Three people play a simplified form of cribbage. The picture cards have been discarded, leaving 40 cards in the deck, with the aces counting as 1's.

Each of the three players is dealt 4 cards. The scoring is as follows:

• Each pair of identical numbers scores 2 points. A given card can be part of more than one such pair. That is, three identical numbers count as three pairs (C(3,2)) and four identical count as six pairs (C(4,2)).
• Each group of cards adding to 15 scores 2 points. Again, a given card can be part of more than one such group, as well as part of one or more pairs.
• If all four cards are consecutive numbers, that counts for four points. Again, this doesn't preclude cards also contributing to a total of 15.
• If any set of three of the cards are consecutive numbers, but not part of a sequence of four, 3 points are added to the score. Again, a given card can be part of more than one such sequence.

As an example, a hand consisting of 3, 4, 5 and 5 would score 8 points, as each of the 3 and the 4 take part in two sets of 3,4,5, and there is one pair.

Another example: a hand consisting of 7, 8, 8 and 8 would score 12 points (six points for the three pairs, and 6 points the three 7+8=15 totals).

In one of the deals, each of the three players had a different score, each equal to the total of the face values of the cards in that hand.

Then just the 12 cards that made up those three hands were reshuffled and dealt to the three players again. This time, the three scores all differed from one another, but for each hand the sum of the face values was a multiple (larger than 1) of that hand's score.

What were the three hands and their scores on that last deal?

In the first hand, as well as the second, the scores differed from one another.

Sorry for the omission. I've corrected the puzzle wording.

Edited on April 16, 2007, 9:11 am

Posted by Charlie on 2007-04-16 09:09:39