[helene_t]

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Which contract would you like to be in at matchpoints? What about 6♣ in a weak field but 7♣ in a stronger field?

Suppose you want to be in a club contract. Any chance of bidding it?

[mr1303]

I wouldn't want to go anywhere near 7C in any field.

I imagine my bidding playing something natural would be:

2C 2D
2S 2NT (nat)
3C 3NT?
4NT (significant extras)? ????

I'm happy with the first 5 bids. After that, it's a bit of a guess.

[mgoetze]

Let's see, for 7♠, we need spades no worse than 4-2 (84%) and ♣J singleton or doubleton (18.5%), that's about 15.6%. Probably not going to be best.

For 6♠, we need spades no worse than 4-2 (84%), ♣J at most tripleton (54.1%). But wait, we can also make it when spades are 5-1 and (14.5%) ♣J singleton or doubleton with the short trumps (conditionally 4.26%). That's already 46%. Better yet, on a non-diamond lead we can also make with ♣J single/double with the long trumps (conditionally 14.64%), up to 48.2%.

Hm OK I suppose that's still not good enough...

Still, if by "weak field" you mean the sort of field where someone might discard a club from Jxxx on the run of the spades... 6♠ might not be so bad.

7♣, let's see. We still need ♣J at most tripleton, which is still 54.1%. But we can also lose to ♣9xxx(x) now. This means we only make on 38.76% of trump breaks, so 7♣ is out. Even in a weak field, opps are not going to manage to drop a trick with ♣9xxx.

6♣, hm. Let's say you get a diamond lead. You need to ruff a spade to get to the ♥K, so you need spades at worst 4-2 (83.97%) and hearts at worst 6-2 (~99.5%). That doesn't seem so great, so we should probably ruff with the ♣10 so that we still we when spades are 5-1 but the short spade hand doesn't have the J. Unfortunately that's only 40% of these cases so we are up to approximately 90% for all this.

Then we still need trumps to break. Since we ruffed with the 10, we need trumps 3-3 or J9 tight. That's only 51.68%, combined with the 90% for no ruffs, I rate 6♣ around 46.5%, slightly worse than 6♠.

So I think at matchpoints I would like to be in 4♠.

[PhilKing]

mgoetze, on 2015-February-07, 08:17, said:

Then we still need trumps to break. Since we ruffed with the 10, we need trumps 3-3 or J9 tight. That's only 51.68%, combined with the 90% for no ruffs, I rate 6♣ around 46.5%, slightly worse than 6♠.

We don't need that much luck. Any 4-2 break is OK as long as the ruff with the ♣10 holds. So it feels like about 75% to me.

[PhilKing]

Possible auction:

2♣-2♦
2♠-2NT
3♣-3NT
5NT-6♣

5NT is "pick a slam" - there is no way I would settle for game with the West hand. With a bit of imagination both players can visualise that the hand may be more or less as it is, and if you take away the spade jack, playing in clubs is even more vital.

Of course, if openers reds are switched we are in the wrong spot. The Kokishian solution is for opener to bid 4♦ over 3NT to show the heart shortage, but that has dangers of its own.

[helene_t]

We bid 7♣ after a Benji 2♦ opening:
2♦-2♥
2♠-3♦
3♠-4♠
4NT-5♣
5NT-6♦
7♣

After hearing one king (which was most likely to be ♦K since that was the suit partner bid), I thought that 7♠ should have a chance but wanted to give p a choice in case she had a void spades, or in case she had ♥K and four small clubs.

Afterwards I thought that we "should" have been in 6♣ which is quite likely to be the par contract, but thinking more about it I realized that it probably isn't the right way to think of optimal matchpoint contracts. 6♣ is a great spot if the other tables also face the choice between 3NT and 6♣, but in practice lots of other tables will be in 4♠, 6♠ or 6NT, and relative to those tables 3NT is more likely to win than 6♣.

[PhilKing]

helene_t, on 2015-February-08, 13:16, said:

6♣ is a great spot if the other tables also face the choice between 3NT and 6♣, but in practice lots of other tables will be in 4♠, 6♠ or 6NT, and relative to those tables 3NT is more likely to win than 6♣.

I tried running it as a a toy game scenario (please note: a toy game often uses silghtly arbitrary assumptions to reach a solution, so do not get too hung up on the assumptions unless they are way off), and 6♣ doesn't come out too well even if it is assumed to be a 75% contract! I treated 3NT and 4♠ as the same, even though 3NT is better - essentially I just treat them as winning when the slam bidders (surely the vast majority of the field) do not succeed. The point is we know 3NT is better than 4♠, but we want to focus on the advisability of bidding 6♣.

My inititial scenario has the 6NT "all the marbles" punt as the clear matchpoint winner. This is a well-known Woolsian situation, but the results were still a bit of a shock. For the purposes of the toy game, the assumptions were as follows:

6♠ and 6NT were regarded as equal contracts (50% for the purposes of the exercise) and would make or go off at the same time.

6♣ was deemed to be a 75% contract, and would make every time the other slams made.

All slams would go off one when they failed. Try not to get bogged down in the details of when this is not true.

3NT is deemed to always make, the number of tricks being irrelevant.

Putting this together, it seems likely that most people will be in slam (say 75%). I ran the numbers for a 4 table duplicate where there are 4 unique contracts and got the following simple scores:

All slams make:

6NT 6
6♠ 4
6♣ 2
3NT 0

Only 6♣ and 3NT make:
6NT 1
6♠ 1
6♣ 6
3NT 4

Only 3NT makes:
6NT 2
6♠ 2
6♣ 2
3NT 6

Since the first scenario occurs half the time in the model and the others a quarter, the scores for the first scenario are doubled leaving these scores:

6NT 15
6♠ 11
6♣ 12
3NT 10

So "going for all the marbles" is the clear winner. The results are somewhat skewed from reality in that there is no field in which a quarter will get to 6♣ but this particular toy game is just analysing this aspect. Using the above assumptions, even if 40% of the field were in game, 6NT would still be a tiny winner over 6♣.

[helene_t]

Nice one, Phil:). But when 6nt makes, 7c usually makes as well. Then again, 7c could go off by more than one.

[helene_t]

Let's try to include 7♣. So there are five tables:

All slams make (50%):
7♣: 8
6NT: 6
6♠: 4
6♣: 2
3NT: 0

Only 6♣ makes, 7♣ is one off (15%):
7♣: 2
6NT: 2
6♠: 2
6♣: 8
3NT: 6

Only 6♣ makes, 7♣ receives a diamond lead and tries to make it and goes several off (10%):
7♣: 0
6NT: 3
6♠: 3
6♣: 8
3NT: 6

No slam makes, 7♣ several off (25%):
7♣: 0
6NT: 4
6♠: 4
6♣: 4
3NT: 8

Total:
7♣: 4.3
6NT: 4.6
6♠: 3.6
6♣: 4.0
3NT: 3.5

Off course, at our club you might as well be in 3NT since since 3NT+4 beats all those who passed Gerber and somehow managed to make only 11 tricks despite spades being 3-3 and East having ♣Jx

[PhilKing]

Yep. The seven club calculations become a little complex, since you are going several down when the clubs are 4-2 without the jack dropping.

The toy scenario suggests that 7♣ would be the right punt if it generally only goes one off when failing. It is only about a 50% contract, but since you have plenty of company when the "normal" slams are also one off, it would then show a huge profit at matchpoints.