WTP?

[cherdanno]

4342 being 3 times as likely as 4351 seems weird.

But it is correct (assuming we ignore the hcp restrictions).

[cherdanno]

But even ignoring that, something seems wrong. Consider:

 

We know partner has 4 spades and 4 diamonds and we know what our own 13 cards are. What are the odds that NONE of partner's 5 unknown cards are diamonds?

 

Can it really be anywhere close to 60%?

 

That is the math problem I tried to solve (roughly) in my head and the answer I got was "definitely not".

 

Either the question I was asking was wrong, I made a mistake when I tried to figure out the answer, or hotShot and twcho are way off the mark.

This is the wrong question.

It's too late for me to think about a good explanation, but maybe the following reasoning is intuitive:

 

Fred's question would be the right question if we knew the first 8 cards partner got dealt were 4 spades and 4 diamonds, and then we watch him getting dealt 5 more random cards. What we actually know is that partner got dealt at least 4 diamonds and 4 spades in total - but maybe he got his 4th diamond only on the 13th card he got dealt, so there was never time to deal him another diamond.

 

(To Fred: I would guess sure you have already thought about this difference - as it is the reason why writing a dealing program with constraints is tricky.)

[twcho]

My figures should be correct if HCP is ignored but I have missed the fact that with balanced shape, the opener will open 1NT with their designed point range and this will reduce the % significantly.

[jdonn]

There are 3 problems I see with the arguments made for bidding 3NT.

 

Small problem: The premise "if we don't have a fit we belong in 3NT" will usually be true but not always.

 

Medium problem: As pointed out by others there are problems with the percentages. A number of shapes were left out and the range of the balanced hands is (11)12-14 and so is surely being overstated.

 

Large problem: The entire argument is bad. In other words, who cares if we usually belong in 3NT? If partner is balanced or we belong in 3NT we can always get there very easily, otherwise it costs nothing at all to investigate alternatives in case they exist. We are at a low level in a game forcing auction so what's the rush, do you have a bus to catch after the hand or something?

[hotShot]

I am not sure how you came up with these numbers, but they feel very wrong to me. For one thing, you seem to be implying that partner will have exactly 4 diamonds over 60% of the time (and I realize that 4242 was a typo - I am not counting that).

 

If that is right I may have to give up bridge.

Well I admit it was nearly 2 a.m. when I set the restrictions for the simulation and I did not consider that HCP might have a relevant impact, when I reduced the problem to hand patterns.

 

When I'm really awake later today maybe I'll find the time for a new simulation taking more facts into account.

[Codo]

We are at a low level in a game forcing auction so what's the rush, do you have a bus to catch after the hand or something?

No need to rush, but if you define 3 NT as balanced, 13-15 HCps partner now knows:

You have exactly 4 hearts, else 4sf. You have at most 4 clubs, else you had answered 2 club to 1 diamond. You have no 4 spades, else yo had raised.

so your shapes are 3424 or 3433 or 3442 (less likely) or 2434 or 2443 (less likely).

 

He can make a pretty good descisson now.

Of course you had liked to have a way to show the position of your High cards too, but this was not possible within the limitations of the given system.

 

There is no need to pass 3 NT with a 4360 hand with the knowledge. If you do, you gamble that the diamonds are runnig. A good bet at mps, but no security at all.

[Fluffy]

my definition of 3NT is I play this :ph34r:

 

I was thinking that often I will have singleton somewhere by bidding 3NT, but thinking deeply it is unlikelly, with 1444 I have big fit in diamonds, and with 4414 I have fit in spades, with 3514 fit in hearts and 3415 should start with 2♣.

[eyhung]

I ran a sim using my definition for a 1D opener, 1H response and passes by the opponents, and threw out all hands with 4+ hearts and 0-3 spades in the opener. Over 100000 hands that fit these criteria, the # of times opener held 4 diamonds was 35732, or 35.73%.

 

Removing the constraints on the enemy hands, reduces the chance of opener holding 4 diamonds to 31.42%.

 

NOTE: This is without eliminating the balanced 18-19 HCP hands -- so in practice the chance of opener holding 4 diamonds is lower. I will rerun my sim eliminating the hands that would rebid 2NT.

[eyhung]

Throwing out 18-19 balanced gives around 25-26% hands with 4 diamonds.

[hotShot]

What is your definition of a 1♦ opener?

It seems to deviate from the SAYC standard where a 1♦ opening almost guarantees 4♦ cards.

Are you sure you did not include hands with 5♠ and 4-5♦?

A lot of people would even open 1♠ holding 5♠ and 6♦.

[Mbodell]

I ran a sim using my definition for a 1D opener, 1H response and passes by the opponents, and threw out all hands with 4+ hearts and 0-3 spades in the opener. Over 100000 hands that fit these criteria, the # of times opener held 4 diamonds was 35732, or 35.73%.

I think you'll get different numbers if you use the 1♥ response versus you use the hand in the OP that is exactly 3424 (and possibly slightly different if you don't make it exactly this 3424 complete with the relevant honors held). Intuitively it feels obvious that if we are more balanced, the odds that partner has only 4 diamonds are higher than if we are something like 3613 (which presumably would reply 1♥ in response).

[fred]

But even ignoring that, something seems wrong. Consider:

 

We know partner has 4 spades and 4 diamonds and we know what our own 13 cards are. What are the odds that NONE of partner's 5 unknown cards are diamonds?

 

Can it really be anywhere close to 60%?

 

That is the math problem I tried to solve (roughly) in my head and the answer I got was "definitely not".

 

Either the question I was asking was wrong, I made a mistake when I tried to figure out the answer, or hotShot and twcho are way off the mark.

This is the wrong question.

It's too late for me to think about a good explanation, but maybe the following reasoning is intuitive:

 

Fred's question would be the right question if we knew the first 8 cards partner got dealt were 4 spades and 4 diamonds, and then we watch him getting dealt 5 more random cards. What we actually know is that partner got dealt at least 4 diamonds and 4 spades in total - but maybe he got his 4th diamond only on the 13th card he got dealt, so there was never time to deal him another diamond.

 

(To Fred: I would guess sure you have already thought about this difference - as it is the reason why writing a dealing program with constraints is tricky.)

Thanks for explaining, Arend. It all makes sense and you are right that I have thought about this difference in the context you guessed.

 

I ran a sim using my definition for a 1D opener, 1H response and passes by the opponents, and threw out all hands with 4+ hearts and 0-3 spades in the opener. Over 100000 hands that fit these criteria, the # of times opener held 4 diamonds was 35732, or 35.73%.

 

Assuming this number is approximately correct, It is good to know that my bridge instincts are still fine, even if I am not very good at a math :)

 

Now that we have that cleared up and I since I won't be quitting bridge, about the actual bridge problem...

 

No matter what the exact numbers are, I would certainly go through 4th suit forcing. While I suppose it is possible this will help the opponents on opening lead, I don't think it is very likely. Meanwhile, not only might we belong in some contract other than 3NT, if we belong in 3NT partner probably should be declarer.

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[hotShot]

I still have no idea where hotShot's and twcho's numbers came from, but I strongly suspect that one reason they go against instinct is that they do not take HCPs into account.

 

If partner has 4342 or 4243, he has 12-14 HCP.

 

If partner has any other distribution, his HCP range is much wider.

 

But even ignoring that, something seems wrong. Consider:

 

We know partner has 4 spades and 4 diamonds and we know what our own 13 cards are. What are the odds that NONE of partner's 5 unknown cards are diamonds?

 

Can it really be anywhere close to 60%?

Now that I'm fully awake, I see the problem.

 

We are looking at the set of hands that have 4+♦ 0-3♥ and 4♠ with the additional restriction that partner holds a 3424 shape.

 

So the number of deals that have 4+♦ is 100%, in ~60% its only 4♦ cards.

 

Deals that have only 3♦ have exactly 4♠4♥32 shape and partner would have raised ♥.

The balanced hands are a little overestimated since the balanced hands with 15-17 HCP would have been opened 1NT. Fortunately the error is small since holding 15 HCP yourself reduces the number of unbalanced 15+ hands that partner could have.

[eyhung]

What is your definition of a 1♦ opener?

It seems to deviate from the SAYC standard where a 1♦ opening almost guarantees 4♦ cards.

Are you sure you did not include hands with 5♠ and 4-5♦?

A lot of people would even open 1♠ holding 5♠ and 6♦.

No hand has 5 spades and 5 diamonds. The opening bids are quite well-tested. And yes, there are hands where opener has 6 diamonds and 5 spades -- in fact any hand with 5-6 in the pointed suits, and does not qualifiy for 2♣ qualifies as a 1♦ opener.

 

"A lot of people" may open 1♠ with 5-6 in the pointed suits, but a lot of people voted for [controversial politician of your choice]. Suffice to say, I don't agree with them. With the rebid being 1♠, I don't think there are many hands that should go out of their way to distort the suit lengths by opening 1S, so all 5-6 spade diamond hands open 1♦.

 

In any case, we are splitting hairs with these definitions. As long as you acknowledge that the constraints are in the ballpark of a reasonable 1♦ opening bid style, 25% and 60% are far apart. Once you realize that opener must hold at least 4 diamonds because of the failure to raise hearts, and most hands with 4 diamonds do not rebid 1♠ thanks to notrump being a higher priority, you will understand that 25% is a far better estimate than 60%. This makes more sense to me as well -- I tend NOT to rebid 1♠ even with weak notrumps that qualify, so it's nice to see that my style only "costs" me a low percentage of the time (less than 25%, since I'd also rebid 1♠ with 4144.)

[eyhung]

I think you'll get different numbers if you use the 1♥ response versus you use the hand in the OP that is exactly 3424 (and possibly slightly different if you don't make it exactly this 3424 complete with the relevant honors held). Intuitively it feels obvious that if we are more balanced, the odds that partner has only 4 diamonds are higher than if we are something like 3613 (which presumably would reply 1♥ in response).

Correct, I reran the sim fixing the responder's hand to the OP's hand, and the chance of opener holding exactly 4 diamonds went up from 25% to 28.6%.

[hotShot]

Once you realize that opener must hold at least 4 diamonds because of the failure to raise hearts, and most hands with 4 diamonds do not rebid 1♠ thanks to notrump being a higher priority, you will understand that 25% is a far better estimate than 60%. This makes more sense to me as well -- I tend NOT to rebid 1♠ even with weak notrumps that qualify, so it's nice to see that my style only "costs" me a low percentage of the time (less than 25%, since I'd also rebid 1♠ with 4144.)

You are comparing 2 numbers that are almost unrelated.

 

Of cause about 25% off all openings are ♦.

 

Twcho and I have been looking at deals that have the ♦ and ♠ length shown by the auction that are restricted by responders known shape. Of these deals 60% have only 4♦ the remaining 40% have more than 4♦.

[eyhung]

Once you realize that opener must hold at least 4 diamonds because of the failure to raise hearts, and most hands with 4 diamonds do not rebid 1♠ thanks to notrump being a higher priority, you will understand that 25% is a far better estimate than 60%.  This makes more sense to me as well -- I tend NOT to rebid 1♠ even with weak notrumps that qualify, so it's nice to see that my style only "costs" me a low percentage of the time (less than 25%, since I'd also rebid 1♠ with 4144.)

You are comparing 2 numbers that are almost unrelated.

 

Of cause about 25% off all openings are ♦.

 

Twcho and I have been looking at deals that have the ♦ and ♠ length shown by the auction that are restricted by responders known shape. Of these deals 60% have only 4♦ the remaining 40% have more than 4♦.

I think it has been quite clear that I am NOT saying that 25% of all possible bridge hands contain exactly 4 diamonds. I do not see why anyone would perceive that unless they had an emotional attachment to erroneous results. Instead, I am saying that ~25% of hands that

 

1) qualify for a 1♦ opening bid under 2/1 and

2) would rebid 1♠ over a 1♥ response

 

contain exactly 4 diamonds.

 

If you would open 1♦ with 4342 and 16 HCP, or you would rebid 1♠ with 4342 and 19 HCP, then you will obviously get different results.

[PhantomSac]

4342 being 3 times as likely as 4351 seems weird.

But it is correct (assuming we ignore the hcp restrictions).

Yeah but as has been said now, those hands would open 1N or rebid 2N a lot so I'm glad to see my inclination to think that was the problem was correct heh.

[aguahombre]

Being less of a wizard at probabilities and percentages, etc., I agree with Roland's last post and stick with the 3NT rebid rather that the 4SF. To paraphrase:

 

I have denied 5+ hearts, denied 4 spades, denied 4 diamonds, shown a balanced hand, and shown around 14 HCP. Why should I be the one trying to figure out what opener has?

 

If I have been able to somewhat describe my size and shape early in the auction, that doesn't mean I am necessarily "in a big hurry", it just means I like to be able to described my size and shape when I can.

 

Opener will know whether he has 4-3-4-2, 5-X-6-y, 4-3-6-0, etc. without my guessing the probability of it, and act accordingly.

[PhantomSac]

Your shape is not nearly as important as your honor location when deciding whether to jump to 3N. For the same reason you would not jump to 3N over a 1D opener with Axx Axx Axx Qxxx even if you played it showing 13-15 balanced, but you would with KJx KJx Qxx KJxx.

 

If you bid 3N partner is going to assume a stiff somewhere and a little extra shape is not enough reason to bid, and he's going to pass. Since there are some hands like that where you will be in the wrong game, it is worth it to investigate that possibility at a low level.

[mike777]

I am not sure how much better off I would be if for me:

 

 

1d=1h

1s=2d(art. gf)

2h(3)=2nt

3d=?

 

rather than:

1d=1h

1s=3nt(14-15 bal.)

 

 

I know pard has 4s, 3h and 5 or 6d and 9/10-17/18 pts.

[jdonn]

I know pard has 4s, 3h and 5 or 6d and 9/10-17/18 pts.

You know a lot more than that.

- Partner doesn't have 9.

- He probably just has 10 if 4360.

- If he has 17-18 he won't pass 3NT anyway.

 

As for his 11-16, you still know a lot more.

- He didn't bid 3♥ or 3♠, showing similar shape but a very strong holding in the suit.

- It's actually an interesting question what 3♣ should mean but in any case he didn't bid it.

- He doesn't have a 4351 with slow values and bad diamonds that would just bid 3NT over 2NT, like Axxx Qxx KJxxx A.

 

Also he knows a lot more about your hand in this case. You will rebid 3♠ over 3♦ (not 3NT!) He can infer that you have spade strength and a single club stopper, and also not lots of heart strength. He is so much better placed to make a decision than he would be over a second round jump to 3NT that it's not even funny.

[mike777]

I know pard has 4s, 3h and 5 or 6d and 9/10-17/18 pts.

You know a lot more than that.

- Partner doesn't have 9.

- He probably just has 10 if 4360.

- If he has 17-18 he won't pass 3NT anyway.

 

As for his 11-16, you still know a lot more.

- He didn't bid 3♥ or 3♠, showing similar shape but a very strong holding in the suit.

- It's actually an interesting question what 3♣ should mean but in any case he didn't bid it.

- He doesn't have a 4351 with slow values and bad diamonds that would just bid 3NT over 2NT, like Axxx Qxx KJxxx A.

 

Also he knows a lot more about your hand in this case. You will rebid 3♠ over 3♦ (not 3NT!) He can infer that you have spade strength and a single club stopper, and also not lots of heart strength. He is so much better placed to make a decision than he would be over a second round jump to 3NT that it's not even funny.

thank you for taking time to respond, very helpful and insightful ,ty.