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The Ultimate Cribbage Hand (Posted on 2006-06-07) Difficulty: 4 of 5
In Cribbage, a hand scores as follows:

  • 2 points for each set of cards that totals 15 (face cards count 10, aces count 1)
  • 2 points for each pair (this means 3-of-a-kind is worth 6 points and 4-of-a-kind is worth 12 points)
  • n points for each maximal straight containing n cards
  • (i.e. a four card straight does not also count as two three card straights)
  • n points for each maximal flush containing n cards
  • (i.e. a four card flush does not also count as four three card flushes)
  • 1 point for the jack of trumps

  • It's easy to show that the best five card hand is J5555, worth 29 points, and, although impossible in an actual game, the best six card hand would be 445566, worth 46 points.

    If the entire deck of 52 cards was considered to be a single cribbage hand, what would be its value?

    See The Solution Submitted by Jer    
    Rating: 3.3333 (3 votes)

    Comments: ( Back to comment list | You must be logged in to post comments.)
    Solution re(4): Part of the solution | Comment 16 of 21 |
    (In reply to re(3): Part of the solution by Eric)

    My program was not taking into consideration that face cards, in addition to actual 10's, count for 10. The corrected central core of the program is:

     IF tot9 = 15 THEN
      ways = 4 ^ den(1) * 6 ^ den(2) * 4 ^ den(3)
      overTot = overTot + ways
      totCt(cdCt9) = totCt(cdCt9) + ways
     ELSEIF tot9 = 5 THEN
      ways = 4 ^ (den(1) + 2) * 6 ^ den(2) * 4 ^ den(3)
      overTot = overTot + ways
      totCt(cdCt9 + 1) = totCt(cdCt9 + 1) + ways
     END IF

    the correction being the bolded 2 instead of 1. (A previous correction had been the 4 ^ den(3) instead of 3 ^ den(3).)

    The new total number of ways of getting 15 is 17264.

    Broken down into how many cards have a given number of ways, that's:

     2    96
     3  1124
     4  3696
     5  5724
     6  4624
     7  1756
     8   240
     9     4

    This agrees with Eric's breakdown except only for 7 cards. Eric lists 576 ways, while my program shows 1756.

    The breakdown of the 1756 ways to use 7 cards to total 15 is as follows

    1's  2's  3's 4's 5's 6's 7's   ways
       1   4   2   0   0   0   0     24
       2   2   3   0   0   0   0   144
       2   3   1   1   0   0   0   384
       2   4   0   0   1   0   0     24
       3   0   4   0   0   0   0      4
       3   1   2   1   0   0   0   384
       3   2   0   2   0   0   0   144
       3   2   1   0   1   0   0   384
       3   3   0   0   0   1   0     64
       4   0   1   2   0   0   0     24
       4   0   2   0   1   0   0     24
       4   1   0   1   1   0   0     64
       4   1   1   0   0   1   0     64
       4   2   0   0   0   0   1     24

    For example, the 24 ways on the first row are the four ways of choosing an ace multiplied by the six ways of choosing two threes.

    If my program is correct the total points from 15's is 17264 * 2 = 34,528 .

    That gives us:

    15's           34,528
    pairs  156
    straights 872,415,232
    flushes   52
    Jack of trump 1

    for a total of  872,449,969 points.

     


      Posted by Charlie on 2006-06-08 09:19:09
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