[PhantomSac]

fred, on Jan 28 2010, 08:52 PM, said:

hotShot, on Jan 29 2010, 01:05 AM, said:

4342 => 25,5%
4243 => 25,3%
4242 => 12,3%
4351 => 7,7%
4153 => 7,6%
4144 =>10,7%
4261 => 2,7%
4162 => 2,7%

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.

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

4054, 4063, 4360, 5x6x are all missing as shapes so I'm sure his numbers are wrong. They are normal hotshot numbers though

[twcho]

Assuming opener will rebid 1♠ on all hands with 4♠ and without 4♥, and excluding the chance of 5+♠6+♦ or 4♠7+♦ (which add up to about 3% anyway), the possibility for each hand shape are:
4342 => 24.3%
4243 => 24.3%
4252 => 14.6%
4351 => 8.5%
4153 => 8.5%
4144 => 9.1%
4261 => 3.6%
4162 => 3.6%
4054 => 1.4%
4360 => 0.9%
4063 => 0.9%
The chance for holding exactly 4♦ is 57.8% which is pretty close to 60%. However, in the case of holding only 4♦, many may choose to rebid 1NT and that will greatly reduce the possibility of only 4 cards in ♦.

[fred]

twcho, on Jan 29 2010, 02:39 AM, said:

Assuming opener will rebid 1♠ on all hands with 4♠ and without 4♥, and excluding the chance of 5+♠6+♦ or 4♠7+♦ (which add up to about 3% anyway), the possibility for each hand shape are:
4342 => 24.3%
4243 => 24.3%
4252 => 14.6%
4351 => 8.5%
4153 => 8.5%
4144 => 9.1%
4261 => 3.6%
4162 => 3.6%
4054 => 1.4%
4360 => 0.9%
4063 => 0.9%
The chance for holding exactly 4♦ is 57.8% which is pretty close to 60%. However, in the case of holding only 4♦, many may choose to rebid 1NT and that will greatly reduce the possibility of only 4 cards in ♦.

My instincts (and some rough math I did in my head) suggest that your 57.8% is also way off the mark even if opener always rebids 1S whenever he has 4 cards in that suit.

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

[PhantomSac]

4342 being 3 times as likely as 4351 seems weird.

[fred]

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%?

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.

Yes, I know my question doesn't eliminate hands with 4-card heart support or hands that would open the bidding 1S, but ignoring these possibilities increase the chances that partner has exactly 4 diamonds. My question also ignores HCPs, but that was intentional as I was trying to get roughly the same answer as hotShot and twcho (or hopefully not roughly the same answer!).

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

[cherdanno]

PhantomSac, on Jan 28 2010, 10:55 PM, said:

4342 being 3 times as likely as 4351 seems weird.

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

[cherdanno]

fred, on Jan 29 2010, 12:09 AM, said:

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]

fred, on Jan 29 2010, 02:52 AM, said:

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]

jdonn, on Jan 29 2010, 03:31 PM, said:

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

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]

eyhung, on Jan 29 2010, 01:32 AM, said:

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]

cherdanno, on Jan 29 2010, 05:45 AM, said:

fred, on Jan 29 2010, 12:09 AM, said:

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.

Quote

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]

fred, on Jan 29 2010, 06:09 AM, said:

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]

hotShot, on Jan 29 2010, 02:40 AM, said:

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]

Mbodell, on Jan 29 2010, 03:44 AM, said:

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%.