<<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.