Background:
Ace-or-King (AoK) cards can be either an Ace or a King, but not at the same time.
In a standard 52-card Bicycle deck, there are already 4 Aces and 4 Kings.
To the deck, we will be adding 4 additional Ace-or-King (AoK) cards.
Question:
What is the chance of drawing 5 cards from this non-standard 56-card deck in which those cards consist of at least one Ace and at least one King? (Order does not matter.)
What I've tried:
I haven't tried much fruitfully. There's too many combinations of mixings of these cards for me to keep track of. I can have a deck consisting of Aces, Kings, AoKs, and Other. Normally, for combinations of cards which can't be used as semi-wildcards, I would use the following logic.
$$
P({\text{at least 1 ace and at least 1 king}}) = 1 – \frac{\binom{52 – 4 \text{ aces}}{5} + \binom{52 – 4 \text{ kings}}{5} – \binom{52 – (8 \text{ aces or kings}}{5}}{\binom{52}{5}}
$$
Build the complement up as a sum of each of the configurations in which an opening hand has none of the cards of each class, excluding the double counts across configurations without both Aces and Kings. Then, I take 1 – the ratio of those configurations with all possible configurations. This would give me the probability of drawing at least 1 Ace and at least 1 King, however, by adding the 4 AoK cards.
There would now be a case in which I draw no Aces and no Kings, drawing one AoK will not be sufficient but two AoKs will make an appropriate hand. There are also the cases in which an AoK could be used for an Ace, but it's needed for a King.
So, I would need the following configurations:
- at least 1 Ace, at least 1 King or AoK
- at least 1 Ace or AoK, at least 1 King
- at least 2 AoK
So would I compute the configurations like above for each of the above and combine all of them? Or is there some simpler model for this kind of thing?
Best Answer
When there is "at least one" in the problem formulation, inclusion/exclusion is usually not far away. There are two conditions to avoid: not drawing any A (or AoK), and not drawing any K (or AoK). Call $C_A$ the set of draws that fail for the first reason, $C_K$ the set of draws that fail for the second reason. Then $$ \#C_A=\#C_K=\binom{56-8}5=\binom{48}5, \quad\text{and} \#(C_A\cap C_K)=\binom{56-12}5=\binom{44}5 $$ And with $U$ the set of all $\binom{56}5$ possible draws, the number of draws that avoid both conditions is $$ \#\bigl(U\setminus(C_A\cup C_K)\bigr) = \#U-\#C_A-\#C_K+\#(C_A\cap C_K)=\binom{56}5-2\times\binom{48}5+\binom{44}5. $$ This includes the $4\times\binom{44}4$ possibilities where there is a single AoK and $4$ cards of no interest in the draw; should one decide to not count those as successes, then $4\times\binom{44}4$ should be subtracted from the above formula.
The numeric outcomes are $1481216$ and $938212$ depending on whether or not the mentioned cases are counted, showing incidentally that these for a quite significant portion of the successes in my initial formula.