[EricK]

The weaker one's requirements for a 1M opener, the more likely it is that you will be dealt such a hand, but the less likley it is that partner will be strong enough to bid a 2/1. But conversely, the stronger your requirements for a 1M bid, the less often you'll be able to bid it, but the more likely partner is to have a 2/1 response.

This suggests that there is a minimum strength for a 1M bid which maximises the probability of an auction starting 1M 2y. Does anybody have any idea of how to work out what this is?

I realise that there is no unique answer to this as it depends on what combined strength you think you need to make a 2/1 and also on exactly how you measure the strength of a hand (i.e. both HCP and distribution). But given answers to these subquestions, a unique answer should exist to the main question.

[FrancesHinden]

I can answer an easier question exactly.

Ignore distribution.

Assume you need assurance of a combined 24+ HCP to make a game forcing 2/1
So if opener's minimum HCP are 11, responder needs 13+, but if opener's minimum is 13, responder needs 11+

Q. What minimum HCP requirement for opener maximises the chance of a 2/1 auction?

A. 13

Opener has minimum 11, responder has 13+ 2.067%
Opener has minimum 12, responder has 12+ 2.256%
Opener has minimum 13, responder has 11+ 2.337%
Opener has minimum 14, responder has 10+ 2.281%

(I would have guessed 12 if asked, so I was mildly interested to see that's the third best answer)

[TylerE]

I would guess it depends on position too. (E.g. if 1st hand passes, partners expected value goes up, to some extent.) That said, requiring a combined 24 might be a bit conservative. In the content of a strong club system where virtually all 11s and some 10s (And that's not even counting the 10-12 NT) are opened, we 2/1 with good 12s, and seem to get into trouble rarely. Do you get to the occasional no-play 3N? Sure...but even those make once in a while.

[helene_t]

I just dealt 100000 deals. For each required HCP for an opening bid, the number of deals in which you have an opening bid and p knows you have at least 24 combined hcps, is:
10: 6618
11: 7261
12: 7474
13: 7219
14: 6669

hey this is quite different from Frances' answers, one of us must be wrong

[Gerben42]

Does Frances' answer include the dependencies that if you are stronger, partner is likely to be weaker? Helene: How about trying "Rule of X greater equal 40" for both?

[TylerE]

Frances' ranges are base+. So like for the 13+ calculation, opener could be on, say 19.

[FrancesHinden]

My answers are

Chance of opener having exactly 11 points, and the pair being 24+ is...

because I thought that was (a simplified version) of what was being asked

So helene and I should have got the same answer and one of us is wrong.
Quite possibly me. I'll wait and see who gets someone else agreeing with them first!

You will get a different answer if you just say chance of opener being 11+, and... but it shouldn't change the ranking, you are just adding in the same set of stronger hands to every answer.

[FrancesHinden]

Gerben42, on Mar 23 2009, 02:00 PM, said:

Does Frances' answer include the dependencies that if you are stronger, partner is likely to be weaker?

Yes

[FrancesHinden]

TylerE, on Mar 23 2009, 01:44 PM, said:

I would guess it depends on position too. (E.g. if 1st hand passes, partners expected value goes up, to some extent.)

True

There is definitely something wrong with helene's numbers, unless they are conditioning on something else.

You have a 12-count in opener opposite 12+ HCP 7.5% of the time, but the chance of a 12-count is only 8% to start with.

I make the overall chance of being 24+ high only 23% for all HCP combinations.

[orlam]

FrancesHinden, on Mar 23 2009, 09:14 AM, said:

My answers are

Chance of opener having exactly 11 points, and the pair being 24+ is...

because I thought that was (a simplified version) of what was being asked

So helene and I should have got the same answer and one of us is wrong.
Quite possibly me. I'll wait and see who gets someone else agreeing with them first!

You will get a different answer if you just say chance of opener being 11+, and... but it shouldn't change the ranking, you are just adding in the same set of stronger hands to every answer.

I thought the question was about the odds of opener having 11+ and responder 13+, versus 12+ and 12+, etc.

I don't see why you and Helene should get the same answers then. To you 12/12+ answer you have to add the 13+/12+ cases. To you 13/11+ answer you have to add the 14+/11+ cases.

[hanp]

It seems to me that both Frances and Helene answered a question correctly, they just answered different questions. I interpreted the question as Helene did.

[FrancesHinden]

hanp, on Mar 23 2009, 02:58 PM, said:

It seems to me that both Frances and Helene answered a question correctly, they just answered different questions. I interpreted the question as Helene did.

OK

Probability that opener has between n and 22 HCP (assuming all 23+ hands are opened something else), and response has at least (24-n)

n = 10.... 6.55%
n = 11.....7.12%
n = 12......7.31%
n = 13......7.11%
n = 14......6.54%

I agree this feels a slightly better question to answer than my original

[hanp]

It is also interesting that now the answer does agree with your intuition. Rightly so I would say.

[Gerben42]

Rule-of-X openers:

20+ vs 21+: 8.21%
19+ vs 22+: 7.83%
18+ vs 23+: 6.95%

20+ vs 20+: 11.31%
21+ vs 19+: 10.97%
22+ vs 18+: 10.11%
23+ vs 17+: 8.83%

Not very surprising. Let's try North is dealer and didn't open, HCP requirements for E and W.

12+ vs 12+: 6.78%
13+ vs 11+: 6.61%
14+ vs 10+: 6.03%

It's kinda safe to assume that for 2/1 GF to come up most, you need to open those hands that partner will force to game with, i.e. opener + opener = game

[Lobowolf]

But responder's hands will be able to take into consideration distribution points depending on whether he knows there is a fit. e.g. a 4441 11-count might pass as opener, but have enough to GF if partner opens 1M and it's not the singleton. So it seems as though if all you want to do is maximize the chances of achieving a 2/1 auction, to some extent lighter openings would help, as they would increase the chances that responder would be able to upgrade, knowing about a fit.

Apologies if I'm overlooking something really basic; this is off-the-cuff.

[FrancesHinden]

hanp, on Mar 23 2009, 04:43 PM, said:

It is also interesting that now the answer does agree with your intuition. Rightly so I would say.

It's probably not very surprising, that whatever you think the minimum combined strength should be for a game forcing auction, it comes up most often when it is equally split between opener and responder.

[kenrexford]

Here's my method:

1. Opener opens whenever he has a plausible opening.
2. Responder bids 2/1 fairly liberally.
3. If we end up with these methods landing us in an occasional BS game contract, we play the socks off the hand and steal the requisite number of tricks.
4. If #3 fails, and we go down, that's the cost of doing business.

In other words, I think the "solution" to this problem is to just not worry about it. It seems to work out OK in the end, anyway. You gain a lot more with frequent 2/1 sequences and lightened opening requirements than you lose from the occasional hopeless game.

[hanp]

FrancesHinden, on Mar 23 2009, 02:24 PM, said:

hanp, on Mar 23 2009, 04:43 PM, said:

It is also interesting that now the answer does agree with your intuition. Rightly so I would say.

It's probably not very surprising, that whatever you think the minimum combined strength should be for a game forcing auction, it comes up most often when it is equally split between opener and responder.

Yes, dburn might be able to give a one-line proof for this fact.

[EricK]

kenrexford, on Mar 23 2009, 07:33 PM, said:

Here's my method:

1. Opener opens whenever he has a plausible opening.
2. Responder bids 2/1 fairly liberally.
3. If we end up with these methods landing us in an occasional BS game contract, we play the socks off the hand and steal the requisite number of tricks.
4. If #3 fails, and we go down, that's the cost of doing business.

In other words, I think the "solution" to this problem is to just not worry about it. It seems to work out OK in the end, anyway. You gain a lot more with frequent 2/1 sequences and lightened opening requirements than you lose from the occasional hopeless game.

You are still answering the question. The only difference is that you are using a lower requirement to force to game than other people.

Just out of interest, you will presumably get to game with, say, an 11 point "plausible" 1♠ opening opposite an 11 point "liberal" 2/1 response even if there is no fit. Would you also get to game if the suits were rearranged in such a way that the auction started 1x 1y instead of 1p 2q? Or do you use the extra room to help stay out of game in that instance?

[EricK]

FrancesHinden, on Mar 23 2009, 07:24 PM, said:

hanp, on Mar 23 2009, 04:43 PM, said:

It is also interesting that now the answer does agree with your intuition. Rightly so I would say.

It's probably not very surprising, that whatever you think the minimum combined strength should be for a game forcing auction, it comes up most often when it is equally split between opener and responder.

Although this is not surprising, it isn't immediately obvious (to me at any rate) that it follows that that will also be the answer to the question I asked.

Can we generalise this method so that we can say the answer to the question "What 3 point NT range maximises the probability that the auction goes 1NT 3NT?" is simply 12-14 (assuming 24 points for game)?