[gwnn]

 nullve, on 2016-February-27, 06:42, said:

But what prevents T -> T' from being a sacrifice more of the aztec kind*, where we'd expect T', but not T, to lead to significant loss or even disaster when used?

* by thinking of the hands involved as people

Because either
a) T' is a very rare call, in which case we're just wasting bidding space and we aren't unloading any problem hands.
b) T' is a reasonably common call, in which case we're causing reasonably common disasters and I don't know how you are going to balance that.

I know this is not technically a logically sound proof but intuitively it seems good enough for me. Or are we back to the "I am pointing out a logical possibility" business here?

[mikestar13]

 nullve, on 2016-February-27, 14:18, said:

I think most RFR proponents would agree that 2/1+WJS has the edge over 2/1+RFR on WJS hands. That the majority also believe it has the edge on RFR hands seems more far-fetched, but even Bobby Levin (http://bridgewinners...haracteristics/) admits that ...

I don't take Bobby Levin as saying that he believes that 2/1+WJS has the edge over 2/1+RFR on the RFR hands; I interpret him to be saying (when playing RFR), that negative inferences (RFR hand not held) when responding something else are even more valuable than positive inferences (RFR hand is held) when RFR is bid.
I believe that this is true at least partially on frequency grounds.

[nullve]

@gwnn: When I asked

 nullve, on 2016-February-24, 08:14, said:

Are there good examples of utilitarian sacrifices in this sense?

I didn't know what the answer was, i.e. I couldn't think of a single convincing example. But since I had only given the terse definition of 'utilitarian sacrifice', I felt the need to give an example anyway, while distancing myself from it by writing e.g. that it's "debatable" whether 2/1+WJS is better than 2/1+RFR on RFR hands.

We seem to agree that it's odd to believe that 2/1+WJS is better than 2/1+RFR on both WJS and RFR hands and still believe that 2/1+RFR is the better system overall. More generally, and given my definition above, I think it's odd (but not necessarily wrong) to believe that S is better than S' on T and T' hands while believing that S' is the better system overall. But such beliefs seem often to be implied when players attempt to justify why they're playing seemingly inferior structures.

[nullve]

 mikestar13, on 2016-February-27, 17:20, said:

I don't take Bobby Levin as saying that he believes that 2/1+WJS has the edge over 2/1+RFR on the RFR hands; I interpret him to be saying (when playing RFR), that negative inferences (RFR hand not held) when responding something else are even more valuable than positive inferences (RFR hand is held) when RFR is bid.
I believe that this is true at least partially on frequency grounds.

You're probably right.

[gwnn]

Ok so we both agree that this is a very odd thing to believe. Why not make that clear in your opening post right away? I don't want to teach you how to post but it helps to know what the opening poster thinks about his own hypothetical scenario when you address it. For example "I know this seems weird and unlikely, but I'm wondering whether there are such cases" is not too much to add to get the ball rolling. Anyway, just my 2 cents. I don't think such cases are likely to occur and I still don't know how they could occur. I think if you do proper house keeping in your system, you would spread out your problems relatively evenly. So the RFR hands would be put in the 2/1+WJS hands in a way that it will bother several sequences a little bit, that is to say, 2/1+WJS would do worse than RFR on RFR hands and a bit worse on "near"-RFR (the hands on which RFR would bring some nice negative inferences). It will recover a bit on WJS hands and near-WJS (the hands on which WJS would bring negative inferences) hands.

Maybe I'm thinking too much of Chebyshev polynomials where you judge an approximation by its worst-case scenario, but I think it pays to make your system suck approximately equally on all likely hands as opposed to making it shine on some and spectacularly fail on the others.

[nullve]

 gwnn, on 2016-February-27, 15:16, said:

Or are we back to the "I am pointing out a logical possibility" business here?

Maybe. But the following might come very to close to being an example of an "aztec" sacrifice that is also a utilitarian sacrifice:

S = Blue Club
T = Blue Club's 7N opening (undefined, I guess, but presumably only used with 13 likely tricks in NT)
T' = 7N opening defined the same way as Blue Club's 2♦ opening, i.e. as 17-24, (4441)

Replacing T with T' frees up the 2♦ opening, so by updating S "the obvious way", we get something like

Aztec Blue Club (our S'):

2♦ = weak preempt not overapping with 2♥+
7N = 17-24, (4441)
other: same as S,

which to me now seems better than regular Blue Club for any reasonable choice of weak 2♦ preempt due to the extremely low frequency of Blue Club's 2♦ opening. So IMO,

* S' is better than S
* S is better than S' on T' hands
* S is at least as good as S' on T hands (but not better, so it's not a perfect example)

[nullve]

Or we can let S be any (hypothetical) system where

2♦ = 432-432-432-5432 exactly
4♠ = AKQJT98765-2-2-2 exactly

and let

T = the 4♠ opening inn S
T' = mandatory 4♠ opening with 432-432-432-5432 exactly

Then after the update, when 2♦ has turned into a reasonably frequent and useful opening,

* S' is better than S (really)
* S is better than S' on T hands (really)
* S is better than S' on T' hands (really)

This post has been edited by nullve: 2016-February-28, 09:46

[gwnn]

I thought the point was that T and T' were the same bid(s)?

[nullve]

 gwnn, on 2016-February-28, 09:30, said:

I thought the point was that T and T' were the same bid(s)?

Yes, sorry. Edited now.

[gwnn]

OK I agree, that's a good example. Well done.

[mgoetze]

 nullve, on 2016-February-28, 07:59, said:

Replacing T with T' frees up the 2♦ opening, so by updating S "the obvious way"

If we know how to do this, why don't we just play "the obvious system"?

[nullve]

 mgoetze, on 2016-February-28, 14:46, said:

If we know how to do this, why don't we just play "the obvious system"?

Fair point, because even in a seemingly simple case like

S = 2/1
T = Weak 2D
T¨= Flannery 2D,

how we choose to update S will clearly depend on our current bidding philosophy. And not just regarding standards for a 1D or 3D opening, because it might not be obvious for all eternity (and all system regulations) that we have to stuff all our T openers into P, 1D or 3D after replacing T with T'.