# Bizzaro Playing Cards [closed]

Here’s a problem a friend of mine posed to me a long time ago, that I was recently reminded of. I figured, with modern computing, there might be some way to figure it out:

Assume that there are two set of 52 cards, 13 of each of four sets. Now, for the purpose of a game, each card of the first deck becomes associated with one card in the other card such that, say, A♠ = 5♣.

The second set follows these conditions: 1.) No card may be it’s own twin 2.) No card may appear twice 3.) Four of a kind in the normal deck may not repeat suits in the second deck. 4.) Any poker run in a normal deck (straight, flush) must be disrupted in the second.

Is there a set of cards that matches these conditions? And if so, what is it?

• This question appears to be off-topic because it is about a puzzle that doesn't have anything to do with an actual game. Try the Puzzling Stack Exchange. – bwarner Jul 11 '14 at 19:50
• Your third and fourth conditions are unclear. From the first two conditions, it seems like you are looking for a partition of the 52 cards into 26 disjoint pairs. However, last two conditions' mentions of the "second deck" make it sound more like you are looking for a bijective mapping from the 52 card deck to another 52 card deck. Basically, do you want to pair cards up and consider those pairs as a single unit, or do you want to replace each card in the first deck with another card in the second deck? – murgatroid99 Jul 11 '14 at 19:58
• There are two decks, each of 52 cards, the pairing those of one deck to another. They are paired, not replaced. Sorry was unclear. – zap Jul 11 '14 at 20:10
• The Puzzling Stack Exchange is a good idea. Thanks! – zap Jul 11 '14 at 20:11
• math.SE is probably an entirely better source than puzzling.SE - though this could easily be a math homework problem (in abstract algebra...) – The Chaz 2.0 Jul 14 '14 at 15:47

Note that (1) and (2) are satisfied by any derangement of the cards. There are probably many "mappings" that satisfy your conditions, but since all you seek is one possible answer, I propose the following solution.

For the values, map them in this way:

A -> 2, 2 -> 4, 3-> 6, 4 -> 8, 5 -> 10, 6 -> Q, 7 -> A, 8 -> 3, 9 -> 5, 10 -> 7, J -> 9, Q -> J, K -> K

Then: A234567890JQK -> 24680QA3579JK

So clearly any straight will not map to any straight. And because any 4 of a kind maps to a 4 of a kind, there won't be any suit duplication (your condition 3).

For the suits, consider some order of the suits [clubs, diamonds, hearts, spades]:

• For values A-4: map to the same suit
• For values 5-8: map to the next suit (e.g. clubs -> diamonds, spades -> clubs)
• For values 9-Q: map to the previous suit
• For K: map to the "opposite suit"

Then any flush (which consists of 5 cards of the same suit) cannot map to a flush, because you cannot obtain more than 4 cards of the same suit if you start with a flush.

Note that changing the suit of the K resolves the issue where K -> K, since it is required that no card be its own "twin".