Maximizing Candy from Raid Bosses: Pinap or Golden Razz?

• The optimal strategy is always using Pinap Berry first and Golden Razz Berry (GRB) last.

• The number of balls left where you switch to GRB is independent of the number of balls you receive

Let

• $p_{1}$ be the catch rate of using GRB

• $c_{1}$ be the number of candies gained of using GRB

• $p_{2}$ be the catch rate of using Pinap

• $c_{2}$ be the number of candies gained of using Pinap

Where $0 < p_{2} < p_{1} < 1$ and $0 < c_{1} < c_{2}$. Then:

• If $p_{1} c_{1} > p_{2} c_{2} $, then the optimal strategy is using GRB in the last $k$ balls, where $$ k = \lceil \log_{1 - p_{1}} ( \frac{ c_{2} - c_{1} }{ c_{1} } \frac{ p_{2} }{ p_{1} - p_{2} } ) \rceil $$

• If $p_{1} c_{1} \leq p_{2} c_{2} $, then it is optimal to always use Pinap.

In any case (using GRB or Pinap), denote the catch rate as $p$ and the number of candies gained as $c$.

Let $X_{n}$ be the number of candies gained with $n$ balls to throw. $X_{n}$ takes two possible values:

\[ X_{n} = \begin{cases} c & \quad \text{with the probability of } p \\ X_{n-1} & \quad \text{with the probability of } 1-p \end{cases} \]

Therefore:

$$E[X_{n}] = pc + (1-p)E[X_{n-1}] $$

At the $n$-th last ball, use GRB if and only if

$$ p_{1}c_{1} + (1-p_{1})E[X_{n-1}] > p_{2}c_{2} + (1-p_{2})E[X_{n-1}] $$

That is (with the assumption that $p_{1} > p_{2}$):

$$ E[X_{n-1}] < \frac { p_{1}c_{1} - p_{2}c_{2} } { p_{1} - p_{2} } $$

• If $p_{1} c_{1} \leq p_{2} c_{2} $, then it is never optimal to use GRB (hence use Pinap only) since

  $$E[X_{n}] \geq 0 \geq \frac { p_{1}c_{1} - p_{2}c_{2} } { p_{1} - p_{2} }$$

• If $p_{1} c_{1} > p_{2} c_{2} $, then the proof continues.

Note that $E[X_{n}]$ increases in $n$. Hence, if it is better to use GRB at the $n$-th last ball, then it must be also better to use GRB at the $(n-1)$-th last, at the $(n-2)$-th last, ..., and at the last ball, since:

$$ 0 = E[X_{0}] < E[X_{1}] < \dots < E[X_{n-1}] < \frac { p_{1}c_{1} - p_{2}c_{2} } { p_{1} - p_{2} } $$

Likewise, if it is better to use Pinap at the $(n+1)$-th last ball, then it must be also better to use Pinap at the at the $(n+2)$-th last, ..., and at the first ball:

$$ \frac { p_{1}c_{1} - p_{2}c_{2} } { p_{1} - p_{2} } \leq E[X_{n}] < E[X_{n+1}] < \dots $$

Suppose using GRB in the last $k$ balls is the optimal strategy. That is, it is better to use GRB at the $k$-th last ball and to use Pinap at the $(k+1)$-th last ball. Thus:

$$ E[X_{k-1}] < \frac { p_{1}c_{1} - p_{2}c_{2} } { p_{1} - p_{2} } \leq E[X_{k}] $$

Since it is always using GRB from the $k$-th last ball to the very last ball, we have:

\[ \begin{cases} E[X_{k-1}] = (1 - (1-p_{1}) ^ {k-1} ) c_{1} \\ E[X_{k}] = (1 - (1-p_{1}) ^ {k} ) c_{1} \end{cases} \]

Plug in and solve the inequality with respect to $k$, and it will give:

$$ k = \lceil \log_{1 - p_{1}} ( \frac{ c_{2} - c_{1} }{ c_{1} } \frac{ p_{2} }{ p_{1} - p_{2} } ) \rceil $$