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"If a player holds a 5 in his hand, he is guaranteed at least two points."

Can someone explain this to me? I can't really figure out how... 72.83.147.28 (talk) 20:13, 21 May 2008 (UTC)[reply]

If you have a five, and you want to score less than two, you can't have a ten (two for fifteen), another five (two for a pair) or indeed any other pair. You can't have a six and a four (six, five and four score two for fifteen and three for a run), you can't have a seven and a three, you can't have a eight and a two and you can't have a nine and an ace. So for a hand of four and a starter card, you have to have the five plus an extra four cards, so you need one of six and four, one of seven and three, one of eight and two and one of nine and ace. This gives sixteen possibilities. However, in each of these cases you can make fifteen from at least one combination of the cards:
Cards Fifteen
5 of diamonds6 of spades7 of hearts8 of clubs9 of diamonds 6 of spades9 of diamonds
5 of diamonds6 of spades7 of hearts8 of clubsAce of diamonds 7 of hearts8 of clubs
5 of diamonds6 of spades7 of hearts2 of clubs9 of diamonds 6 of spades9 of diamonds
5 of diamonds6 of spades7 of hearts2 of clubsAce of diamonds 6 of spades7 of hearts2 of clubs
5 of diamonds6 of spades3 of hearts8 of clubs9 of diamonds 6 of spades9 of diamonds
5 of diamonds6 of spades3 of hearts8 of clubsAce of diamonds 6 of spades8 of clubsAce of diamonds
5 of diamonds6 of spades3 of hearts2 of clubs9 of diamonds 6 of spades9 of diamonds
5 of diamonds6 of spades3 of hearts2 of clubsAce of diamonds 5 of diamonds6 of spades3 of heartsAce of diamonds
5 of diamonds4 of spades7 of hearts8 of clubs9 of diamonds 7 of hearts8 of clubs
5 of diamonds4 of spades7 of hearts8 of clubsAce of diamonds 7 of hearts8 of clubs
5 of diamonds4 of spades7 of hearts2 of clubs9 of diamonds 4 of spades2 of clubs9 of diamonds
5 of diamonds4 of spades7 of hearts2 of clubsAce of diamonds 5 of diamonds7 of hearts2 of clubsAce of diamonds
5 of diamonds4 of spades3 of hearts8 of clubs9 of diamonds 4 of spades3 of hearts8 of clubs
5 of diamonds4 of spades3 of hearts8 of clubsAce of diamonds 4 of spades3 of hearts8 of clubs
5 of diamonds4 of spades3 of hearts2 of clubs9 of diamonds 4 of spades2 of clubs9 of diamonds
5 of diamonds4 of spades3 of hearts2 of clubsAce of diamonds 5 of diamonds4 of spades3 of hearts2 of clubsAce of diamonds
Hope that makes sense. TimR (talk) 17:52, 22 May 2008 (UTC)[reply]

Highest Combined Score

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I have added a paragraph on the highest score for both players in a single deal. I state that 105 is the highest known score. Can anyone find a higher one? Sicherman (talk) 16:02, 3 November 2012 (UTC)[reply]

Yes. The counterexample is two paragraphs up. In the maximum score for the dealer hand where the dealer gets 78 points, the non-dealer pegs 12 and then scores 20 for a total of 32 points - 110 points combined for both players in a single deal. --Noren (talk) 18:38, 18 July 2014 (UTC)[reply]

Removed a section

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I took out a section incorrectly claimed that a 55-0 was the best possible score for a shutout. I was tempted to edit it to show a higher scoring version but I could also be wrong about the best possible hand ... and this was unreferenced anyhow.

  • In a 2-person game, dealer can theoretically shutout his opponent while scoring 58 points. Play could proceed as shown:
Bob holds 7TQK cut
card
4
T Q K 7 Hand: 7TQK + 4 for 0 Total: 0
discards 56 10 25 10 22
Alice holds 5566 5 6 5 6 Hand: 5566 + 4 for 24 Total: 58
discards 56 15 for 2 31 for 2 15 for 2 28; 1 for last + 3 for a run of 3 Crib: 5566 + 4 for 24

--Noren (talk) 07:10, 27 August 2014 (UTC)[reply]

59-0 is possible.

Bob holds 67TQ cut
card
4
T 7 Q 6 Hand: 67TQ + 4 for 0 Total: 0
discards 46 10 22 10 21
Alice holds 5566 5 6 5 6 Hand: 5566 + 4 for 24 Total: 59
discards 55 15 for 2 28; 1 for last + 3 for a run of 3 15 for 2 27; 1 for last + 2 for a pair Crib: 4556 + 4 for 24

--Noren (talk) 07:28, 27 August 2014 (UTC)[reply]

60-0 is possible, not sure how I missed this.

Bob holds 67TQ cut
card
4
T 7 Q 6 Hand: 67TQ + 4 for 0 Total: 0
discards 46 10 22 10 21
Alice holds 4556 5 6 5 4 Hand: 4556 + 4 for 24 Total: 60
discards 55 15 for 2 28; 1 for last + 3 for a run of 3 15 for 2 25; 1 for last + 3 for a run of 3 Crib: 4556 + 4 for 24

--Noren (talk) 05:12, 28 July 2018 (UTC)[reply]

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Median of Crib point distribution

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The median of the crib point distribution assuming random discards to the crib is 4 because the probability that the crib is worth 4 or less is about 56% and the probability that the crib is worth 4 or more is about 66%. Should we add

Median = 4

to the list of descriptive statistics of the crib point distribution?

Irchans (talk) 04:34, 3 February 2024 (UTC) irchans[reply]