The description of the September tournament mentions the first successful Grand Open in our tournaments. The tournament was nonSynchron so the hand was only played once. This type of tournament generates more hands in total, so Grands Open are more likely to occur there.
The hand, held by Chris as Vorhand, was
J: CS
C: 
S: ATQ
H: A
D: ATK7
I believe this hand to fall into the fourthbest group for in terms of probability of success for Vorhand, at 64.7%.
Earlier on we have had two candidates in Synchron tournaments. The more recent was an unsuccessful attempt by Mike in July this year:
J: CSH
C: AK987
S: A
H: A
D: 
This is in the fifthbest group at 61.0%, and when the clubs didn't split Mike scored a record 480. (Mike was Hinterhand so the odds were slightly worse.) I am surprised this hasn't yet been mentioned on this board!
Before that we had the best candidate to date in February 2008. Five players held as Vorhand
J: CHD
C: A
S: 
H: AT87
D: AT
Three players made Grand Hand with schneider announced, and two (Mike and Thomas) tried Grand Open. Despite being in the secondbest group with a 84.4% success probability they found a 30 heart break and went off for 384.
We are still waiting for a hand in the top group (100%), but since I believe these hands occur only about 42 times in a million for Vorhand (and only 9 times per million for the other two seats combined) that isn't all that surprising. We have probably played about 6000 hands of Synchron and half that number dealt at the table. (The website has 100 past events listed up to September. Three of these had no winner, so we are somewhere near our 100th tournament now. So far we have had 30 different winners.)
Exercise for the reader: what hand is most likely to make Grand Open without being absolutely certain? Which seat is it in, and what is the probability of success? (Unlike all the hands above I believe it to be the right side of the odds.)
Patrick
Grand Open

 Posts: 137
 Joined: Mon Apr 23, 2012 9:26 pm
Re: Grand Open
I would go for something like CSH, A, AT, AT, AT in Mittelhand or Hinterhand. It loses if the nonVorhand opponent has a void in one of the suits plus the missing jack.
The chance that the missing jack is in the right place is 10/22. The chance of this player having a void in any one of the Grand's twocard suits is then (12×11×10×9×8)/(21×20×19×18×17) and for the singleton ace suit it is (12×11×10×9×8×7)/(21×20×19×18×17×16). Putting this together, the probability of losing seems to be about 0.4545 × (3×0.0389 + 0.0170) = 0.0608, so you have about a 94% chance to win. Before I did the calculation I expected the success rate with this hand to be higher. It is in fact very slightly higher, mainly because on some potentially defeatable hands Vorhand will have to guess which of two suits to lead, not knowing what is in the skat.
The chance that the missing jack is in the right place is 10/22. The chance of this player having a void in any one of the Grand's twocard suits is then (12×11×10×9×8)/(21×20×19×18×17) and for the singleton ace suit it is (12×11×10×9×8×7)/(21×20×19×18×17×16). Putting this together, the probability of losing seems to be about 0.4545 × (3×0.0389 + 0.0170) = 0.0608, so you have about a 94% chance to win. Before I did the calculation I expected the success rate with this hand to be higher. It is in fact very slightly higher, mainly because on some potentially defeatable hands Vorhand will have to guess which of two suits to lead, not knowing what is in the skat.

 Posts: 75
 Joined: Tue Jun 05, 2012 4:26 pm
Re: Grand Open
When Mittelhand holds CSH, A, AT, AT, AT there are 39130 hands out of 646646 in which Hinterhand holds DJ and at least one void.
These divide up as follows:
A failure rate of just over 5% is with the odds for bidding Grand Open.
However there are still safer hands other than the perfect ones, with better than 99% success rate.
Patrick
These divide up as follows:
Code: Select all
Hand type No. of Number of skats leading to: Effective
hands Defeat 1/2 guess 1/3 guess Miss hands
a. Two voids 195 66 195.00
b. A void and a singleton, neither in clubs 4950 41 15 10 3637.50
c. A singleton club and a void 810 46 10 10 625.91
d. A void in clubs and a singleton 675 39 12 15 460.23
e. A void not in clubs and two doubletons 2700 56 10 2427.27
f. A void in clubs and two doubletons 300 51 15 254.55
g. A void not in clubs and a doubleton 19500 56 10 18022.73
h. A void in clubs and a doubleton 3000 51 15 2659.09
i. A void and three tripletons 7000 66 7000.00
Total 39130 35282.27
Probability 0.056165 0.050642
However there are still safer hands other than the perfect ones, with better than 99% success rate.
Patrick

 Posts: 324
 Joined: Tue Apr 24, 2012 8:49 am
 Location: Kent, UK
Re: Grand Open
Middlehand holding CHD and all seven of one suit.
Again, 10/22 probability of CJ being in the right hand. Chances of this defender being void in any one of declarer's void suits is (12*11*10*9*8*7*6)/(21*20*19*18*17*16*15), all of which comes out at 99.07%. There's also the additional possibility of FH holding two five card suits and having to guess.
I woke up at four o'clock in the morning thinking about this, so it'd better be right;.
Again, 10/22 probability of CJ being in the right hand. Chances of this defender being void in any one of declarer's void suits is (12*11*10*9*8*7*6)/(21*20*19*18*17*16*15), all of which comes out at 99.07%. There's also the additional possibility of FH holding two five card suits and having to guess.
I woke up at four o'clock in the morning thinking about this, so it'd better be right;.

 Posts: 75
 Joined: Tue Jun 05, 2012 4:26 pm
Re: Grand Open
Mike,
Well done! My sample hand was
J: CHD
C: ATKQ987
Given Mittelhand's exposed hand there are 646646 possible holdings for Hinterhand. Of these 2002 consist of SJ plus nine red cards, so that a spade lead would defeat the contract. Similarly there are 2002 hands with each of the other two voids, making 6006 in total.
This suggests a probability of failure of 6006/646646 = 0.009288 (99.0712% success).
Among the 6006 hands there are 126 where Hinterhand is 720 in the three suits outside clubs (21 possible doubletons times 6 permutations of the suits). These leave Vorhand and the skat with seven cards in one suit and five in another. There are 21 skats out of a possible 66 which give Vorhand two fivecard suits. Vorhand will guess wrong half the time, so the 6006 hands on which Hinterhand is expecting a ruff must be reduced by
(1/2) x (21/66) x 126 = 20.045.
This raises declarer's probability of success to 99.0743%.
Patrick
Well done! My sample hand was
J: CHD
C: ATKQ987
Given Mittelhand's exposed hand there are 646646 possible holdings for Hinterhand. Of these 2002 consist of SJ plus nine red cards, so that a spade lead would defeat the contract. Similarly there are 2002 hands with each of the other two voids, making 6006 in total.
This suggests a probability of failure of 6006/646646 = 0.009288 (99.0712% success).
Among the 6006 hands there are 126 where Hinterhand is 720 in the three suits outside clubs (21 possible doubletons times 6 permutations of the suits). These leave Vorhand and the skat with seven cards in one suit and five in another. There are 21 skats out of a possible 66 which give Vorhand two fivecard suits. Vorhand will guess wrong half the time, so the 6006 hands on which Hinterhand is expecting a ruff must be reduced by
(1/2) x (21/66) x 126 = 20.045.
This raises declarer's probability of success to 99.0743%.
Patrick