5
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ It is not clear to me if you count draws with one AoK card and nothing else of interest as successful. I would say both clauses of you conjunction (at least one A) and (at least one K) are true, so this should be a success. But some people would maybe give a different meaning to "and" here.In any case you can easily count the number of such occurrences, and add or subtract that number if you have used a method that supposes an opposite ruling of these cases than the arbiter has decided. $\endgroup$
    – Marc van Leeuwen
    Jun 25 at 5:06
  • $\begingroup$ 1 AoK can only satisfy either 1 Ace or 1 King, but not both at the same time. Thus, if you don't have any natural Aces or natural Kings, you will need 2 AoKs to be a satisfactory hand. $\endgroup$
    – Axoren
    Jun 25 at 5:09
  • $\begingroup$ I really don't know how I'd really clarify that further without implementing some sort of intermediate step where you have to replace AoKs with either one or the other, but that's effectively what's happening. Were it possible for 1 AoK to satisfy both at the same time, this would be a MUCH simpler problem. $\endgroup$
    – Axoren
    Jun 25 at 5:10
  • 1
    $\begingroup$ It's pretty simple, you already know all the ways of doing it without a new card, so you only have to consider among the new hands that are created, which means you have at least one a-or-king, and to generate a failure you need to not hit another ace, king, or ace-or-king. Details in my answer $\endgroup$
    – Alan
    Jun 25 at 5:13

2 Answers 2

Reset to default
4
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ Just to be clear, you're using $\#$ as a cardinality marker here, right? It took me a bit to follow the logic, but I see how you're removing the double counts now. By organizing the configurations into sets, you were able to better describe the relationships of their cardinalities through some few known sets. Might be a simpler approach than what I was planning to do for some of my other cases. $\endgroup$
    – Axoren
    Jun 25 at 5:36
  • $\begingroup$ Yes, read $\#$ as "number of" (which I find more evocative than suggesting an absolute value of a set). My method is classic inclusion/exclusion, here for a set of just two conditions. OP seems to also be using inclusion/exclusion, but only for the traditional deck case. $\endgroup$
    – Marc van Leeuwen
    Jun 25 at 5:58
2
$\begingroup$

Well, you can start with the known case of the 52 card deck, and add to that the extras that come in the 56 card deck. You have $\binom {56} 5-\binom{52} 5$ new hands you are adding in which at least 1 of the new cards is drawn. Of those, we know at least 1 is an ace-or-king, so we can consider the remaining 4 cards. A hand is a success if at least one of them is an ace, a king, or an ace-or-king, thus there are $4+4+3=11$ possible choices for a successful card, and your failure cards start at the 2 through queen of each suit, or $44$ cards. In order to generate a failure, you need 4 of these bad cards, or $\binom {44} 4$. So the additional failure options you get by adding the 4 cards is $\binom {44} 4$, add those to the existing failure options, then change the total to $\binom {56} 4$

$\endgroup$
5
  • $\begingroup$ This is a very interesting perspective, to focus on the way the extra configurations arise from adding the cards! I'm going to think a bit deeper on this as I try to digest your answer. This paradigm might help to solve quite a number of additional cases I'm considering beyond just this one! $\endgroup$
    – Axoren
    Jun 25 at 5:16
  • $\begingroup$ So, if I'm understanding correctly, if I were then to consider adding additional 4 AJKs (count as A, J, or K), rather than taking a step back and considering 8 wild cards, I could chain this reasoning by then adding an additional $\binom{44}{4}$ failed hands to the complement case and dividing by the new total deck configurations? $\endgroup$
    – Axoren
    Jun 25 at 5:22
  • $\begingroup$ Sorry, I cannot decipher the procedure for computation you are trying to describe; for instance are you counting successes or failures? It would be easier if you just stated a formula that describes the final result rather that a method to arrive at such a formula. $\endgroup$
    – Marc van Leeuwen
    Jun 25 at 5:30
  • $\begingroup$ @MarcvanLeeuwen From what I can see, if I was understanding correctly, he's describing how to compute the additional failure configurations. These can be added to the formula I provided in my "What I've tried so far" section, where I compute the complement and deduct those configurations from the total to get the successes (and then the ratio of success to total) $\endgroup$
    – Axoren
    Jun 25 at 5:38
  • $\begingroup$ It would be good if the full formula were laid out, and the result shown. $\endgroup$
    – true blue anil
    Jun 25 at 8:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.