A more complete analysis of how to preserve your money in Silksong

In the first article of the “blog daily until XMass” series we got to a very simple expression linking the probability of dying on a runback to a boss and the best choice on whether to convert the rosaries (in-game currency) to necklaces (losing some value, but persisting after death). That analysis didn’t consider that players might have to fight the same boss multiple times before they can finally proceed. We do this now.

The problem requiring us to extend the analysis is that if a player dies within an arena, when they attempt to retrieve their cocoon (that is, all the dropped currency), the arena battle starts again. So, even if the player retrieves their currency, they might still die and have to execute the same runback again.

Let’s denote with \(q\) the probability that the player dies when encountering the boss. From the previous article, we already have \(p\) as the probability to die on the runback. The possible events are:

1. win the boss on first fight (with probability \(1-q\))
2. lose on the first fight (with probability \(q\)) and then
   1. die on runback, losing money (with probability \(p\))
   2. reach the boss again (with probability \(1-p\)) and then repeat this scenario

We can represent all of this by the following state machine.

From here, the probability to end in the green state is

\[P_w = \left(1-q\right) + q\left(1-p\right)\left(1-q\right) + q^2\left(1-p\right)^2\left(1-q\right) + \ldots\]

That is,

\[P_w = \frac{1-q}{1-q+pq}\]

And, the probability to end on the red, bad state is:

\[P_r = qp + q^2\left(1-p\right)p + q^3\left(1-p\right)^2p + \ldots\]

Which shortens to

\[P_r = \frac{pq}{1-q+pq}\]

Like in the previous article, we can save our \(R\) rosaries by repeating a conversion process that costs \(c\) to preserve \(v\), a number of \(k\) times.

With probability \(P_w\), we win and recover all the unconverted rosaries, for a value of \(R - ck + vk\). With probability \(P_r\), we die and only the converted \(vk\) rosaries are left. So, the expected value of performing \(k\) conversions is:

\[\mathbb{E}\left(k\right) = \frac{pq}{1-q+pq}vk + \frac{1-q}{1-q+pq}\left(R - ck + vk\right)\]

Just like before, we group by \(k\):

\[\mathbb{E}\left(k\right) = k\left(v - \frac{1-q}{1-q+pq}c\right) + \frac{1-q}{1-q+pq}R\]

Which means that the sign of \(v - \frac{1-q}{1-q+pq}c\) tells us if we need to convert everything, or convert nothing. From here, we just need estimates for \(p\) and \(q\), based on the skill of the player.