y5nw

y5nw is a player on the server.

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Notes

Note: This section is more or less written based on my own knowledge. Feel free to suggest corrections in my talk page.

Event chance calculation

For a one-shot event that fires with a probability of $p$ checked at an interval of $T$ (e.g. tree growth), the expected amount of time $t$ it takes for the event to fire is $E(t)={\frac {T}{p}}$.

Note that, in some definition tables in Minetest (e.g. ABMs), the probability is specified using the reciprocal of $p$.

Calculations: Let $q:=1-p$ be the probability of the event not firing at the next check.

${\begin{aligned}E(t)&=\sum _{n=1}^{\infty }(q^{n-1}(1-q))(Tn)=(1-q)T\sum _{n=1}^{\infty }q^{n-1}n=(1-q)T\sum _{n=1}^{\infty }\sum _{m=1}^{n}q^{n-1}\\&=(1-q)T\sum _{n=1}^{\infty }\sum _{m=n}^{\infty }q^{m-1}={\frac {(1-q)T}{q}}\sum _{n=1}^{\infty }\sum _{m=n}^{\infty }q^{m}\\&={\frac {(1-q)T}{q}}\sum _{n=1}^{\infty }\left(\sum _{m=0}^{\infty }q^{m}-\sum _{m=0}^{n-1}q^{m}\right)={\frac {(1-q)T}{q}}\sum _{n=1}^{\infty }\left({\frac {1}{1-q}}-{\frac {1-q^{n}}{1-q}}\right)={\frac {(1-q)T}{q}}\sum _{n=1}^{\infty }{\frac {q^{n}}{1-q}}\\&=T\sum _{n=0}^{\infty }q^{n}=T\cdot {\frac {1}{1-q}}={\frac {T}{p}}\end{aligned}}$