<p>Note that (1) and (2) are satisfied by any <a href="http://en.wikipedia.org/wiki/Derangement" rel="nofollow">derangement</a> 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.</p>

<p>For the values, map them in this way:</p>

<p>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</p>

<p>Then: A234567890JQK -> 24680QA3579JK</p>

<p>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).</p>

<p>For the suits, consider some order of the suits [clubs, diamonds, hearts, spades]:</p>

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

<p>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.</p>

<p>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".</p>

CC BY-SA 3.0
2014-07-11T20:30:58.857Z
2014-07-11T20:30:58.857Z
Hao Ye
<p>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:</p>

<p>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♣.</p>

<p>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.</p>

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

CC BY-SA 3.0
2014-07-11T19:50:05.230Z
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
CC BY-SA 3.0
2014-07-11T19:58:18.007Z
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
CC BY-SA 3.0
2014-07-11T20:10:38.510Z
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
CC BY-SA 3.0
2014-07-11T20:11:08.297Z
The Puzzling Stack Exchange is a good idea. Thanks!
zap
CC BY-SA 3.0
2014-07-14T15:47:38.240Z
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
<playing-cards>
Bizzaro Playing Cards
zap
# Bizzaro Playing Cards

(Asked by: username_0 on 2014-07-11)

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?

**Tags:** playing-cards

> **Comments:**
> **username_1**: 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.  
> **username_2**: 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?  
> **username_0**: There are two decks, each of 52 cards, the pairing those of one deck to another. They are paired, not replaced. Sorry was unclear.  
> **username_0**: The Puzzling Stack Exchange is a good idea. Thanks!  
> **username_3**: math.SE is probably an entirely better source than puzzling.SE - though this could easily *be* a math homework problem (in abstract algebra...)  

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## 1 answers

### Answer by username_4 on 2014-07-11. Score: 2

Note that (1) and (2) are satisfied by any [derangement](http://en.wikipedia.org/wiki/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".


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