## Problem No. 12, 2002

### The problem

I first heard about our new problem a year ago, but I have no idea about the source. I have modified it slightly, and dare say it is even better now.

The following ten clues have all the information you need to know the full deal (all 52 cards).

1. All players have 10 high card points.
2. All players have one queen.
3. South and West are void in the same two suits.
4. Neither king-queen of diamonds nor king-queen of clubs are in the same hand.
5. The red aces are in different hands.
6. North has no kings.
7. South has no jacks.
8. East has the heart jack.
9. North's spot cards are all even.
10. West has no spot cards of another value than any of East's.

### Solution

How do you solve such a problem? By logical thinking, of course. One of our solvers, Ingemar Assarsjö of Sweden, has shown how it can be done by a simple step-by-step reasoning, and since I can't explain it better myself, I'll let Ingemar do it. Over to him!

If you use the conditions in the right order, you can find the solution by reasoning like this.

1. All players have 10 high card points.
2. All players have one queen.

The only possible combinations of 10 high card points including one queen are A-A-Q, A-K-Q-J, A-Q-J-J-J-J and K-K-Q-J-J.

For all high card points of the deck to be present, the four hands has to be
a) A-A-Q, A-A-Q, K-K-Q-J-J, K-K-Q-J-J
or
b) A-A-Q, K-K-Q-J-J, A-K-Q-J, A-K-Q-J
or
c) A-K-Q-J, A-K-Q-J, A-K-Q-J, A-K-Q-J

3. South and West are void in the same two suits.

This means that South and West have all 26 cards from two of the suits – and that North and East shares the 26 cards from the two other suits.

6. North has no kings.

North has to have A-A-Q. Since North and East shares two suits, East has to have K-K-Q-J-J.

7. South has no jacks.

South and West also shares two suits, and since South can't have any jacks, South has to have A-A-Q, while West has K-K-Q-J-J.

8. East has the heart jack.

North and East share hearts and another suit.

5. The red aces are in different hands.

Since North and South have two aces each, North and East must share hearts and a black suit.

4. Neither king-queen of diamonds nor king-queen of clubs are in the same hand.

The players with K-K-Q-J-J can't have both clubs and diamonds, since that would mean K-Q of either clubs or diamonds. And if South and West can't have clubs + diamonds, North and East can't have hearts + spades. But North and East should have hearts plus a black suit, so North and East has to have hearts + clubs.

The players with K-K-Q-J-J can't have K-Q in clubs or in diamonds. Therefore, East has K-Q-J of hearts, and West has K-Q-J of spades.

This is what we have so far:

 – A – A Q K Q J – – K Q J K J – – K J A – A Q –

9. North's spot cards are all even.

North shall have ten more cards, and there are exactly ten even numbered spot cards in hearts and clubs. So, North shall have all of them. East gets the remaining spot cards in those two suits.

 – A 10 8 6 4 2 – A Q 10 8 6 4 2 K Q J – – K Q J 9 7 5 3 K J – – K J 9 7 5 3 A – A Q –

10. West has no spot cards of another value than any of East's.

The remaining odd spot cards in spades and diamonds are exactly the same as East's. West therefore gets the same valors on his spot cards as on East's. And the full deal looks like this:

 – A 10 8 6 4 2 – A Q 10 8 6 4 2 K Q J 9 7 5 3 – – K Q J 9 7 5 3 K J 9 7 5 3 – – K J 9 7 5 3 A 10 8 6 4 2 – A Q 10 8 6 4 2 –

If you have followed us so far, you realize that all conditions were necessary and that there cannot be any duals. Thank you for your help, Ingemar!

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