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?

  • 1
    The player in question would also receive two for 31. – The Chaz 2.0 Nov 9 '18 at 11:22

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

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.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.