Ggtkuk

Albert, Fred, Larry, Ron and Patrick are playing cards at Albert’s house with a regular 52 card deck that is missing all the 2s. Patrick has 10 cards, none of which are higher than 6. Ron has 13 cards, but has no royalty or love. In Fred’s hand are 9 cards, none of which are the same number, but all are the same color and nothing is higher than 11. Larry has 15 cards, each of a different shape and spaced 5 apart. Five of his cards equal to 56 and a king can only be found in his hands. Only 2 people have a 7 and 3 have an 8. One person is exclusive and holds all of the same. Lonely is one man with only one card to his name. So, what’s the pattern to the name of this game, that’ll crack the code to reveal the hand of the lonely man, and the owner of the game?

Albert
Patrick
Ron
In
Larry
Five
Only
One
Lonely
So

Either that's a coincidence or Im guessing that's the pattern that cracks the code? Pretty interesting.

Partial

As it stands, it seems fairly trivial that the "lonely man" is

Albert

since

He's the only one whose hand isn't listed, and as the others hands total 47 out of a possibly 48, he must have the remaining card.

In terms of working out the actual hands, Larry must have

4 Kings (stated), at least 3 Queens (nobody else has higher than 11, except possibly Albert). It's not clear what "spaced 5 apart means"

Partial solution.

Assumptions:
Love means Hearts.
Royalty means 10-11-12-13-1 as in Royal Flush and not 11-12-13 which are called Court Cards.


Summarise the direct information:
Patrick holds 10 cards up to $6$ in all suits.
Ron holds 13 cards up to $9$ but no Hearts.
Fred holds 9 cards up to $11$, no duplicate rank in Red or Black.
Larry holds 15 cards including 4 Kings.
Albert holds 1 card.


Is Ace high or low? That matters because of "higher than $6$" etc.

If Ace is high, then Patrick, Ron and Fred hold none, Albert perhaps one, so Larry has 3 or 4.

In that case Larry must be holding at least 3x$1$, 3x$10$, 3x$11$, 4x$12$ and 4x$13$ which is 17 cards, less one that Albert might hold, which is still more than 15.

Which means that
Ace is low.


Summarise again:
Patrick holds 10 cards from $1, 3, 4, 5, 6$. All suits.
Ron holds 13 cards from $3, 4, 5, 6, 7, 8, 9$. No Hearts.
Fred holds 9 cards from $1, 3, 4, 5, 6, 7, 8, 9, 10, 11$. No duplicates. Red or Black.
Larry holds 15 cards including 4 Kings. All suits.
Albert holds 1 card.


Fred must hold at least one of the $10$ and $11$, because he has 9 out of 10 possibilities.

So Larry holds - all possibly one less because of Alberts's card:

3x$10$, 4x$11$, 4x$12$ and 4x$13$ (14 or 15 cards), or

4x$10$, 3x$11$, 4x$12$ and 4x$13$ (14 or 15 cards), or

3x$10$, 3x$11$, 4x$12$ and 4x$13$ (13 or 14 cards).

There is a strange clue about Larry "spaced 5 apart", so let's assume that his highest other card is ($10$ - $5$) = $5$.


One clue is "only 3 people have a $8$", who now must be Ron, Fred and Albert.

This changes Larry's high-value cards to one out of:

3x$10$, 4x$11$, 4x$12$ and 4x$13$ (15 cards), or

4x$10$, 3x$11$, 4x$12$ and 4x$13$ (15 cards), or

3x$10$, 3x$11$, 4x$12$ and 4x$13$ (14 cards).

but since Larry has at least one lower card, it must be the last one:
Larry holds 1x$5$, 3x$10$, 3x$11$, 4x$12$ and 4x$13$. Some unknown suits.
Albert holds an $8$. Unknown suit.
Fred holds both the $10$ and $11$.


Now look at the $9$s, it must be Ron who holds 3x$9$ and Fred holds $9$ of Hearts.

So Fred's two suits are Hearts and Diamonds.


Another clue is "only 2 people have a $7$", who are Ron and Fred.

So it must be Ron who holds 3x$7$ and Fred holds $7$ of Hearts.

Partial table of cards held

       Cards   1  3  4  5  6  7  8  9 10 11 12 13

 Patrick  10   3              -  -  -  -  -  -  -

 Ron      13   -              3  2  3  -  -  -  -  No Hearts

 Fred      9   1              1  1  1  1  1  -  -  Hearts & Diamonds 

 Larry    15   -  -  -  1  -  -  -  -  3  3  4  4

 Albert    1   -  -  -  -  -  -  1  -  -  -  -  -

This is as far as I have got, and there are some unsolved clues still.