Maximising the probability of 1x 2y in 2/1GF

[EricK]

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.

Huh? I thought that was exactly the question you asked.

I don't think it is :)

 

24 points come up most often when they are split 12/12. But how does directly imply that in order to maximise the probability that

 

1. dealer has an opening bid

AND

2. responder has enough to know that the partnership has at least 24 points

 

you need to set the minimum range of the opening bid to 12?

 

It may well be true. But it isn't obviously true to me.

 

That much should be obvious, because if it were obvious I wouldn't have asked the question in the first place :)

[rbforster]

24 points come up most often when they are split 12/12. But how does directly imply that in order to maximise the probability that

 

1. dealer has an opening bid

AND

2. responder has enough to know that the partnership has at least 24 points

 

you need to set the minimum range of the opening bid to 12?

 

It may well be true. But it isn't obviously true to me.

In fact it isn't true if you remember to count passed hand 2/1 auctions (which I guess would still be game forcing assuming you have the same high opening bid standards in 3rd/4th as 1st/2nd).

 

If you look at the 3 most likely ways to have a 24 HCP game, you've got:

 

A. 11-13

B. 12-12

C. 13-11

 

-- If you open starting at 11, you'll open all of ABC and force to game only for case A (1/3 of cases)

 

-- If you open starting at 12, you'll open B & C in 1st seat, and only have a 2/1 auction for B. You'll also open A in 3rd seat, but not have a 2/1 auction. (1/3 of cases)

 

-- if you open starting at 13, you'll have a 2/1 GF auction for A & C, where opener passes in case A and later bids 2/1. Unfortunately, you pass out case B! (2/3 of cases)

 

So here's an example where starting at 13 maximizes the probability of getting a 2/1 auction (twice the chances of other systems in likely marginal cases), although it's not clear that maximizing a 2/1 auction is really what you want if you have to pass out some games to do it! At least if you open "light", you'll get to a part score or an invitational sequence and those tend to score somewhat better than pass outs when you've got game going values.

[dburn]

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.

He might, but he begs to observe that he is no mathematician. As far as he can make out, there are essentially three important results in probability theory:

 

The Law of Large Numbers, which states that things will probably happen about as often as you would expect;

 

Bayes's Theorem, which states that if you observe for long enough how often things happen, you can probably work out how likely they are to happen again; and

 

The Central Limit Theorem, which states that both of the above theorems are probably true, or at any rate not demonstrably false.

 

Beyond that he will not venture, except to observe that if he had to make game with 24 hcp between the hands, he would rather have them divided 12-12 than 24-0. Luckily (so to speak) if Frances's pivot tables - which he would trust with his life - are to be believed, he will more often than not be in this relatively happy position.

[MarkDean]

Oops, I'm getting different answers to Helene again.

 

Just looking at HCP, I make

 

Opener 10-12 + responder 14-20 = 4.27%

Opener 11-13 + responder 13-19 = 4.81%

Opener 12-14 + responder 12-18 = 5.09%

Opener 13-15 + responder 11-17 = 5.09%  (but a slightly lower 5.09%)

Opener 14-16 + responder 10-16 = 4.78%

Opener 15-17 + responder 9-15 = 4.21%

 

Note that my numbers are not from simulations, they are exact calculations. The only risk is that I am mis-reading my pivot table (or answering the wrong question).

 

If you want to maximise the number of auctions that go 1NT-3NT you should not play a 3-point range.  If you assume that responder looks for slam if the combined point count could be 33 or higher, then playing 12-15 1NT opening you will have a combined 24-32-count 5.75% of the time.

I got these numbers - no pivot tables - that concur with Frances' calculations.

 

10-12 0.042667213
11-13 0.048051471
12-14 0.050875245
13-15 0.050847073
14-16 0.047792177
15-17 0.042121701

 

except that the 13-15 figure rounds down to .0508 .

I matched cascade's numbers. I cannot believe there are so many other nerds on here.

[mikestar]

I agree that if your sole goal is to maximize the occurrence of 1NT-3NT auctions, then a wider range opening is to be preferred to a three point range. But the wider range also increases the chance of 1NT-2NT-P as well. If you aren't going to game it's much better to play in 1NT than 2NT--2NT never gains and loses big if seven tricks are the limit.

 

I rather like 12-14 with invitations only on a good 11 or really bad 12. I have played 12-14 with no power invites at all--it's playable. The occasional missed 14-11 game is bad but the total absence of 2NT-1 results compensates for most of these.