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Playing cribbage with 4 platers
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Playing cribbage with 4 platers
Posted by gjamies1 (VIP) 2 Jun 2017 4:56am
    


I know on here we only play singles cribbage, but in many clubs in the UK we play pairs cribbage with 4 players. I have searched the internet, but cannot find the answer to a problem that recently arose on scoring and I am hoping to find the answer here as I am sure doubles cribbage must be played the world over.

The problem. The cards are dealt out, 5 cards each and each one discards one card to the crib leaving each player with 4 cards. The cards are then played as follows :-
Player 1 plays an 8
Player 2 plays a 4
Player 3 plays a 5
Player 4 plays a 3 scores 3
Player 1 plays a 2 scores 4
Player 2 plays an ace scores 5 no problems so far
Player 3 plays another 5 and scores 5, is this correct?
Player 4 plays another ace and scores 5 is this correct?
Runs in cribbage cause the most problems when scoring and I am sure that all cards on the table when face up are "live" and can count in runs. Help from this knowledgeable source would be appreciated.



Not correct
Posted by GoDiva (TD) 6 Jun 2017 12:27pm
    


Player 3 would not score 5 as the run going backwards is broken by the first 5, so does not reach the 4. ie it goes 5A235 then the 4, which does not count towards a run.



Diva is correct
Posted by jethrotulll (VIP) 23 Jun 2017 2:35pm
    


Runs must remain unbroken, but can be in any order. Diva is correct. You are missing the 4 that player 3 needs to complete their run. Also playing an A, 5 A will not score anything except possibly a GO.



Thanks
Posted by gjamies1 30 Jun 2017 11:32am
    


Thanks for the comments, and appreciate them. However, if player 3 plays a 6, second time round, that run would count as 6, so why is it that replacing the first 5 with his second 5 it does not count?



Now it will work
Posted by jethrotulll (VIP) 6 Jul 2017 12:56am
    


Here is your new example: 8,4,5,3,2,A,6,A. In this example here are your scoring runs: (4,5,3) (4,5,3,2)(4,5,3,2,A) (4,5,3,2,A,6)-Playing another Ace at the end does not do you any good. In the first example you used the 5 played, then 3,2,A,5 (5,3,2,A,5) That is NOT a run when the second 5 is played.--there is no 4 in the run) If you change your second 5 to a 6 then it works. Hope that helps!



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