# Probability involving cards, using negative binomial and hypergeometric distributions

I'm working on the following problem, and I don't really know how to approach it! It has 3 subsets: You randomly take one card at a time from a deck, with replacement. How many 'draws' are needed to guarantee a 50% chance of having exactly 2 kings (in this case, it can be the same king drawn twice).

My problem is this - I have no idea how to deal with that "exactly" 2 kings! If it weren't there, I believe I would simply set up the negative binomial with 2 successes and common probability $\frac{1}{13}$, and then just take the sum of the binomial distributions from x=1 to x=[whatever gives me 50%]. So it would be set up like this: $\sum$$(x-1)\choose(2-1)$$(1-\frac{1}{13})^{x-2}(\frac{1}{13})^2$, with x from 0 to ??, where ?? is the number that makes this yield 50% (I came up with 22, I think).

However, I don't think this is correct, since on 22 draws I may obtain my 2nd success early on, and then proceed to draw another king, which would not fulfill the requirements! Any thoughts here? I tried subtracting off the probability that 3 or 4 kings are drawn, but at higher numbers this yielded a negative probability, which certainly doesn't make sense!

The second question is basically the same, only this time sampling without replacement. I tried setting up the hypergeometric distribution with 4 type 1 objects (kings), 48 type 2 objects (not-kings), with X (the number of type one objects chosen) equal to 2. So I get [$4\choose{2}$$48\choose{k-2}$]/$52\choose{k}$, where K is the number of cards drawn. I figured I would just increase K until I got a probability >.5, however this doesn't appear to ever happen, presumably because as K gets larger the chance of 2 rather than 3 kings gets low? I figure I'm probably setting this up incorrectly, I'd really appreciate some input!

The last part involves taking cards out of the deck, without replacement, until only face cards (J,Q,K,A) are left in the deck. What is the probability that the ace of hearts is still in the deck? I figure I could probably do 16 iterations of hypergeometric (chance that all 16 are left, chance that 15 are left, chance that 14 are left...) and then multiply each probability by the chance that the ace of hearts is among those cards (so 1 for 16 face cards, (15/16 for 15 face cards etc...), but this seems kinda clunky and I'm not really sure if I'm even setting things up right on account of the previous two questions! Thanks so much for your help, I greatly appreciate it!

Edit: It should be noted, the first two parts of this question are impossible; the probability never reaches 50% in either of those cases.

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Yes, the $k$ was not a good choice of name, with $n$ you would not have made the mistake. The system does not like long strings of comments. I will delete most of mine, since you now know the contents. I suggest you do the same. – André Nicolas Dec 14 '12 at 21:29
I also had one more question (apologies for putting this in the comments!). For the final part, I need to find out the probabilities of 16...0 facecards being left in the deck; I thought I would want to use a hypergeometric distribution (since I'm sampling w/out replacement) but I don't see how to set it up properly. Is this incorrect? – Juliaaa Dec 14 '12 at 21:49
@Juliana: The experiment is incompletely described. I do not know what taking out cards until only face cards are left means. So an external referee tells us the first time that only face cards are left? If so, I can certainly get an expression for the answer. The crude approach is cases. (i) if endcard not a face card, we get probability $1$ that Ace of Hearts is gone, so case (i) contributes $(36/52)(0)$. (ii) endcard is face but next to last is not. This contributes $(16/52)(36/51)(1/16)$. And so on. If you run into trouble ask (but I will be gone for an hour or so). [Cont] – André Nicolas Dec 14 '12 at 22:13
[Cont] But there seems no point to discussing your first two questions, we just never get probability of exactly $2$ Kings anywhere near $1/2$. – André Nicolas Dec 14 '12 at 22:14
I think I am a bit confused about what the fractions represent in case (ii); (1/16) is the chance the face card is the king of hearts, but how does (36/51) denote the chance that the last card isn't a face card? And what does (16/52) represent? Thanks again! I really appreciate it :) – Juliaaa Dec 14 '12 at 22:42

Added: Please see the end for a very simple way to solve the problem.

We assume that cards are removed from the deck until only face cards are left. (So a referee tells us when to stop.) We want the probability that at that time, the Ace of Hearts is still in the deck.

It is clear that if the referee lets us go through the whole deck, because the last card isn't a face card, the Ace of Hearts is gone.

There are $16$ individual cases to examine: (i) the last card is a face card, but the one before is not; (ii) the last $2$ cards are face cards, but the one before is gone; (iii) the last $3$ cards are face cards, but the one before is not. Continue in this way until case (xvi), the last $16$ cards are face cards.

Case (i): The probability that the last card is a facecard (F) is $\frac{16}{52}$. Given it is an F, the probability the previous card is NF is $\frac{36}{51}$. So the probability we end with NF, F is $\frac{16}{52}\cdot\frac{36}{51}$. Given that this has happened the probability the last card is the Ace of Hearts is $\frac{1}{16}$, for a probability of $\frac{16}{52}\cdot\frac{36}{51}\cdot\frac{1}{16}$.

Case (ii): The probablity the last card is F is $\frac{16}{52}$. Given that, the probability the previous card is an F is $\frac{15}{16}$. Given that, the probability the previous card is NF is $\frac{36}{50}$. Given all that, the probability that the Ace of Hearts is still in the deck is $\frac{2}{16}$. So our case (ii) probability is $\frac{16}{52}\cdot\frac{15}{51}\cdot\frac{36}{50}\cdot\frac{2}{16}$.

Case (iii): The probability that the last $3$ cards are $F$, but the fourth from the end is NF, is $\frac{16}{52}\cdot\frac{15}{51}\cdot\frac{14}{50}\cdot\frac{36}{49}$. Given this has happened, the probability the Ace of Hearts is among the last $3$ is $\frac{3}{16}$. So the probability of case (iii) is $\frac{16}{52}\cdot\frac{15}{51}\cdot\frac{14}{50}\cdot\frac{36}{49}\cdot \frac{3}{16}$.

I hope the pattern is clear. One has to add up all of these. There is undoubtedly a much simpler way to view things. Certainly one can express the products above in terms of binomial coefficients. That part is not very interesting, but I think there is a way to avoid summing.

A much much simpler way: Don't worry about all the face cards except the Ace of Hearts. Confine attention to these $37$ cards.

The Ace of Hearts survives precisely if it is the last card among the $36$ non-face cards and the Ace of Hearts. Since any order of these $37$ cards is just as likely as any other order, the probability the Ace of Hearts is last among them is $\dfrac{1}{37}$.

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Makes sense to me! I put it into a sum as follows: $\Large\sum_0^{16}\frac{\frac{16!}{(16-x)!}}{\frac{52!}{(52-x)!}}\cdot\frac{36}{‌​(52-x)}\cdot\frac{x}{16}$ which yields 1/37, which makes a fair amount of sense (i think!) – Juliaaa Dec 15 '12 at 1:16
@Juliana: Good, the expression looks right, I had started to type it out in that style but decided to make it more concrete. There is, however, a big problem, which I will try to deal with when I have tine. The nature of answer screams "There is a much simpler way of seeing this!" – André Nicolas Dec 15 '12 at 1:34
@Juliaaa: Got it! Please see edited post for a very short solution. I am keeping the old ugly one to show how I missed the obvious. – André Nicolas Dec 15 '12 at 2:04