Four hands of 13 cards each were dealt from a standard deck - one hand for each of the players Spott, Duce, Trey, and Nave - for a game in which aces are high.

Duce was dealt one card of each denomination, and the same number of cards of a certain suit as Trey.

Trey's hand had no red cards and no face cards.

Spott's hand had neither hearts nor spades, nor any card lower than a five.

Nave held no black cards and no aces, but he had at least as many jacks as any of the others.

Spott had the same number of cards as Duce in a certain suit and in that suit, Spott's cards were all higher than Duce's.

Can you reconstruct the contents of all four hands?

Is that same as "nobody has more jacks than Nave"?

No.

Denis, yes.

Jema, this was created by Virginia McCarthy - one of my favorite constructors.

One more question; is that same as:
Spott had no 2's, 3's, or 4's, and had at least one 5 ?

Spott had no 2's, 3's, or 4's. It's safe to say he had at least one 5 or that clue would have been worded differently.

Pandora,

I know which player has to have each of the 4 Jacks and 3 of the Queens, Kings, and Aces. I have gone over the clues and my results a dozen times and cannot see another logical placement/elimination.

Can this puzzle be solved entirely logically?

Color me purple with frustration.

I have the first 4 clues done, buy the fifth is a real tough one.

Tough but good puzzle.

This puzzle can be solved entirely logically. No if/then guessing needed. I just solved it again and took notes this time. If nobody solves this in a few days, or even after it's solved, we can do the steps one at a time.

Toughie! This is what I get...I think.

Duce: ♥A, ♦234567, ♠JQK, ♣8910
Trey: ♠2345678910A, ♣234
Spot: ♦8910QKA, ♣567JQKA
Nave: ♥2345678910JQK, ♦J

&*!@&%#(%#!!!!!!!!!!!!!!!
I took this: "Duce was dealt one card of each denomination"
to mean 4 of a kind....like 4 7's

...now I see how silly that assumption was

Nice one, Cue; a real 4 banks in the side pocket!!

Brilliant solve, Cue!

Denis, I know you'd have gotten it if you'd realized what that first clue meant.

Curly, Jema - did you get tripped up in the same place?

Yes Sue. Great puzzle anyway. Thanks.XX

It would have made it easier, anyway:
we can tell right off the bat that Duce has J-Q-K spades,
Nave has J-Q-K hearts and Spott has J-Q-K clubs....
plus Duce has Ace hearts, so none of the other Aces;
and Ace diamonds is forced to Spott, Ace spades to Trey.

Again, nice puzzle.

Pandora,

No, Clue 1 didn't trip me up. I think I understood all the clues, and I have the same solution as Cue. But I do have a question about how to make one decision using only logic.

After determining logically which player had each of the 4 Jacks and 3 each of the Queens, Kings, and Aces, I figured the next step would be to determine which of Spott's 2 suits was the common one with Duce and how many cards of that suit each player held.

I still can't see how that can be determined without some if/then. I can't see a strictly logical reason for it to be only Diamonds and 6 cards.

After determining that, I did solve the rest of the puzzle using only logic.

<<I figured the next step would be to determine which of Spott's 2 suits was the common one with Duce and how many cards of that suit each player held.>>

Exactly. Here are my sketchy notes from that point:

Spott has to have at least 4 clubs. Duce & Trey have at most 9 clubs to split evenly. Either they each have 4 clubs and leave Spott with 5 clubs or they each have 3 clubs and leave Spott with 7 clubs. Spott and Duce can't have the same nuber of clubs, hearts or spades, so they must have the same number of diamonds. If Spott has 5 clubs then he has to have 8 diamonds and Duce can't also have 8 diamonds. Therefore Spott has 7 clubs, Duce and Trey each have 3 clubs, Spott has 6 diamonds and Duce has 6 diamonds.

Sure hope this helps!

Jema, I think the suits can be logically determined:

Duce and Trey = clubs
- has to be clubs or spades (Trey only has black cards)
- ALL spades are held by Duce and Trey (Spott and Nave have none)
- only way they can have same number is 6 1/2 each
- so has to be clubs

Duce and Spott = diamonds
- has to be diamonds or clubs (Spott has only diamonds and clubs)
- if clubs, then Duce, Trey and Spott will have the same number
- we know that Nave has no clubs
- so only way Duce/Trey/Spott can have same number is 4 1/3 each
- so has to be diamonds

On the number of diamonds being 6:
- only 6 works, as you say
- then the number of clubs held by Duce/Trey are "logically" 3 (as you say)
- so this appears to require some "if/but" (must try 2 to 5 before!)
I'm still trying to find "something logical"...

What cards are left?

This is the point where things get tricky. All of the known cards are shown in red.

Spott only has two possible suits to make up his 13-card hand. Spott has to have an odd number of Clubs because Duce and Trey have matching Club counts, an even sum, and we'll need an even and an odd number to make 13. That makes Spott's possible Club count either 9 or 7. It can't be 11 because there aren't that many Clubs and it can't be 5 because he would need 8 Diamonds and that isn't possible. Since the Club count for Spott is 9 or 7, then the Club count for Trey and Duce is either 2 or 3.

Since Trey's hand is only made up of two suits then he'll need an odd and even number to get a total of 13 from the two suits. We know he either has 2 or 3 Clubs so the only workable solution is that he has 3 because he doesn't have 11 Spades to make 13 cards in his total hand. With Trey and Duce each taking 3 Clubs, that leaves Spott with 7 Clubs and 6 Diamonds.

Therefore, Spott and Duce have 6 diamonds each from which Spott takes the highest 6 cards.

Sorry it took me so long to reply - had to take care of other things yesterday and then get some sleep.

Thanks Pandora, Denis, and Cue. Denis, you explained so clearly how the common suits between Deuce and Spott and Deuce and Trey could be determined logically. As were your explanations for logically determining the number of Diamonds and Clubs in Spott's hand, Pandora and Cue.

to the 3 of you.