# Genshin Impact Probability: For Material Crafting, Should You Pick the 10% Double Output or the 25% Refund? Plenty of people online have already computed the expected value, but no one seems to have computed the variance, so I'll fill that in here. ## 1. Setup In Genshin Impact, **three low-level materials can be crafted into one high-level material**. We denote this as $a \xrightarrow{\text{转化}}\dfrac{b}{3}$ To keep units consistent, we uniformly **use b for both input and output**, assuming both sides start with initial material a, which is equivalent to $\dfrac{b}{3}$. There are two talents, **as shown below** ![The two talents](/img/原神概率.zh-cn-20240523115001685.webp) - One has a 10% chance of double output. > Each time you use a materials, i.e. $\dfrac{b}{3}$ material, there is a 10% chance of directly producing $\dfrac{2}{3}b$ material. - The other has a 25% chance of refunding some low-level materials. > Each time you use a materials, i.e. $\dfrac{b}{3}$ material, there is a 25% chance of returning $\dfrac{1}{3} \times \dfrac{1}{4}b$ material. In terms of a **random variable**, that is: $$X_{总产出}=\frac{b}{3}+Y_{额外产出}$$ ## **2. Computing Expectation and Variance** > **Plenty of people online have already computed the expected value, but no one seems to have computed the variance, so I'll fill that in here as well.** ### **(1) Expectation and Variance of the 10% Double Output** **1. Expectation** This is a **binomial discrete distribution**. 10% chance of double output, 90% chance of ordinary output. As in Table 1: ![Binomial discrete distribution](/img/原神概率.zh-cn-20240523115047136.webp) $$ \mathrm{E}\_{{双倍}}[X]=\mathrm{E}\_{{双倍}}[\frac{b}{3}+Y\_{额外}]=\frac{b}{3}+np=10\% \times\frac{2}{3}b+90\% \times\frac{1}{3}b $$ So investing a materials is equivalent to investing $\dfrac{b}{3}$ material, and the expectation of the 10% double talent is $E_{\text{双倍}}(\dfrac{b}{3})=\dfrac{11 b}{30} $ . **2. Variance** $$ \mathrm{Var}[X]=\mathrm{Var}[\frac{b}{3}+Y\_{额外产出}]=\mathrm{Var}[Y\_{额外产出}] $$ $$ \mathrm{Var}[Y]=np(1-p)=\frac{b}{3}\times0.1\times0.9=\frac{3}{100}b $$ So investing a materials is equivalent to investing $\dfrac{b}{3}$ material, and the variance of the 10% double talent is $Var_{\text{双倍}}(\dfrac{b}{3})=\dfrac{1}{100}b^2$. ### (2) Expectation and Variance of the 25% Refund **1. Two ways to compute the expectation** **Method 1: geometric series** $$ \begin{align*} E_{\text{返回}}\left(\frac{b}{3}\right) &= \frac{b}{3} + \frac{b}{3} \times 25\\% \times \frac{1}{3} + \ldots + \frac{b}{3} \times \left(\frac{1}{12}\right)^n \newline &= \frac{b}{3} \times (1+{\frac{1}{12}}+{\frac{1}{12}}^2+...+{\frac{1}{12}}^n) \newline &= \frac{b}{3} \times \frac{1 \times (1-{\frac{1}{12}}^n)}{1-\frac{1}{12}} \newline \end{align*} $$ $$ \lim_{n\rightarrow + \infty}E_{\text{返回}}\left(\frac{b}{3}\right)=\frac{b}{3} \times \frac{1 \times (1-{\frac{1}{12}}^n)}{1-\frac{1}{12}}=\frac{4b}{11} $$ The illustration is as follows: ![Tree diagram](/img/原神概率.zh-cn-20240523115108123.webp) We can verify whether this classification has covered all cases by summing all the probabilities: $$ \lim_{n\rightarrow +\infty}75\\%+...+{(25\\%)}^{n-1}75\\%=1 $$ All the corresponding probabilities sum to 1, which to some extent verifies that this classification is correct. **Method 2: recursion** **Because the refunded materials are refined again, a recursion forms, so we get:** $$ E_{\text{返回}}\left(\frac{b}{3}\right)=\frac{b}{3} +E_{\text{返回}}\left(\frac{b}{3}\right)\times 25\% \times \frac{1}{3} $$ The answer is the same as above: So investing a materials is equivalent to investing $\dfrac{b}{3}$ material, and the expectation of the 25% refund talent is $E_{\text{返回}}(\frac{b}{3} )=\frac{4b}{11}$. **2. Variance** The following approach to computing the variance comes from *[A More Refined Calculation of Whether to Pick the 10% Double or 25% Refund for the Material Crafting Talent](https://zhuanlan.zhihu.com/p/656792523)* Now we continue by introducing a recurrence, and we have the following recurrence. $$\mathrm{E}(a_k)=\mathrm{E}(a_{k-1})\times 25\% \times \dfrac{1}{3}$$ > Meaning: > > Before the $k$-th craft, you have $a_k$ low-level materials in hand. > > The condition for being able to do the $k$-th craft is: the $(k-1)$-th craft triggered the effect and refunded some low-level materials. Using the law of total expectation, this is $$ \mathrm{E}(a_k)=\mathrm{E}[\mathrm{E}[a_k| a_{k-1}]]=\mathrm{E}[\frac{1}{12}a_{k-1}]=\frac{1}{12}\mathrm{E}[a_{k-1}] $$ $$ \mathrm{E}[X]=\sum_1^{k}a_k=\sum_1^{k}\frac{1}{12^k}a_0 $$ > This expression can be expanded as a recursion or as a geometric series — the two methods for computing the expectation above. After introducing conditional expectation, next we introduce the variance decomposition formula: $$ \mathrm{Var}[X]=\mathrm{E}[X^2]-[\mathrm{E}(X)]^2 $$ Rearranging and adding **conditional expectation**, we can get two expressions involving $\mathrm{E}(X|Y)$. $$ \mathrm{Var}[\mathrm{E}(X|Y)]=\mathrm{E}\left\\{\left[\mathrm{E}(X|Y)\right]^{2}\right\\}-\left\\{\mathrm{E}\left[\mathrm{E}(X|Y)\right]\right\\}^{2}=\mathrm{E}\left\\{\left[\mathrm{E}(X|Y)\right]^{2}\right\\}-\left[\mathrm{E}(X)\right]^{2} $$ $$ \mathrm{E}[\mathrm{Var}(X|Y)]=\mathrm{E}\left\\{\mathrm{E}(X^{2}|Y)-[\mathrm{E}(X|Y)]^{2}\right\\}=\mathrm{E}(X^2)-\mathrm{E}\left\\{\left[\mathrm{E}(X|Y)\right]^2\right\\} $$ Adding the two expressions above gives the variance decomposition formula: $$ \mathrm{Var}(X)=\mathrm{Var}[\mathrm{E}(X|Y)]+\mathrm{E}[\mathrm{Var}(X|Y)] $$ Substituting $a_k, a_{k-1}$ into this formula: $$ \mathrm{Var}(a_k)=\mathrm{Var}[\mathrm{E}(a_k|a_{{k-1}})]+\mathrm{E}[\mathrm{Var}(a_k|a_{k-1})] $$ where $$ \mathrm{Var}[\mathrm{E}(a_k|a_{{k-1}})]=\mathrm{Var}[\frac{1}{12}a_{k-1}] $$ $$ \mathrm{E}[\mathrm{Var}(a_k|a_{k-1})]=\mathrm{E}[np(1-p)]=\mathrm{E}[np(1-p)]=\mathrm{E}\left[\frac{a_{k-1}}{16}\right] $$ We get: $$ \begin{aligned}\mathrm{Var}\left[a_k\right]=\frac{3}{4} \frac{a_0}{12^k}+\frac{1}{12^2} \mathrm{Var}\left[a_{k-1}\right]\end{aligned} $$ Since $\mathrm{Var}(a_0)=0$, using the recursion we get the general term: > I tossed it to GPT to compute and then verified it; it also satisfies the 144-fold relationship between variance terms: $$ \mathrm{Var}[a_k]=\frac{9}{143}a_0\left(1-\frac{1}{144^k}\right)=a_0\times \frac{1}{16\times 144^{k-1}} $$ $$ \mathrm{E}\left[a_k^2\right] = \mathrm{Var}[a_k] + \left(\mathrm{E}\left[a_k\right]\right)^2 =9\frac{k{a_0}}{12^{2k}} + \frac{a_0^2}{12^{2k}} $$ Then we set the subsequent recursion to fuse $l$ times $$ \begin{aligned} &\mathrm{E}\left[a_k a_\ell\right]=\mathrm{E}\left[\mathrm{E}\left[a_k a_\ell \mid a_\ell\right]\right]\\\\ &=\mathrm{E}\left[a_\ell \mathrm{E}\left[a_k \mid a_\ell\right]\right]\\\\ &=\frac{1}{12^{k-\ell}}\mathrm{E}\left[a_\ell^2\right]\\\\ &=\frac{9a_0^2l+a_0^2}{12^{k+\ell}} \end{aligned} $$ This is to compute the covariance: $$ \mathrm{Cov}(a_\ell, a_k) = \mathrm{E}\left[a_\ell a_k\right] - \mathrm{E}\left[a_\ell\right]\mathrm{E}\left[a_k\right]=\frac{9a_0^2\ell}{12^{k+\ell}} $$ Substituting to compute the variance: Because Genshin Impact crafts three into one: $$ X_n=\sum\limits_{k=0}^n \frac{a_k}{3} $$ We get the variance formula: $$ \mathrm{Var}\left[X_n\right] =\frac{1}{9}\left(\sum_{k=0}^n \mathrm{Var}\left[a_k\right]+2 \sum_{\ell The part below was expanded by GPT; I can sense that the problem lies in the variance accumulation part — there can't be a standalone 1 term that can't be canceled out, because then it wouldn't converge. $$ \begin{align*} \text{Var}[X_n] &= \frac{1}{9} \Bigg( \frac{9}{143} a_0 \left[ (n+1) - \frac{144}{143} \left( 1 - \frac{1}{144^{n+1}} \right) \right] \\\\&\quad + 2 \cdot \frac{9a_0^2}{11} \sum_{\ell=0}^{n-1} \frac{\ell}{12^{2\ell}} \left( 1 - \frac{1}{12^{n-\ell}} \right) \Bigg) \end{align*} $$Next, let n tend to the limit. Based on the code plot, eyeballing it, the limit should be around 0.009b, i.e. around 0.027a. ## 3. Conclusion and Computer Simulation Fitting with code, it also matches the computed result: - The 10% has a slightly higher expectation, but within 1000 materials the difference in expectation isn't large. - The 25% has smaller variance fluctuation; it just needs more crafts, so the Mora cost is a bit higher. - Considering Genshin Impact's 999 material cap, the expected high-level materials differ by only about 2-3. ![100 times](/img/原神概率.zh-cn-20241225140147025.webp) ![1000 times](/img/原神概率.zh-cn-20241225140222239.webp) ![800-1000 times](/img/原神概率.zh-cn-20241225140244046.webp) Experiment code ```python import random import matplotlib.pyplot as plt import numpy as np # set matplotlib's font so it supports Chinese display plt.rcParams['font.sans-serif'] = ['SimHei'] # used to display Chinese labels correctly plt.rcParams['axes.unicode_minus'] = False # used to display the minus sign correctly # function that simulates crafting high-level items def synthesize_items(low_items, strategy): high_items = 0 while low_items >= 3: 合成结果 = random.random() if strategy == 1: if 合成结果 < 0.1: high_items += 2 # additionally gain one high-level item else: high_items += 1 elif strategy == 2: if 合成结果 < 0.25: low_items += 1 # refund one low-level item high_items += 1 low_items -= 3 return low_items, high_items # simulate many crafts and compute the mean return and variance def simulate_synthesis(low_items, strategy, trials=10000): results = [] for _ in range(trials): _, high_items = synthesize_items(low_items, strategy) results.append(high_items) mean = np.mean(results) variance = np.var(results) return mean, variance # plot the charts def plot_results(means, variances, strategies): plt.figure(figsize=(14, 6)) # plot the mean comparison chart plt.subplot(1, 2, 1) for strategy, data in means.items(): plt.plot(data, label=f'策略{strategy}均值') plt.xlabel('低级道具数量') plt.ylabel('高级道具均值') plt.title('均值比较') plt.legend() # plot the variance comparison chart plt.subplot(1, 2, 2) for strategy, data in variances.items(): plt.plot(data, label=f'策略{strategy}方差') plt.xlabel('低级道具数量') plt.ylabel('高级道具方差') plt.title('方差比较') plt.legend() plt.tight_layout() plt.savefig('results.png') # save the chart as an image file # main function def main(): num_items = range(1, 1001) # range of the number of low-level items invested trials = 100000 # number of simulations strategy1_means = [] strategy1_variances = [] strategy2_means = [] strategy2_variances = [] for items in num_items: mean1, variance1 = simulate_synthesis(items, 1, trials) mean2, variance2 = simulate_synthesis(items, 2, trials) strategy1_means.append(mean1) strategy1_variances.append(variance1) strategy2_means.append(mean2) strategy2_variances.append(variance2) plot_results({ 1: strategy1_means, 2: strategy2_means }, { 1: strategy1_variances, 2: strategy2_variances }, [1, 2]) if __name__ == '__main__': main() ``` For more discussion, see: [Genshin Impact Probability](https://zhuanlan.zhihu.com/p/648950822?) --- > Author: [滑翔闪](https://github.com/hzp2333/hzp2333.github.io) > URL: https://blog.huaxiangshan.com/en/posts/genshin/