hotShot

What is the lower limit for partner to transfer over 1NT? The way I play it, it can be very low, even after opps interference. So I don't think it's mandatory to jump.

You got it.

You sung your song, now let them have it. Just pass.

I took 24 because hrothgar used that number. Seems I have been lucky. 30 board tourneys: Number of tourneys: 33333 The mean HCP NS: 20.00167601676 The variance: 0.76401016443091 The standard deviation: 0.87407674973706 Minimum: 16.333333333333 Maximum: 23.5 18 Board tourneys Number of tourneys: 55555 The mean HCP NS: 20.00167601676 The variance: 1.2653701780311 The standard deviation: 1.1248867400903 Minimum: 15.388888888889 Maximum: 24.5 128 Boards Number of tourneys: 7812 The mean HCP NS: 20.001689108103 The variance: 0.1810962450686 The standard deviation: 0.42555404482697 Minimum: 18.4609375 Maximum: 21.515625

Perhaps South can just count! South knows that this is a first time partnership, System on over 2♣ is unusual and the probability to find 5♥ opposite your own 5 card length is only about 6%. After 1NT 2♣(♣ and a higher) 2♦ Pass the higher suit seems to be ♦ or ♠. RHO passed, there is no need to rush in 3♥ now. 2♥ 2♠ pass 3♣ Now South knows that West hast 4-5+♣ and 4-5♠ and 3-5 red cards. A 3-0 split for the missing 3♥ is about 10% so you can assume that West has at least 1♥ 90% of the times. Now South counts that there are 5-6♠, 7♥, 7-8♦, and 5-6♣ left for North and East. If East had ♠ why did East correct to 3♣? If East does not have 3+♠ than North has 3+♠. If North has 3♠ and 5♥ East would have 2-3♠, 2♥ and 8-9 minor cards. Why did not East raise 2♣ at once or doubled 2♦ holding 4-5cards there? So South has (good) reasons to suspect that North forgot the agreement.

While I assumed that the curve should be symmetrical, because distributional functions usually are, my laziness suggested to get that as result of the simulation instead of thinking it trough.

I agree completely. On the other hand, how come the min and max are not integers? Because it's the average of the 24 boards tourney. In theory you are right,if I would simulate a infinite number of deals the result should be symmetrical.Now take a coin and throw it. Both sides are equally likely, but only one side can be up. So to get symmetry you have to throw it a 2nd time, and there is a 50% chance you will get the same side up. If the same side comes up again you will have to throw another 2 times to have a chance for symmetry again. So whenever you set a limit lower than infinite for the the number of tries/simulations, you will find deviations from the symmetry. Because the other side is not independent, because the sum has to be 40.

To finalize the topic I simulated about 114 years of 7 24 board tourneys a week. Number of tourneys: 41666 The mean HCP NS: 20,00165502648 The variance: 0,95194261701304 The standard deviation: 0,97567546705503 Minimum: 15,958333333333 Maximum: 24,458333333333

Obviously 2NT was suggesting a no-trump contract.

Just to add a little facts: 1000000 random deals average HCP NS 20.0017 std. deviation 4.77 Distribute as follows: Comb. HCP NS count 1 5 2 14 3 66 4 200 5 424 6 862 7 1812 8 3446 9 5837 10 9494 11 14744 12 21488 13 29799 14 38694 15 49086 16 58605 17 68200 18 75410 19 80372 20 81950 21 80357 22 75905 23 68310 24 59259 25 49104 26 39165 27 29382 28 21065 29 14607 30 9471 31 5893 32 3430 33 1887 34 930 35 450 36 198 37 52 38 21 39 5 40 1

It is possible and probably useful to combine splinter and J2NT. Over a 1M opening you can use 3m (and 3♥ over 1♠) as mini-splinter forcing to 3M to show unbalanced invitational hands, a direct 3M for balanced invitational hands, J2NT to show (semi)- balanced GF hands and have 4m splinter to show unbalanced GF hands. I prefer the 3 level answers to J2NT to be shortage, so that we can stop in 3NT if both sides have (semi)balanced hands. And yes 1m - 3M can be used as splinter.

If you are a good pair among average pairs you can score a lot when you have the weaker (HCP) hands: - because you're better in defending, leading and signaling - because you'll know when to preempt or sacrifice (and when opps did not bid enough) - because you'll make the most out of the few good hands you get And it is more fun, because it's actually a challenge. Everybody can cash the gifts of weak leads, poor defense skills holding strong hands.....

You can draw the same 3 cards in 3 different orders, so this triple has a population of 3. The question is, is 20 the number of different triples (thats what I think) or does it include the population of each triple.

There is a lot of water in the ocean to cover with surveillance. The pirates operate from small fisher boats. There is almost no way to tell which are normal fishermen and which are pirates, before they start to attack. Somalia is an independent country, navies from other nations have no right to operate there. This is especially true for ground forces that "invade" the country. So you can spend a lot of money, cause a lot of negative propaganda just to save the investment of some greedy shipowners who try to save money by traveling a shorter, but more risky route.

I think you are missreading the table, there are 20 different 3-2 combinations (distributing the 5 cards on a 3 and 2 card stack), and I don't know where you get the 32 from. There are 70 4/3 combinations.

How about these? http://forums.bridgebase.com/index.php?showtopic=2542&hl= http://forums.bridgebase.com/index.php?showtopic=3278&hl= http://forums.bridgebase.com/index.php?showtopic=3555&hl= http://forums.bridgebase.com/index.php?showtopic=3385&hl= http://forums.bridgebase.com/index.php?showtopic=9893&hl=

My impression was that GIB is taking every jump as strong and forcing.

1) pass, you'll get 3 tricks and if partner has anything worth a trick opps are down. Your partner is likely to be void or single in ♥, still he had no bid over 4♥. 2) Opps already found a good lead, so I expect them to keep that level. If opps can prevent you from entering dummy, theres a good chance you'll go down. So you could make, but ....

I guess the probability for this is (something below 1)^(number of boards) where "something below 1" is the value for 26 combined points of the distribution curve. The result must be quite small.

So if we had 2 good games, I can predict that the computer will give us weak hands next board?

Are they playing 5cM or 4CM?

Upps

Perhaps I can give you the idea (note that this are not all possible hands partner can hold). You know that partner promised 5+♥ and LHO promised 5+♠. So partner and RHO share 3♠ cards. The odds are: partner [space] RHO [space] [space] prob. [space] partners most likely shapes [space]3 [space] [space] [space] [space] [space]0 [space] [space] 11% [space] [space] 35(23), 35(14) [space]2 [space] [space] [space] [space] [space]1 [space] [space] 39% [space] [space] 2533, [space]25(42) [space]1 [space] [space] [space] [space] [space]2 [space] [space] 39% [space] [space] 15(43) [space]0 [space] [space] [space] [space] [space]3 [space] [space] 11% [space] [space] 0544* *(I'll assume that partner would dbl 1[sp] with this shape) If partners shape is one of 35(23) or 2533 we don't have a fit and everything depends of partner having ♠ stopper. If partners shape is 34(14), 25(42) or 15(43), my 3♣ bis is forcing so I can never be sure, if partners 3♦ bid promises 3 or 4 cards (we might play 3♦ on a 4-3 fit and miss our 4-4♣ fit) and we can't play ♣ below the 4 level. So the bidding choices are: 3♣ forcing => partner bids 3♦ with or without ♦ fit who knows or have to play 4♣ (perhaps in a 4-3 fit) 2NT => without a ♠ stopper could lead to 3NT down because of the missing ♠ stopper pass => partners forcing 2♥ bid

To me it's not obvious what a slow 2♥ would suggest.

I'll pass, and that is non forcing. My 4♠ cards make it unlikely for partner to have good ♠ stopper, so I don't see a good NT contract on our side. They also suggest that opps might not have a fit, which suggests that we won't have a fit either. I hope that partner can ruff some ♠ with his small trumps, entering my hand in the minors and makes his ♥ honors. With a little luck thats enough for 7-8 tricks. With a little more luck, opps will balance in 2♠.....