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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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    $\begingroup$ 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). $\endgroup$ – JMoravitz Jan 5 '17 at 22:53
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    $\begingroup$ @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. $\endgroup$ – Wildcard Jan 5 '17 at 23:06
  • $\begingroup$ 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) $\endgroup$ – soul Jan 5 '17 at 23:10
  • $\begingroup$ For the four aces, the calculation is $\frac {\binom {50}{6} }{ \binom {54}{10}}=\frac {15890700}{23930713170}=\frac {70}{105417} \sim 0.000664$ $\endgroup$ – lulu Jan 5 '17 at 23:23
  • $\begingroup$ For computing these kinds of probabilities, the hypergeometric distribution is your friend. $\endgroup$ – amd Jan 6 '17 at 8:08

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