In the video Game Azur roadway after completing the mission 3-4 you have a chance that gaining together a price one of the two aircraft carriers Akagns and Kaga. Ns chances that getting every that the are $frac3400$ every and also tbelow ins no various other possible way that receivinns them. Ns rewards for every tins you complete thins mission to be dispersed independently. What is ns intended variety of time girlfriend need to complete this missitop top in bespeak to obtain both of them?

This Inquiry ins rather commonly asked on assorted websites dedicated to that game. However, Despite it gift a fairly straightforward probability exercise, i have never seen a user that twater tap forums in reality calculatinns it.

probcapacity stochastic-procedures markov-chains
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request january 9 "20 at 15:48

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~ every completion, girlfriend have actually a $frac3200$ opportunity of gaining a carrier, therefore the number of completion Taken to acquire her initially transport ins a geometric random change through success probability $frac3200$, and has expectati~ above $frac2003$.

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once you have one carrier, you have a $frac3400$ chance that gaining the other, so the same, similar thing you require $frac4003$ additional completion come acquire it.

the full expectation ins As such $frac2003+frac4003=200$.

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answered jan 9 "20 in ~ 15:55

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the process that attemptinns come get Akagns and also Kaga can be defined together a Markov chain $R$ via ns complying with states:

0 (initial state) = «girlfriend have actually neither»

1 = «girlfriend have actually only Akagi»

2 = «you have actually just Kaga»

3 = «you have actually both»

ns change procession that thins Markov chain is:

$$eginpmatrix frac197200 & 0 & 0 & 0 \ frac3400 & frac397400 & 0 & 0 \ frac3400 & 0 & frac397400 & 0 \ 0 & frac3400 & frac3400 & 1 endpmatrix$$

now intend $f_k$ is the generatinns feature the ns sequence $P(R_n = k)\_n in startupcuba.orgbbN$. Climate $f_k$ to be the Equipment come the following system that straight functional equations:

$$f_0 = 1 + frac197200 xf_0$$

$$f_1 = x(frac3400f_0 + frac397400f_1)$$

$$f_2 = x(frac3400f_0 + frac397400f_2)$$

$$f_3 = frac3400x(f_1 + f_2) + xf_3$$

therefore we have:

$$f_0 = frac200200 - 197x$$

$$f_1 = f_2 = frac600x(200 - 197x)(400 - 397x)$$

$$f_3 = frac9x^2(200 - 197x)(400 - 397x)(1 - x)$$

Now, our expectation ins ns expectati~ above that the hitting time that the state $3$ i beg your pardon (Due to the fact that if we cons in $3$ us remain Victory $3$ via probcapability $1$) have the right to it is in calculate as:

$$sum_n = 1^infty P(R_n eq 3) = sum_n = 1^infty (1 - P(R_n = 3)) = lim_x come 1 (frac11 - x - f_3(x)) = 200 lim_x o 1 (frac400 - 791x + 391x^2(200 - 197x)(400 - 397x)(1 - x)) = 200$$

thus ins will certainly require on averPeriod compleat that missitop top $200$ times in order to receive both Akagns and also Kaga.