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So I'e done a bit of searching and haven't found an answer or a source I can re-learn the probability equations so here's hoping for an answer.

I'm curious if you have a deck of 54 cards (jokers) and you draw 10 cards, what are the chances of drawing 4 aces?

What about 4 aces and 1 or 2 jokers?

Cards are not being replaced but we can assume a fresh 'perfectly' shuffled deck. The other cards do not matter.

I would appreciate the math as it would be helpful to understand how to perform this myself, it's just been ~15 years since I've needed to use any of this.

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    It is easiest to work with "hands" here where order within the hand is irrelevant. There are $\binom{54}{10}$ different hands of size ten possible. $\binom{50}{6}$ of those have all four aces (*and possibly some jokers*). $\binom{48}{6}$ of them have four aces and no jokers, $\binom{48}{5}\cdot 2$ have four aces and one joker, and $\binom{48}{4}$ have four aces and both jokers. Take the ratio of number of possible "desired" hands over the total number of hands possible to get the probability (*in an equiprobable setting such as this one*). – 2017-01-05
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    @JMoravitz, could you please post that as an answer? Even though brief, it *does* answer the question and IMO it is more clear than the only posted answer so far. – 2017-01-05
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    Yeah. The other was deleted anyway. I'm just not sure how to break those down into a percentage based chance. I found it at about 2% myself but I just have no way of knowing if I'm actually correct. Also not sure how I could apply it differently (e.g. 2 aces of the 10 cards, etc) – 2017-01-05
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    For the four aces, the calculation is $\frac {\binom {50}{6} }{ \binom {54}{10}}=\frac {15890700}{23930713170}=\frac {70}{105417} \sim 0.000664$ – 2017-01-05
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    For computing these kinds of probabilities, the [hypergeometric distribution](https://en.wikipedia.org/wiki/Hypergeometric_distribution) is your friend. – 2017-01-06

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