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.