Multiplying the results of a hypergeometric distribution to get a total probability For a (trading) card game I would like to determine the probability of a specific hand from a deck of cards. I can determine the probability of a single card occurring any number of times in an initial hand, but I don't understand how to determine a probability of a specific hand of 7 cards.
Would I multiply the probabilities of a given card with the other cards in this "golden" hand, or is there another method I am missing?
 A: Let us assume: 52 card deck, shuffled 7 times so now random, one player dealt 5 cards, then two additional cards. The probability of being dealt two aces, one king, and two other cards would be:
$P$($2$ $Aces$, $1$ $King$, $2$ $Other$) = $\dfrac{\binom{4}{2}\binom{4}{1}\binom{44}{2}}{\binom{52}{5}}$ = $\dfrac{6*4*946}{2,598,960} = 0.0087358.$
Now, two more cards are dealt, but the probability of this hand is conditional on the first hand. Let us consider next the likelihood of being dealt two aces (assuming 4 $Aces$ and 1 $King$, two other cards discarded, yields the $Golden Hand$) given we already have two aces, one king, and two other cards (not an ace and not a king):
$P$($2$ $Aces$$_{second hand}$ | $2$ $Aces$, $1$ $King$, $2$ $Other$$_{first hand}$) = $\dfrac{\binom{2}{2}\binom{42}{0}}{\binom{44}{2}}$= $\dfrac{1*42}{946} = 0.04439746.$
So that the probability of first being dealt $2$ $Aces$, $1$ $King$, $2$ $Other$ $and$  $then$ being dealt $2$ $Aces$ on the second hand would be: 
$0.0087358$ * $0.04439746$ = $0.00038785$
Does this reply help?
