My friends and I were playing a game -- as part of this game you can draw cards to gain a benefit. Now, over the course of this we got into a debate over the probability of various things occurring.
What led to the debate was the game listed drawing an Ace as a lower benefit than drawing a pair.
Let me explain the rules and then I'll get into what I did and what problem I am having.
The rules are that you may draw a minimum of 6 cards (up to a maximum of around 12). How many cards you specifically may draw are dependent upon other variables. It is not something you choose as much as it is predetermined. Once you draw your allotted cards you make the best hand you can with 5 of the cards (any 5 of the cards you drew).
The listed hands you can make are here: https://pastebin.com/jkRF9jEf
Now for the probabilities. I decided the approach I would take is pretty standard (1 minus the probability of not getting the hand to see getting at least 1).
For the Ace, I knew it would be $\frac{4}{54}$ (Jokers are included and count as wildcards but have a penalty for using). However, each card you draw would reduce the total cards remaining and as such it would be $\frac{4}{54-x}$ where $x$ is the number of cards drawn. Taking the probability of it not happening would be $1-\frac{54-4-x}{54-x}$ or $1-\frac{50-x}{54-x}$. Again, $x$ is the number of cards drawn. However, that would be the probability of drawing an ace on the next draw after drawing $x$ cards. I'm pretty sure you're suppose to take it to the power of the number of trials which would make it $1-(\frac{50-x}{54-x})^x$ and this would be the probability if you're going to draw $x$ cards that at least 1 would be an Ace.
For pairs I took a similar approach: $1-(\frac{54-4x}{54-x})^x$
Now, I'm not entirely sure that is correct. I remember from back when I took statistics something about ${n \choose k}p^k(1-p)^{n-k}$ but that is giving very different results. I tried using $1-\prod_{n=1}^x{\frac{50-n}{54-n}}$ for both pairs and Aces but this is giving even more different results.
Further, I wanted to apply this to the more complex hands listed in the pastebin but am lost as to how to approach them.
Thus, my question: What am I doing wrong and how can I apply the right method to the more complex hands?
As for why I think none of these approaches are correct or that I implemented them incorrectly:
If we draw 2 cards, the probability of getting at least 1 Ace would be
$(\frac{4}{54})(\frac{3}{53})+(\frac{4}{54})(\frac{50}{53})+(\frac{50}{54})(\frac{4}{53})$
Which would be 2 Aces, no Ace and 1 Ace, 1 Ace and no Ace. The probability of getting at least 1 Ace would then be approximately 0.144 but none of the mentioned methods actually produce that result.
Additional update: It would appear I've solved it for the first two (Aces and pairs).
For Aces: $1-\prod_{n=1}^{x-1}{\frac{50-(n-1)}{54-(n-1)}}$
For pairs: $1-\prod_{n=1}^{x-1}{\frac{54-4n}{54-n}}$
For Jacks or better I'm struggling. If you were drawing 1 card the probability would be zero. If you're drawing two cards it would be, I think, $4(\frac{16}{54})(\frac{3}{53})$ because you'd have to get 1 of the four cards (Ace, Jack, Queen, King), then you would have to draw 1 of the remaining 3 cards for the first card drawn but there are 4 ways this could play out (draw Ace first, draw King first, draw Queen first, draw Jack first). However, after that, I'm not sure how it would play out.