Can anybody calculate this?
#21
Posted 2007-September-06, 22:00
Out of the last N hands actually played on BBO, how many made 12 tricks? How many 13?
#22
Posted 2007-September-06, 22:20
For example, there are a lot of hands where there are three finesses (or two finesses and a suit break, etc) where if everything works you make six, if nothing works you fail in four. Obviously you want to be in game on these hands and not slam, but the double dummy analysis will say that 1/8 of these are slam hands, 3/4 game hands, and 1/8 partscore hands.
a.k.a. Appeal Without Merit
#23
Posted 2007-September-07, 01:17
awm, on Sep 7 2007, 05:35 AM, said:
While it's true that the double-dummy nature of the analysis can help the defense too, I suspect that on slam deals (where the defenders will rarely have the lead) double-dummy creates an artificial declarer advantage.
I think we have had this discussion of bias in DD results. AFAIR the conclusion was a slight bias in favor of defenders due to the lead, but you may be right that that is less of a concern for slam deals. Not sure though. For my feeling, many slams (especially small slams) turn out to be settable with a specific lead that is hard to find. Maybe this is an artificat of slams being more post-mortemed than non-slams.
But generally I agree with you. Sometimes you have a double fit and only one of the two denominations makes game, sometimes game needs to be rightsided, sometimes 3NT is the only game with a 8- or 9-card major fit etc. All this makes for a lot of DD games that would never be bid in practice.
#24
Posted 2007-September-07, 02:22
Quote
Par 0 deals are quite rare, Thomas Andrews has some on his site.
#25
Posted 2007-September-07, 02:45
jtfanclub, on Sep 6 2007, 11:00 PM, said:
Out of the last N hands actually played on BBO, how many made 12 tricks? How many 13?
I have just over 13000 of the hands I have played on BBO recorded:
By declarer:
13 tricks: 310 times (~2%)
12 tricks: 951 times (~7%)
11 tricks: 1940 times (~15%)
10 tricks: 2632 times (~20%)
9 tricks: 2746 times (~21%)
8 tricks: 2084 times (~16%)
7 tricks: 1337 times (~10%)
6 tricks: 612 times (~5%)
5 tricks: 262 times (~2%)
4 tricks: 104 times (~1%)
3 tricks: 22 times
2 tricks: 4 times
1 trick: 0 times
0 tricks: 2 times
passed out: 125 times (~1%)
#26
Posted 2007-September-07, 02:55
George Carlin
#27
Posted 2007-September-07, 05:34
#28
Posted 2007-September-07, 09:06
It might change the defense if the defenders are defending 7 instead of 6.
You would lead an ace against 7, you might not against 6.
In a 7 level contract you might be forced to take a certain line, where in 6 you would take another, guaranteeing the contract for the certain loss of 1 trick.
As for 33 HCP slams, those are not that common in my experience, usually shape contributes some points.
>I guess my (anecdotal) experience is that there are a fair number of boards where I don't bid slam and slam is lousy looking at my hand and partner's, but things lie very favorably and slam ends up making. It feels like there are probably more of these than hands where slam is actually good in fact! But these are basically "unbiddable" slams because if you bid them you will also end up bidding a much larger number of slams which are equally lousy and where the cards don't like favorably...
I agree.
You make 6 or 7 because of working jacks and 3-3 splits, adn 2 finesses being right.
Pard made 12 on 3NT becaus eopps lead a suit other than Diamonds allowing a finesse, and then later a squeeze. No way to bid that
#29
Posted 2007-September-07, 09:20
helene_t, on Sep 7 2007, 02:17 AM, said:
Don't forget all the contracts that require dropping a singleton king or taking a deep finesse that would never be found in practice.
There are some cancellations, though. DD won't bid 75% contracts that require either of 2 finesses, but they're both off, while it *will* bid 25% contracts when both finesses are on.
#30
Posted 2007-September-07, 09:28
There's a basic asymmetry
It's very common to be able to make 2 or 3 overtricks in game, EVEN ON BEST DEFENCE, but not want to be in slam.
It's much less common to find a hand where you want to be in slam single dummy but it is off double dummy.
The "10%" figure feels roughly right on my empirical experience, although if anything very slightly too low.
#31
Posted 2007-September-07, 10:19
I have the percent that between the 2 hands there are a fixed number of HCP.
I agree on the concept that HCP aren't the only determinant, but here are the HCP layouts that I came up with.
HCP Percent % Cummulative (those HCP or higher)
0 0.0000513% 100.0000000%
1 0.0004846% 99.9999487%
2 0.0019990% 99.9994641%
3 0.0063868% 99.9974652%
4 0.0179859% 99.9910784%
5 0.0426910% 99.9730925%
6 0.0927186% 99.9304015%
7 0.1856480% 99.8376830%
8 0.3411376% 99.6520350%
9 0.5877062% 99.3108974%
10 0.9546726% 98.7231912%
11 1.4634633% 97.7685186%
12 2.1239452% 96.3050553%
13 2.9433742% 94.1811101%
14 3.8833877% 91.2377359%
15 4.8918758% 87.3543482%
16 5.9072842% 82.4624724%
17 6.8310913% 76.5551882%
18 7.5662308% 69.7240969%
19 8.0467824% 62.1578661%
20 8.2221673% 54.1110837%
21 8.0467824% 45.8889163%
22 7.5662308% 37.8421339%
23 6.8310913% 30.2759031%
24 5.9072842% 23.4448118%
25 4.8918758% 17.5375276%
26 3.8833877% 12.6456518%
27 2.9433742% 8.7622641%
28 2.1239452% 5.8188899%
29 1.4634633% 3.6949447%
30 0.9546726% 2.2314814%
31 0.5877062% 1.2768088%
32 0.3411376% 0.6891026%
33 0.1856480% 0.3479650%
34 0.0927186% 0.1623170%
35 0.0426910% 0.0695985%
36 0.0179859% 0.0269075%
37 0.0063868% 0.0089216%
38 0.0019990% 0.0025348%
39 0.0004846% 0.0005359%
40 0.0000513% 0.0000513%
Algorithm used:
I cycled through every possible combination of A's, K's, Q's, and J's using "for" loops. I added their HCP and determined how many cards where left to fill in the other cards. Multiplying the number of combinations of the remaining cards times the number of combinations that could made.
Then add up each of these for all layouts with those HCP's.
Ex.
For 9 HCP, using four cards, there are 304 possible combinations.
For the other 22 cards out of 36 with no point value, there are 3796297200 possible combinations.
So for 9 HCP using 4 cards, there are 1.154074E12
BebopKid (Bryan Lee Williams)
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