Arcane Collector

Probability Level 4

Sealed packs of the collectible card game Arcane: The Congregation have a $\dfrac{1}{8}$ chance to contain one of the 15 distinct mythic cards, chosen at random. If a pack contains a mythic card, each of the 15 mythic cards is equally likely.

If each pack of Arcane: The Congregation is independent of all other packs, what is the expected value of the number of packs needed to obtain at least one copy of each mythic card? Round your answer to the nearest integer.

Note : The coupon collector problem page contains an approximation that makes the solution easier to compute, and the approximation is sufficient to obtain the correct answer.

The answer is 398.

2 solutions

Andy Hayes
Jul 4, 2016

Relevant wiki: Expected Value - Problem Solving

This problem is like a coupon collector problem , except there is only a $\frac{1}{8}$ chance that each pack is relevant. Let $M$ be the event that a pack contains a mythic card.

Let $X$ be the discrete random variable representing the number of packs needed to open to obtain at least one copy of each mythic rare card.

$\text{E}[X\mid M]$ can be approximated with the following formula with $n=15$ and $\gamma$ the Euler-Mascheroni constant:

$\text{E}[X\mid M]\approx n(\ln{n}+\gamma)+\frac{1}{2}\approx 15(\ln{15}+0.577216)+0.5\approx 49.779$

Now $\text{E}[X]$ can be calculated:

$\text{E}[X]=\frac{\text{E}[X\mid M]}{P(M)}=8\text{E}[X\mid M]\approx 398.232$

The expected number of packs needed, rounded to the nearest integer, is $\boxed{398}$ .

Incidentally, the exact answer is $\text{E}[X]=8\times 15\times H_{15}\approx 398.187479$ , where $H_{15}$ is the $15^\text{th}$ harmonic number .

I am afraid this is all rather muddled. You have $X$ equal to the total number of packs needed to get a full set, and $M$ the event that a pack contains a mythic card. Which pack? Unless you define this, the quantity $E[X|M]$ has no meaning. If $M$ was the event that the first pack chosen contained a mythic card, then $E[X] \; = \; \tfrac{1}{8} E[X|M] + \tfrac{7}{8} E[X|M']$ and we can write $E[X|M'] = 1 + E[X]$ , but $E[X|M]$ is not the quantity $15H_{15}$ that you seem to want it to be. We now have $E[X|M] = 1 + E[Y]$ , where $Y$ is the number of packs required to get a full set, given that you have one mythic card already.

What you are basically saying is that it takes $8$ times as long to collet a full set as it would if there were mythic cards in every pack. This is true, and the factor of $8$ comes out of all the terms in my proof. Your argument needs tidying up to work properly.

Mark Hennings - 4 years, 10 months ago

Mark Hennings
Aug 2, 2016

Let $X_j$ be the number of packs needed to collect a new mythic card, given that you have already collected $j-1$ different cards. Then the total number of packs required is $T = X_1 + X_2 + \cdots + X_{15}$ , and hence the expected number of packs is $E[T] \; =\; \sum_{j=1}^{15} E[X_j]$ If you have got $j$ different mythic cards already, then the probability that any pack contains a new mythic card is $\frac18 \times \frac{16-j}{16} = \frac{16-j}{120}$ , and hence the distribution of $X_j$ is geometric, with $X_j \sim \mathrm{Geo}\big(\tfrac{16-j}{120}\big)$ . Thus $E[X_j] \; = \; \frac{120}{16-j}$ and hence $E[T] \; = \; 120 \sum_{j=1}^{15}\frac{1}{16-j} \; =\; 120H_{15} = 398.187....$ making the answer $\boxed{398}$ .

Arcane Collector

The answer is 398.

2 solutions

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