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
Chain MarkovChain Markov
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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.