0
$\begingroup$

I'm trying to get more into stats and I had a problem that I couldn't wrap my head around.

It's a solitaire card game in which you play 4 hands. Each hand consists of putting 4 cards on the table. There are four unique decks to choose your cards from. The decks are Jack (J), Queen (Q), King (K), and Aces (A), one of each suit.

With each hand, you start by taking a card from the top of one of the decks and place it on the table. Next, you take another card from the top of the deck and place it to the left or the right of the first card. Again, the third card comes off the top of one of the decks, and goes to the left of the other two cards, in the middle of them, or to the right of them. The same thing happens with the fourth card.

You are able to choose the same pile twice in a hand if you like. Here is a sample turn,

Choose from Aces

Table - |A|

Choose from Queens, place to the left of the Ace

Table - |QA|

Choose from Kings, place in the middle

Table - |QKA|

Choose from Queens, place to the right of the King

Table - |QKKA|

So I'm trying to find the number of possible ways to play a game. I know that when playing, you can choose from 4 piles, then place the card in 1, then 2, then 3, then 4 places. That's pretty easy to calculate. But my problem is how to take into consideration the limited size of each deck. If I choose to play all four Aces in the first hand, then I will only have 3 decks to choose from in the next hand.

Any insight would be greatly appreciated.

Sidenote - This is an abstraction of the game "Miracle Merchant" on mobile devices if anyone was wondering what it was based on.

$\endgroup$
3
  • $\begingroup$ "Jack (J), Queen (Q), King (K), and Aces (A), one of each suit" — Those are face cards, not suits. $\endgroup$ – Kodiologist Mar 8 '18 at 0:38
  • $\begingroup$ Sorry, that wasn't very clear. In each pile, there is a card of each suit. There are piles of Jacks, Queens, Kings, and Aces. Hope that clears it up. $\endgroup$ – Kristoffer Anderson Mar 8 '18 at 1:01
  • $\begingroup$ If I answered your question to your satisfaction, you can accept my answer by clicking the check mark under the voting arrows. $\endgroup$ – Kodiologist Mar 22 '18 at 4:25
0
$\begingroup$

I see three ways to interpret "the number of possible ways to play a game".

(1) The number of possible layouts after you've drawn every card

This is just 16! = 20,922,789,888,000, the number of permutations of 16 distinct objects (presuming we count cards of the same face but different suit as distinct).

(2) As (1), but counting cards of the same suit as identical

This is $\frac{16!}{4!^4} = 63,063,000$.

(3) The number of possible sequences of moves in a complete game, given a fixed deck

You'll need to draw 4 cards from each of 4 piles, so your sequence of drawing choices will be a permutation of 4 groups of 4 identical elements, just like (2). You have 1 choice for where to put the first card, 2 choices for the second, 3 for the third, and so on, for a total of 16! positions as in (1). Multiplying (1) and (2) together yields 1,319,453,898,706,944,000,000.

$\endgroup$

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.