\tag{R1}\\
\text{MEFD}_{1}(m,d,v)&amp;=\frac{1}{d}(1+\text{MEFD}(m-1,30,v-1))\\&amp;+\frac{d-1}{d}\text{MEFD}(m,d-1,v-1)\tag{R2}\\ 
\text{MEFD}_{2}(m,d,v)&amp;=\frac{1}{d}\text{MEFD}(m-1,30,v))\\&amp;+\frac{d-1}{d}\text{MEFD}(m,d-1,v)\tag{R3}
\end{align*}$$

Equation (R1) says that the maximum expected fishing days is the larger of the expected fishing days yielded by two strategies, John selects the next day and John does not select the next day. John will proceed with the best strategy in the remaining days.

Equation (R2) compute the maximum expected finishing days when John has selected the next day. The probability that the next day will be a fishing day is $\frac1d$, in which case John will proceed with the best strategy for the remaining $m-1$ months and $v-1$ vacations day. The probability that the next day will not be a fishing day is $\frac{d-1}d$, in which case John will proceed with the best strategy for the remaining $d-1$ day and $v-1$ vacation days.

Equation (R3) compute the maximum expected finishing days when John has decided to work for the next day. The probability that the next day will be a fishing day is $\frac1d$, in which case John will proceed with the best strategy for the remaining $m-1$ months and $v$ vacations day. The probability that the next day will not be a fishing day is $\frac{d-1}d$, in which case John will proceed with the best strategy for the remaining $d-1$ day and $v$ vacation days.

Note "John needs to inform his boss in order to get time off from work at least a day in advance" is interpreted as something like if John wants to take tomorrow off, he can just inform his boss tonight, at which time he must have known whether today is a fishing day. If it is, he will not, of course, select any remaining day in the current month as a vacation day. That is why whenever the next day will be a fishing day, we will only consider the remaining $m-1$ months in the above recurrence relations.