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CONDITIONS

Imagine, that we have five standard 52-card decks. We will now mix all these five decks into a one (afterwards "big deck", 260 cards in total). Now our task is to turn all the cards open, one card at a time, until we can form "four of a kind" from these cards that have been turned open ("dealt").

QUESTION:

What is the maximum number of these cards to be dealt (in worst case scenario), where we still can not form any "four of a kind" that consists of:

1) any four cards that have the same numeric value (example "8♥ 8♥ 8♣ 8♣" or "J♥ J♣ J♣ J♣");

2) four completely suited cards (example "K♥ K♥ K♥ K♥" or "7♣ 7♣ 7♣ 7♣").

3) four completely offsuited cards (example "K♥ K♦ K♣ K♠" or "7♣ 7♥ 7♦ 7♠").

The suite of those cards that doesn't form a four of a kind is irrelevant (they can be both suited or offsuited).

My anwsers

1) any four of a kind

We simply run three of a kinds for all 13 different numeric values (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K and suit doesn't matter). We now get a total of 39 cards.

$$ (3 \cdot 13) = 39 $$

The 40th card to be dealt needs to gives us one four of a kind no matter what card it is. So my anwser is 39 (40th card will give us four of a kind - suited or offsuited).

Note that the number of decks used is irrelevant for the assigment (assuming that we use at least one complete deck of cards).

2) four of a kind - suited

Here we need to avoid any four of a kind that consists of the same suit. What were gonna do is to deal three cards from all suits per numeric value (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K) - three of each, because the whole idea is to avoid the 4th. Now we have a total of 12 cards per numeric value like this

A♥ A♥ A♥, A♦ A♦ A♦, A♣ A♣ A♣, A♠ A♠ A♠ and 2♥ 2♥ 2♥, 2♦ 2♦ 2♦, 2♣ 2♣ 2♣, 2♠ 2♠ 2♠ ect.

As there are 13 different numeric values (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K), there is a maximum of $12 * 13 = 156$ different cards without any four of a kind: same suit. Let's deal these cards described above and see where we stand.

The next one, let's make it a six of clubs (6♠), is the fourth six of clubs dealt, so the result is 156. No matter what the card to be dealt is, it will inevitably be the 4th of that specific suite and value. Since the 157th card dealt will always give us four of a kind that consists of four same suit and value, we have now reached the maximum.

Note that for this task there must be at least four decks used. Any added deck after four doesn't effect at all; the limit comes from the fact that there are only 13 different numeric values per deck.

3) four of a kind - offsuited

We deal all cards from all 5 decks, except one suit (in this example we exclude spade ♠):

A♥ A♥ A♥ A♥ A♥, A♦ A♦ A♦ A♦ A♦, A♣ A♣ A♣ A♣ A♣ and 2♥ 2♥ 2♥ 2♥ 2♥, 2♦ 2♦ 2♦ 2♦ 2♦, 2♣ 2♣ 2♣ 2♣ 2♣ ect.,

so we got a total of 15 individual cards for each numeric value. As we have 13 different numeric values in five decks, and all the spades will be taken out of the deck, we are left with a total of 195 cards. There is also an opposite way to do this. First let's calculate

$$ X \cdot Y = 65 $$

Where X is the number of spades per deck (13) and Y is the number of the decks used (5). We now substitute 65 from original 260 cards and our anwser is 195. We can also do this also by

$$ (A \cdot B) \cdot C = 195 $$

where A is number of one suit per deck, B is number of suits minus one suit (not including one suit) and C is the number of decks used. The anwser to this is 195 cards, that still won't make any offsuited four of a kind. Changing the number of decks used makes a big difference in the math (four decks will do $(13 * 3) * 4 = 156$ for example). Here is the previous equation again:

$$ (A + B + C) \cdot X = 195 $$

Where A is number hearts per value in 5 decks (5), B is number of diamonds per value in 5 decks (5), C is number of clubs per value in 5 decks (5) and Y is number of different numeric values in a card deck (13).

Given that we now have used all hearts, diamonds and clubs in the deck, the next card dealt, 196th in order, must be spade, because there are no other cards left in our big deck. The 196th card creates the first four of a kind specified in article 3).

Note that the more decks we use, the more cards of the same suite we can deal. If we use only one deck, the result is the same as in article 1): 39 cards - 40th will make the first four of a kind, offsuited of course up to three decks, because there just cannot be four ace of spades in three standard 52-card decks.

Am I right with these?

P.S. a lot of edits are made to improve the structure and readability of the question.

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    $\begingroup$ I think all your numbers are right. You certainly have tight arguments that the maximum number is in each case at least what you say it is. However, one should argue in each case that it really is the maximum, using an explicit pigeonhole argument. This will be easy for the first problem, and a little more unpleasant for the third. $\endgroup$ – André Nicolas Dec 27 '15 at 0:05
  • $\begingroup$ Thank you for your kind comment! I will delve into that pigeonhole argument and maybe then complete my anwser. $\endgroup$ – Kasperi Koski Dec 27 '15 at 0:55
  • $\begingroup$ You are welcome. Good quality work, with clear explanations. $\endgroup$ – André Nicolas Dec 27 '15 at 1:01

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