Methodology

Enhancing tyrant equips is a random process which can be modeled by a Markov chain. One might begin by thinking that this is a simple birth-death chain on the state space $\{G,0,1,2,...,n\}$ (where $n$ is your "goal" number of stars), where $G$ represents the "graveyard" state (boom) and the other states correspond to the number of stars on your equipment. However, "chance time", which grants you 100% success rate if your last two attempts resulted in drops, makes this process non-Markovian. In order to apply classical Markov chain theory, we have to add some more states: $i^+$ will represent the state reached by dropping from $i+1$ to $i$ stars, and $i^*$ will represent the state reached by dropping twice in a row down to $i$ stars. Note that there is no reason to consider the state $0^+$ , since no chance time is awarded for failing at $0$ . We set up our chain to be absorbing at $G$ and $n$ . So our transition matrix $P$ is given by

$P(G,G)=P(n,n)=1$
$P(0,0)=1-p_0$
$P(i,i+1)=P(i^+,i+1)=p_i$
$P(i,G)=b_i$
$P(i,i-1)=P(i^+,i-1)=1-p_i-b_i$
$P(i^*,i+1)=1$

where $p_i$ is the probability of transitioning from $i$ to $i+1$ stars in one step (without chance time), and $b_i$ is the probability of transitioning from $i$ to $G$ in one step (i.e. one-step boom rate starting at $i$ ). The rates in GMS can be found here.

Now that we have our transition matrix set up for our transient chain with absorbing states $G$ and $n$ , we can apply basic Markov chain theory to get the numbers that we want.

1. [Hitting probability of $n$ ] This can be obtained by taking $P^k(0,n)$ for sufficiently large $k$ . $P^k(0,n)$ is the probability of being at $n$ stars after $k$ attempts, starting at $0$ . The interpretation is that after taking a sufficiently large number of steps, you will either have reached your goal of $n$ stars, or your equipment was destroyed, so $\lim_{k\rightarrow\infty}P^k(0,n)$ is precisely the probability of reaching $n$ stars before destruction.

2. [Expected number of booms] This is just a standard geometric random variable on $\{0,1,2,\ldots\}$ with success probability equal to $\rho_n:=\lim_{k\rightarrow\infty}P^k(0,n)$ . The expected value of this random variable is $(1-\rho_n)/\rho_n$ .

3. [Expected amount of mesos] Set $M=55,832,200$ , which is the cost per enhance attempt. The expected amount of mesos to get to $n$ stars is equal to $M$ times the expected number of steps to reach $n$ . Here we're saying that if your equipment is destroyed before reaching $n$ , we just start over from $0$ , until we finally reach $n$ . The expected number of steps, starting at $0$ , to get absorbed to either $G$ or $n$ , is given by the sum

$e_B:=\sum_{s\notin\{G,n\}}Q(0,s)$ where $Q$ is the fundamental matrix, which is given by $Q:=\sum_{k=0}^\infty P^k=(I-P)^{-1}$ . Now each time we hit a boundary (either $G$ or $n$ ) after starting at $0$ , on average we will have taken $e_B$ steps, and we have a $\rho_n$ probability of being at $n$ . By Wald's identity (or probably even by some weaker fact), the expected number of steps to reach $n$ is simply $e_B \mathbb{E}[R_n]$ , where $R_n$ is equal to the number of boundary hits until reaching $n$ . $R_n$ has geometric distribution with success rate $\rho_n$ on $\{1,2,3,\ldots\}$ , so $\mathbb{E}[R_n]=1/\rho_n$ and hence the expected number of steps to reach $n$ is $e_B/\rho_n$ . Multiplying this number by $M$ gives us the average meso cost to reach $n$ stars.

4. [Variance of amount of mesos] This is computed in similar fashion to the expected amount of mesos. The only difference is we use Wald's second identity to give us a formula for the variance, and from there it is straightforward, so I will not go into details.

Remark: Some of you might be wondering about the "+ Enhancement Success Rate" from landing the star in the middle strip. Did I implement that in my computations? Yes and no. The truth is... the whole thing is a scam and doesn't do anything. Nexon implemented it just to troll people into caring about it, but in reality it does absolutely nothing other than provide marginal entertainment to those of you who enjoy it. It's up to you whether you want to believe me or not, but I know this as a fact.

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