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During cribbage play, (three) players play (for example) 7-9-8, last player scores 3, and the next player plays a 7 and claims that he scores 3 points for the last three cards. Is that double-counting and not allowed, or is it fair?

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    The player in question would also receive two for 31. – The Chaz 2.0 Nov 9 '18 at 11:22
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That's completely valid. One player scoring a run doesn't stop those cards being used in a following run.

  • Perhaps it would complete the answer to show examples, where a run is lengthened on each successive card. In theory one could have the extreme of "run for 7" followed by "run for 7" followed by "run for 7 and 2 for 31". – Nij Nov 11 '18 at 18:52
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According to the rules at Bicycle's site the pegging rules state:

For adding a card that forms, with those just played: [...] a sequence of three Peg 3

There is no requirement that the cards are played in order, which is of course why 7-9-8 is a 3 point sequence.

Next note:

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.

The example given is 8-7-7-6. Playing the 6 here does not form a run as the 2nd 7 disrupts the run with the 8.

In your scenario 7-9-8-7 there is a sequence 9-8-7 formed which is valid for 3 points.

The easy way to remember this is you peg the number of cards in the run. 9-8-7 is a run, so away you go with 3 points. If I play AA and you play A I don't get to say "Hey, wait a second, I already scored the points on that pair". You'd say so what? Now I have AAA and that's 6 points. Same idea here.

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