You start with a standard deck of cards. Each card is assigned a numeric value from 1 to 52 as follows:

A through K of Clubs = 1 through 13
A through K of Diamonds = 14 through 26
A through K of Spades = 27 through 39
A through K of Hearts = 40 through 52

Each player draws two cards and calculates the cube of the sum of the values of the two cards. Each player then selects five or less non-zero digits from their answer to form their Poker hand. Hands are evaluated solely on the digits, 1 is low and 9 is high, and there are no suits involved. For example: One player draws the 4 of Clubs (4) and the 6 of Diamonds (19); 4 + 19 = 23, 23^3 = 12167, the player has a pair of 1's. A second player draws the Ace of Dimaonds (14) and the Ace of Clubs (1); 14 + 1 = 15, 15^3 = 3375 and she wins the hand with a pair of 3's.

Given these rules, what is the best possible Poker hand a player can have and how many possible combinations of cards will yield that hand?

As I interpret the rankings for this rather odd "poker" I find no four or more of a kind, and no interpretation of a flush (same suit), so offer what I guess is the next best: a full house.

There are six ways to reach a sum of 92, and that yields a full house of three 8s and two 7s.  The cube is 778688.  Pairs would be 40+52, 41+51, 42+50, 43+49, 44+48, 45+47.

Other guesses at the rules?

Posted by ed bottemiller on 2009-08-11 17:22:48