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A standard deck of cards has 4 suits, with 13 cards per suit. 2 different colored decks (red and blue) are shuffled together to form a single 104-card deck. Red suits are distinct from blue suits. Some cards are picked up randomly from the deck without replacement.

a) What is the expected number of suits picked up when 30 cards are randomly chosen?

b) How many cards should be picked up so there is at least a 99% chance of getting all 8 suits?

From How many ways can $b$ balls be distributed in $c$ containers with no more than $n$ balls in any given container? we can use $$N(c,s,13) = N(c-1, s, 13)+N(c, s-1, 13)-N(c-14, s-1, 13).$$ to recursively get number of ways to divide $c$ cards into $s$ suits, where each suit has between $0$ and $13$ cards inclusively.

But the question asks for expected value of the number of suits. So we need to know the distribution. I was thinking for something like $(5, 6, 7, 12, 0, 0, 0, 0)$ we can first find ways to get positive number of cards in 4 suits (via stars and bars), then times by $\frac{8!}{4!}$ to account for the all the ways colors can be assigned. But for $(4, 4, 4, 5, 5, 8, 0, 0)$ we have to do $\frac{8!}{3!2!2!}$ for the permutation with duplicates, that is a different treatment is need for each type of "repeat structure".

It also feels like I need to apply multinomial or hypergeometric distribution which might mean iterating through all the partitions but accounting for 13 cards max. That will work for at least part a) but it does not appear to be the spirit of the question.

okzoomer
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1 Answers1

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(a) Define $$X_i = \begin{cases} 1 \qquad \text{if at least one card of suit } i \text{ is included in the sample} \\ 0 \qquad \text{otherwise} \end{cases}$$ for $1 \le i \le 8$. Then $$P(X_i = 1) = 1 - \frac{\binom{91}{30}}{\binom{104}{30}}$$ So by linearity of expectation, the expected number of suits is $$E \left( \sum_{i=1}^8 X_i \right) = \sum_{i=1}^8 E(X_i) = 8 \left( 1 - \frac{\binom{91}{30}}{\binom{104}{30}} \right)$$ (b) We want to find the probability that all $8$ suits are represented when $n$ cards are drawn without replacement from the deck. We will use the Principle of Inclusion and Exclusion.

Let's say a sample of $n$ cards has "property $i$" if there are no cards from suit $i$ for $1 \le i \le 8$, and let $S_j$ be the total of all the probabilities for the samples that have $j$ of the properties, for $1 \le j \le 8$. So $$S_j = \binom{8}{j} \frac{\binom{104-13j}{n}}{\binom{104}{n}}$$ By Inclusion / Exclusion, the probability that a sample has none of the properties, i.e. has no void in any of the suits, is $$p_0 = 1 - S_1 + S_2 - S_3 + \dots + S_8$$ By experiment, we find that the least value of $n$ with $p_0 > 0.99$ is $n= 40$, with $p_0 = 0.0.991416$

awkward
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  • that's amazing! i got 39.3 cards (so round up to 40 cards) for part b. is that correct? I solved for $P(X_i = 1)^8 > 0.99$ replacing $30$ with $n$ or would they not be independent? – okzoomer Nov 19 '21 at 14:31
  • also $39$ is very close to $\frac{104}{e} \approx 38.3$, not if that's a coincidence or not – okzoomer Nov 19 '21 at 14:40
  • @okzoomer For (b) you need a different approach. I edited my answer to include the answer to (b), using PIE. – awkward Nov 19 '21 at 15:47
  • thanks again. is that supposed to be $\binom{8}{j} \frac{\binom{104-13j}{n}}{\binom{104}{n}}$? also bit confused about the 100 and 110 but 104 would make more sense? – okzoomer Nov 19 '21 at 15:57
  • @okzoomer Oops! Yes, corrected. – awkward Nov 19 '21 at 15:59
  • thanks for being patient with me ... just wondering are the 100 and 110 supposed to be 104? just making sure i understood your solution – okzoomer Nov 19 '21 at 16:04
  • @okzoomer I should be thanking *you* for being patient, because yes, those numbers should be $104$, and it makes a difference in the final answer. I hope it's straight now. – awkward Nov 19 '21 at 22:10