Consider laying out 10 sets of 4 cards plus a turn-up, from a single deck of cards, forming what might appear to be 10 cribbage hands including a different turn-up for each. (Two cards will be left over.)

How could these sets be arranged to maximize the sum of the scores of these hands?

http://en.wikipedia.org/wiki/Cribbage may provide you with some guidance.
Note: Each hand has its own turn-card so a single jack in each hand could be the right jack.

A total of 205 points can be achieved with the following
ten hands:

 A, A, 2, 2, 3 = 16
 A, A, 2, 2, 3 = 16
 4, 4, 4, 4, 3 = 20
 6, 6, 6, 6, 3 = 24  
 7, 7, 8, 8, 9 = 24
 7, 7, 8, 8, 9 = 24
10,10,10,10, 5 = 20
 J, J, J, J, 5 = 21
 Q, Q, Q, Q, 5 = 20
 K, K, K, K, 5 = 20
===================
Total           205    

The two cards left over are two 9s.

Edited on July 18, 2006, 9:14 pm

Posted by Dej Mar on 2006-07-18 19:43:42