Suppose we have a standard 52 card deck from which we draw five cards. What are the chances of drawing one or more Aces and one or more Kings? How do I calculate this?
I know that we can calculate the chance of drawing one or more Ace by using cumulative hypergeometric distribution (Population 52, Subpopulation 4, Sample Size 5). This gives us a result of 34.116%.
How do I then find the odds of drawing one or more kings as well?
I mistakenly thought I could "set aside" the Ace, reducing the Population and Sample Size by 1, then calculate the chance for drawing one or more kings out of the rest (28%). Multiplying the two would find the total odds. However, the results off this - 9.765% - is well under the 10% that I continuously get out of a game simulator (multiple tries of 10 million hands).
So then, how do I calculate the above question, and - more directly - what is the correct way to calculate two different cumulative hypergeometric distributions that occur simultaneously?
Let $A$ represent the event that you have at least one ace in your hand.
Let $B$ represent the event that you have at least one king in your hand.
You ask how to calculate $Pr(A\cap B)$.
To do this, let us try to calculate the complementary event and apply some common identities.
$Pr(A\cap B)=1-Pr((A\cap B)^c)=1-Pr(A^c\cup B^c)=1-Pr(A^c)-Pr(B^c)+Pr(A^c\cap B^c)$
Each of these terms at the end are relatively easy to calculate once you recognize what they represent. $Pr(A^c)$ represents the probability that there are no aces in your hand. $Pr(A^c\cap B^c)$ represents the probability that there are no aces and there are no kings in your hand, etc...
Running the calculations, we arrive at a final answer of:
$$Pr(A\cap B)=1-\dfrac{\binom{48}{5}}{\binom{52}{5}}-\dfrac{\binom{48}{5}}{\binom{52}{5}}+\dfrac{\binom{44}{5}}{\binom{52}{5}}\approx 0.1001785$$