Skip to main content
9 events
when toggle format what by license comment
Sep 2, 2013 at 14:05 comment added john mangual perhaps you can study bidding in Bridge hands?
Aug 3, 2013 at 1:28 review Close votes
Aug 7, 2013 at 3:04
Apr 4, 2013 at 6:33 history edited Raphael
edited tags
Dec 29, 2012 at 5:09 comment added Aryabhata This is a very vague question. What does "to be further processed" mean (and the 'best' method probably depends on it, don't you think)? In any case, have you consider looking at the papers behind GIB (the double dummy solver by Matt Ginsberg)?
Dec 28, 2012 at 5:39 comment added Peter Shor In the extreme case where one person has all the clubs, another all the hearts, a third all the spades, and the fourth all the diamonds, then the number of trick combinations is $(13!)^4$, even though it's trivial to figure out who wins the hand.
Dec 27, 2012 at 23:34 answer added Sean timeline score: -1
Dec 27, 2012 at 13:08 comment added Yuval Filmus There are too many of them, in fact at least $13! \approx 6 \times 10^9$, which is still somewhat reasonable but probably a substantial underestimate. (I get $13!$ since at each trick, the player who leads can choose any of her remaining cards.)
Dec 27, 2012 at 12:32 review First posts
Dec 27, 2012 at 12:35
Dec 27, 2012 at 12:12 history asked facimus CC BY-SA 3.0