User:Guy Bukzi Montag/sandbox

''Nothing to see here. It's only my playground for future blog posts.''

Everything on this page is a draft and will be deleted when the final post is published.

''If you do see something here, it's probably a blog post in the making. Feel free to comment, corrections are welcome!''

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With this formula we are able to answer the question of the beginning: We have $$n = 10$$ boxes and want $$d = 1$$ engine of which each has a probability of $$P(E) = 0.08835$$, so the probability of getting at least 1 engine from these boxes is



\begin{align} &\sum_{k = 1}^{10} 0.08835^k \cdot \left (1 - 0.08835 \right )^{10-k} \cdot \binom{10}{k}\\ \\ \\ &= \left (\frac{1}{10} \right )^2 \cdot \left (\frac{9}{10} \right )^{30} \cdot \binom{32}{2} + \left (\frac{1}{10} \right )^3 \cdot \left (\frac{9}{10} \right )^{29} \cdot \binom{32}{3} + ... +\left (\frac{1}{10} \right )^{32} \cdot \left (\frac{9}{10} \right )^{0} \cdot \binom{32}{32}\\ \\ \\ &= 84.36 %. \end{align} $$

Boxes
Porsche Rookie Box
 * Description: "Grants 5 cards for the Porsche Cayman GT4!"
 * Observed:
 * 1 High-Grade Engine:0:0
 * 2 Common Tech, probably 1 Early and 1 Initial Tech
 * 2 Class B parts (random)
 * 2 Class B parts (random)

Porsche Advanced Box
 * Description: "Grants 6 cards for the Porsche Cayman GT4!"
 * Guessed:
 * 1 High-Grade Engine:0:0
 * 1 Mid-Tech:0:0
 * 4 Common Tech or Class B parts (random)
 * 4 Common Tech or Class B parts (random)

Porsche Professional Box
 * Description: "Grants 10 cards for the Porsche Cayman GT4! You're guaranteed to get a High-Grade Engine:0:0!"
 * Observed:
 * 2 High-Grade Engines:0:0
 * 1 Mid-Tech:0:0
 * 1 Common Tech
 * 6 Class B parts (random)
 * 6 Class B parts (random)

Bla
Drop Rates Explained: 5th Anniversary (V)

This is part IV of a series on the maths behind drop rates.


 * Part I: Practical questions
 * Part II: Extracting probabilities
 * Part III: More than one box
 * Part IV: Box inside a box
 * Part V: The Asphalt 8 Box Calculator
 * Part VI: All formulas in a nutshell (tba)

This part continues the explanations of part IV

Bla.

If you have questions or find mistakes: Feel free to comment!

Formula collection
Here are all the formulas from the previous parts in a nutshell.

(Part II, Boxes with guaranteed items):
 * Probability of a random item in a box with guaranteed items


 * $$P(D) = \frac{\operatorname{E}[D]}{1 - \operatorname{E}[G]}$$

$$\operatorname{E}[G]$$: expected value of all guaranteed items $$G$$ (sum of all guaranteed percentages in the box info) $$P(D)$$: probability of $$D$$
 * $$\operatorname{E}[D]$$: expected value of the desired random item $$D$$ (percentage in the A8BoxInfoAnniversaryBundle.png box info)

(Part III, Permutation):
 * Number of possibilities (permutations) when buying boxes with random items


 * $$(d+1)^n$$

$$d$$: number of random items of a kind granted per box
 * $$n$$: number of boxes

(Part III, Combination):
 * Number of combinations of a random item in a set of boxes


 * $$\binom nk$$

$$k$$: number of items in the combination
 * $$n$$: number of boxes

(Part III, Combination):
 * Number of possible combinations of a minimum number of a desired item from several boxes


 * $$\sum_{k = d}^n \binom{n}{k}$$

$$n$$: number of boxes
 * $$d$$: number of desired items

(Part III, Probability):
 * Probability of one combination in a set of boxes


 * $$(P(D))^d \cdot (1 - P(D))^{n-d}$$

$$d$$: number of desired items $$n$$: number of boxes
 * $$P(D)$$: probability of getting the desired item $$D$$

(Part III, Probability):
 * Probability of getting a minimum number of a desired item from several boxes


 * $$\sum_{k = d}^n (P(D))^k \cdot (1 - P(D))^{n-k} \cdot \binom{n}{k}$$

$$d$$: number of desired items $$n$$: number of boxes
 * $$P(D)$$: probability of getting the desired item $$D$$

Statistical test
Did the calculations withstand the real numbers of items found in opened boxes?

Appendix
Mercedes-Benz SE:

The amount of boxes needed to get at least 1 (one!) blueprint with a probability of 90 % is


 * $$n \geq \frac{\ln(1 - 0.9)}{\ln(1 - 0.1)} = 21.85$$.