Expected number of times you have to complete the mission 3-4 in Azur Lane in order to get both Akagi and Kaga?

Question

In the video game Azur Lane after completing the mission 3-4 you have a chance of getting as a reward one of the two aircraft carriers Akagi and Kaga. The chances of getting each of the are $\frac{3}{400}$ each and there is no other possible way of receiving them. The rewards for each time you complete this mission are distributed independently. What is the expected number of times you have to complete this mission in order to obtain both of them?

This question is quite frequently asked on various websites dedicated to that game. However, despite it being a quite simple probability exercise, I have never seen a user of those forums actually calculating it.

Answer 1

After every completion, you have a $\frac{3}{200}$ chance of getting a carrier, so the number of completions taken to get your first carrier is a geometric random variable with success probability $\frac{3}{200}$, and has expectation $\frac{200}{3}$.

Once you have one carrier, you have a $\frac{3}{400}$ chance of getting the other, so similarly you need $\frac{400}{3}$ extra completions to get it.

The total expectation is therefore $\frac{200}{3}+\frac{400}{3}=200$.

Answer 2

The process of attempting to get Akagi and Kaga can be described as a Markov chain $R$ with the following states:

0 (initial state) = «You have neither»

1 = «You have only Akagi»

2 = «You have only Kaga»

3 = «You have both»

The transition matrix of this Markov chain is:

$$\begin{pmatrix} \frac{197}{200} & 0 & 0 & 0 \\ \frac{3}{400} & \frac{397}{400} & 0 & 0 \\ \frac{3}{400} & 0 & \frac{397}{400} & 0 \\ 0 & \frac{3}{400} & \frac{3}{400} & 1 \end{pmatrix}$$

Now suppose $f_k$ is the generating function of the sequence $\{P(R_n = k)\}_{n \in \mathbb{N}}$. Then $f_k$ are the solution to the following system of linear functional equations:

$$f_0 = 1 + \frac{197}{200} xf_0$$

$$f_1 = x(\frac{3}{400}f_0 + \frac{397}{400}f_1)$$

$$f_2 = x(\frac{3}{400}f_0 + \frac{397}{400}f_2)$$

$$f_3 = \frac{3}{400}x(f_1 + f_2) + xf_3$$

Thus we have:

$$f_0 = \frac{200}{200 - 197x}$$

$$f_1 = f_2 = \frac{600x}{(200 - 197x)(400 - 397x)}$$

$$f_3 = \frac{9x^2}{(200 - 197x)(400 - 397x)(1 - x)}$$

Now, our expectation is the expectation of the hitting time of the state $3$ which (because if we come in $3$ we stay win $3$ with probability $1$) can be calculated as:

$$\sum_{n = 1}^{\infty} P(R_n \neq 3) = \sum_{n = 1}^{\infty} (1 - P(R_n = 3)) = \lim_{x \to 1} (\frac{1}{1 - x} - f_3(x)) = 200 \lim_{x \to 1} (\frac{400 - 791x + 391x^2}{(200 - 197x)(400 - 397x)(1 - x)}) = 200$$

Thus it will require on average compleating that mission $200$ times in order to receive both Akagi and Kaga.