[Ant590]

Hi all,

I've been playing with a player whose style differs from my own. For instance, he doesn't like opening 1NT with a 5-card major, whereas I do routinely. Our latest divergence is that he recently said that he prefers not to open 1NT with a small doubleton major. As well as explaining the implications this had in terms of messing up the a rebid structure, I thought I'd make a small simulation to bring home the rarity of it being a problem.

Scenario: We open 15-17NT with a small doubleton spade. Partner does not bring a spade stop, and the opponents can cash the first five tricks in spades. I approximated this in code as this as partner's spades are at most Q/J high (but not both), and at least one opponent has 5 spades.

Results:
No restriction on responder's HCP: the opponents can cash five spades 25% of the time
(Responder is 10+: the opponents can cash five spades 9% of the time)

BUT:
I then thought to drive home the point, I would compare for when opener has Jx or Qx
No restriction on responder's HCP: the opponents can cash five spades 27% of the time
(Responder is 10+: the opponents can cash five spades 4% of the time)

So, if I don't constrain responder's HCP, the opponents are *more* likely to be able to cash the first five in a suit that we hold Jx/Qx, compared to xx. This *must* be an error in my code, right? I've run a few times with the same result, so I don't think it is a sampling error...

My code (deal 3.18):

sdev nbalanced
sdev stopper
sdev oppsrun

main {
	reject unless {[hcp north]>=15 && [hcp north]<=17}
	reject unless {[balanced north]}
	#1 reject unless {[hcp north spades]>0 && [hcp north spades]<=2 && [spades north]==2} 
        #2 reject unless {[hcp north spades]==0 && [spades north]==2} 
        nbalanced add [spades south]
	reject unless {[hcp south spades]<=2}
	stopper add [spades south]
	reject unless {[spades east]>=5 || [spades west]>=5}
	oppsrun add [spades south]
	accept
}

deal_finished {
  puts "With small doubleton spade=[nbalanced count])"
  puts "With no stopper=[stopper count]"
  puts "opps can cash first 5=[oppsrun count]"
}

[helene_t]

Well if you hold Jx or Qx, partner is less likely to hold a queen or jack, than if you hold xx? Isn't that what you are saying?

A better approximation might be: Opps can cash five spades if one of them has five AND our combined spades HCPs are <3.

Anyway, I am not sure this argument is so great, because:
- If RHO has five spades, LHO is unlikely to lead it (while he might lead it if we have a more informative auction starting with a minor suit opening, in particular if we give RHO room to bid his spades).
- If LHO has five spades he might bid them over out 1NT.
- Even if partner has a spade stopper it might not be enough.
- Having Qx or Jx reduces the amount of points we have in the other suits and therefore makes it more likely that one spade stopper is not enough (or that we go down on a 4-4 break).
- Qx might be a stopper if LHO has AKxxx and underleads.
- Even if our minor suit opening can pinpoint the lacking spades stopper, we might find ourself in a gf with no game making.
- Partner's Qxx may be a stopper, especially if we have Tx or p has QTx.
- Partner's QJ tight is not a stopper.

If your partner thinks he doesn't need the inference that a minor suit opening denies 15-17 balanced, I think you should play Gerben's big notrump system:
1m.....1NT = 12-15
1m.....2NT = 16-17
1m.....3NT = 18-19
1NT=omnibus (semi) GF
2♣=weak

Especially if you play some kind of step responses to the forcing opening it is really great to have one extra room of bidding space. Having 2♣ in your preempt structure is even better, especially if you can play it as some two-suited hand instead of natural.

[FrancesHinden]

I am not very good at writing or reading dealer code, so I may have misunderstood what you have written, but the extract of your code seems to leave the South hand seems to be undefined other than at most 2 HCP in spades? Can ersponder have length in spades even if no honours?
I am also not sure if you mean 'can cash 5 spades off the top' or 'can cash 5 spades with some entry help' e.g. LHO has Jxx and RHO has AK9xx and dummy has Q10x then you don't have a stop as long as LHO can lead them twice, but they can't take 5 spades tricks directly.


Anyway, if you give opener a low doubleton, then you have a stop any time responder has Qxx and the AK are onside, or Q10x and AJ/KJ is onside (or just the jack if there is no entry), or J109x which is always a stop, or lots of layouts when partner has queen to length and they are blocked. Or something like Q98x.

If you give opener Qx then any time partner has neither the queen or the jack they have 5 cashers. (nearly any time; 98xx in dummy may block the suit)

In summary, give responder the Q or the Jack to length and opener nothing and there's some chance of a stop. Give opener Qx or Jx and responder nothing, and there is no chance of a stop.
This effect may be out-doing the advantage of having Qx opposite Jxx / Jxx opposite Qx.

[Lorne50]

Maybe a better way to convince him is to ask how many top players refuse to open 1N with a small doubleton. I think the answer is none which strongly suggests he is on the wrong wavelength here.

As far as a simulation goes you need to factor in how often you reach a better spot or worse spot opening a suit and how often the oppo do not cash when they could which are both difficult to do in simulations.

[jogs]

The main reason for opening 1NT with a small doubleton is to avoid rebid problems.

Qxx, AKxx, AQJx, xx

1♦ - 1♠
?

What do you bid now?