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Two players kept the same hands: 2 threes, 1 four, 1 five. The play went as follows (count is in brackets, scoring was quickly beyond us)

  • Player 1 led with a 3 [= 3]
  • Player 2 followed with a three [= 6]
  • Player 1 followed with a three [= 9]
  • Player 2 followed with a three [= 12]
  • Player 1 followed with a four [= 16]
  • Player 2 followed with a four [= 20]
  • Player 1 followed with a five [= 25]
  • Player 2 followed with a five [= 30]

How does one score this round? (Counting the crib and hands are irrelevant to this, just the play, please.)

1
  • 3
  • 3 scores 2 for the pair
  • 3 scores 6 for the triple
  • 3 scores 12 for the quadruple
  • 4 scores nothing
  • 4 scores 2
  • 5 scores nothing
  • 5 scores 2.
  • And then that last play also scores 1 for being the last card played to that series, since the opponent has no cards to play and must say "go" at that time.

None of the plays scores a run, as there are other (foreign) cards that interrupt the runs.

Bicycle rules:

it is important to keep track of the order in which cards are played to determine whether what looks like a sequence or a run has been interrupted by a "foreign card." Example: Cards are played in this order: 8, 7, 7, 6. The dealer pegs 2 for 15, and the opponent pegs 2 for pair, but the dealer cannot peg for run because of the extra seven (foreign card) that has been played.

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  • So, in other words, double runs can't be scored during play? (And instead of playing the second 4, it would have meant another point to have played the 5 for a three-card run...)
    – Kyle
    Feb 15 at 18:35
  • Exactly right. The double runs only score in the counting of hands (and then only as a shorthand way of counting all the combinations of runs; they're equivalent). Feb 15 at 19:18
  • Thanks very much! It has been eons since I've played, and my opponent is new to the game. We were trying to add up all the double, etc. runs and got boggled!
    – Kyle
    Feb 15 at 20:37

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