How to calculate the variance of this probability model?

Question

There is such a rule in the game "Genshin Impact"： Every three materials a can be fused into one material b.

Suppose the amount of original material are a, which is the corresponding $\dfrac{b}{3}$

There are two talents to choose from:

One is double output with a 10% chance:

This is the binomial distribution,like the table

left	center	right
X	$\dfrac{2b}{3}$	$\dfrac{b}{3}$
P	10%	90%

$$ E_{\text{double}}(\dfrac{b}{3})=0.1\times \dfrac{2b}{3} + 0.9 \times \dfrac{b}{3} = \dfrac{11b}{30} $$

$$ Var_{\text{double}}=E(X^2)-(E(X))^2 =(\dfrac{2b}{3})^2\times 0.1+ (\dfrac{b}{3})^2\times 0.9 -(\dfrac{11b}{30})^2=\dfrac{1}{100}b^2 $$

The other is 25% chance to return part material a:

Every time a material is used, that is, $\dfrac{b}{3}$ material, there may be a 25% probability to return $\dfrac{b}{3} \times \dfrac{1}{3} $ material.

$$ E_{\text{return}}\left(\frac{b}{3}\right) = \frac{b}{3} + \frac{b}{3} \times 0.25 \times \frac{1}{3} + \ldots + \frac{b}{3} \times \left(\frac{1}{12}\right)^n = \frac{b}{3} \times (1+{\frac{1}{12}}+{\frac{1}{12}}^2+...+{\frac{1}{12}}^n) = \frac{b}{3} \times \frac{1 \times (1-{\frac{1}{12}}^n)}{1-\frac{1}{12}} $$ $$ \lim_{n\rightarrow + \infty}E_{\text{return}}=\frac{b}{3} \times \frac{1 \times (1-{\frac{1}{12}}^n)}{1-\frac{1}{12}}=\frac{4b}{11} $$

Regarding the variance, I would like to handle it like this

$$\lim_{n\rightarrow + \infty}E_{\text{return}}({\frac{1}{9}}b^2)={\frac{1}{9}}b^2 \times \frac{1 \times (1-{\frac{1}{12}}^n)}{1-\frac{1}{12}}=\frac{4b}{11}\times \frac{b}{3} =\frac{4}{33}b^2$$

However,,,,,,

$$Var_{\text{return}}(\frac{b}{3})=E(X^2)-(E(X))^2=\frac{4}{33}b^2-(\frac{4b}{11})^2=-\frac{4}{363}b^2$$

May I ask where is wrong, how to find the correct variance? Thanks for your reply and help!

Answer 1

I get this

$$ E_{\text{return}}\left(\frac{1}{9}b^2\right) = \sum_{i=1}^{n-1} \frac{3}{4^{i}} \times \frac{1}{9}b^2 \left(\frac{1-\left(\frac{1}{3}\right)^i}{1-\frac{1}{3}}\right)^2 $$

this is my mind map， We just need to add up the red part

Answer 2

Let $(a,b)$ denote the amounts of materials A and B one has. I'll treat $a$ and $b$ as continuum quantities and assume that fusion always operates on the total amount of A one has. Talent 1 processes $(a,b)$ into $(0,b+a/3)$ with probability $90\%$ and into $(0,b+2a/3)$ with probability $10\%$. Talent 2 processes $(a,b)$ into $(0,b+a/3)$ with probability $75\%$ and into $(a/3,b+a/3)$ with probability $25\%$. However, if one has chosen talent 2, then one keeps applying it until no more A remains.

Now suppose you begin with one unit of A and zero of B. With talent 1, you obtain $1/3$ units of B with probability $9/10$ and $2/3$ with probability $1/10$. The mean and variance of this distribution are$$ \begin{align} \mu_1 &= \frac{9}{10}\cdot\frac{1}{3} + \frac{1}{10}\cdot\frac{2}{3} = \frac{11}{30},\;\text{and}\\ V_1 &= \frac{9}{10}\cdot\left(\frac{1}{3}\right)^2 + \frac{1}{10}\cdot\left(\frac{2}{3}\right)^2 -\mu_1^2 = \frac{1}{100}. \end{align} $$ On the other hand, with talent 2 you obtain $1/3$ units of B with probability $3/4$, you obtain $1/3 + 1/9$ units with probability $1/4 \times 3/4$, etc. The mean and variance of this distribution are$$ \begin{align} \mu_2 &= \frac{3}{4}\cdot\frac{1}{3} + \frac{1}{4}\cdot\frac{3}{4}\left(\frac{1}{3}+\frac{1}{9}\right) + \cdots ,\;\text{and}\\ V_2 &= \frac{3}{4}\cdot\left(\frac{1}{3}\right)^2 + \frac{1}{4}\cdot\frac{3}{4}\left(\frac{1}{3}+\frac{1}{9}\right)^2 + \cdots - \mu_2^2. \end{align} $$ If we only cared about $\mu_2$ it would be simpler to think of the first sum above as obtaining $1/3$ units with probability $1$ plus an additional $1/9$ with probability $1/4$, plus an additional $1/27$ with probability $1/16$, etc., which yields $\mu_2 = 4/11$. This is slightly below $\mu_1$, so talent 1 is better on average. However, to obtain the variance we need to evaluate the sums$$ \frac{1}{3}+\frac{1}{9} + \cdots + \frac{1}{3^n} = \sum_{k=1}^n \frac{1}{3^k} = \frac{1-3^{-n}}{2}. $$ Therefore$$ \begin{align} \mu_2 &= \frac{3}{4} \sum_{n=1}^\infty \frac{1}{4^{n-1}} \frac{1-3^{-n}}{2} ,\;\text{and}\\ V_2 &= \frac{3}{4} \sum_{n=1}^\infty \frac{1}{4^{n-1}} \left(\frac{1-3^{-n}}{2}\right)^2 - \mu_2^2. \end{align} $$ I'll omit the algebra details, but one may verify that the first sum above simplifies to $\mu_2 = 4/11$, in agreement with the simpler approach, and that$$ V_2 = \frac{12}{4235}. $$ Therefore talent 1 yields $\mu_1 = 0.3667$ units of B on average (per unit of A) with a standard deviation of $0.1000$ units, whereas talent 2 yields $\mu_2 = 0.3636$ units with a standard deviation of $0.0532$ units. The lower standard deviation of talent 2 could be viewed as compensation for its lower mean.