hold back,,, or go for it?

[han]

In case somebody is interested, here is the calculation for 45/7. I'm going to write C(n,m) for the number of ways to choose m cards from n. If you are not familiar with this then I can't help you here.

 

There are 4 diamonds out, 6 hearts and 16 black cards for a total of 26.

 

The number of ways to get exactly 3 hearts and 0 diamonds (and 10 black cards) is therefore C(6,3)*C(16,10), while the number of ways to get 2 hearts and 2 diamonds (and 9 black cards) is C(6,2)*C(4,2)*C(16,9). When you write this out in terms of factorials and divide the second fraction by the first almost all terms cancel and you get 45/7.

[han]

It reflects that, imo, this is a very, very poor situation for any simulation.

I agree with everything you said but I do think that a simulation could be useful. Just the double dummy results (if correct) for 5D vs 4H would already tell us something more than we know now. While this will not tell us what we should bid, it will be meaningful information.

[foo]

I'm still cranking the math, but using the actual 4H hand and this hand, I can tell you that GOPs most likely shapes are

=5314, =6313, =6214, =6304 in that order.

If you are as literate with simple probabilities as you are with computer programming then you must have noticed that these claims are way off. A simple computation shows that partner is more than 6 times as likely (45/7) to be 2-2 in the red suits than 3-0.

 

I'm not going to check the rest of your numbers.

I'm not sure where you are getting your numbers, but they certainly are not taking into account that conditional probability is at work here.

 

For example, the simpler analysis that starts with two known hands

Given the two known hands

x.AQJTxxx.Tx.Axx and xxx..AQJ98xx.KJ9

What are GOPs most likely shapes?

 

S's: x.xxx.?.? (AKQJT98xx) 9/26 unused

H's: AQJTxxx..?.? (K98xxx) 6/26 unused

D's: Tx.AQJ98xx.?.? (Kxxx) 4/26 unused

C's: Axx.KJ9.?.? (QT8xxxx) 7/26 unused

 

Now let's say we want to create the most likely shapes for a 3rd hand given the above.

A simple procedure to do so would be to iteratively "pull" a card from the most likely suit until we have taken 13 cards.

So the 1st card would be a S (8+6+4+7= 25 left).

So would the 2nd. (7+6+4+7= 24 left)

Then a S+C (6+6+4+6= 22 left)

Then a S+H+C (5+5+4+5= 19 left)

and again S+H+C (4+4+4+4= 16 left)

We have only 3 spaces left and 4 equally probable suits they can come from, so

the last 3 cards can be any of (S+H+D, S+H+C, S+D+C, H+D+C)

 

A shorthand for describing the most likely "pulls" of cards for GOP is

9, 8, 7|7, 7, 6|6|6, 6|6, 6, 5|5|5, 5|5, 5, 4|4|4|4, 4|4|4, 4|4

ssscshcshc + (hdc =5314, or shd =6313 or sdc =6214 or shc =6304)

 

Once we know the most likely shapes, straight combinatorics gives us the likelihood of a specific pattern of honors and "x" 's in each suit.

 

This leads to the single most likely hand pattern for GOP to have being

HHxxx.xxx.x.Qxxx or HHxxx.Kxx.x.Qxxx (with equal weight of 60*10*3*20)

 

As I said before "HH" is just as likely to be "AK" as "QJ" here from the set (A,K,Q,J)

 

I'm certainly fallible, but I believe the logic as to the conditional probabilities is correct here.

[han]

I'm still cranking the math, but using the actual 4H hand and this hand, I can tell you that GOPs most likely shapes are

=5314, =6313, =6214, =6304 in that order.

If you are as literate with simple probabilities as you are with computer programming then you must have noticed that these claims are way off. A simple computation shows that partner is more than 6 times as likely (45/7) to be 2-2 in the red suits than 3-0.

 

I'm not going to check the rest of your numbers.

I'm not sure where you are getting your numbers, but they certainly are not taking into account that conditional probability is at work here.

 

For example, the simpler analysis that starts with two known hands

Given the two known hands

x.AQJTxxx.Tx.Axx and xxx..AQJ98xx.KJ9

What are GOPs most likely shapes?

 

S's: x.xxx.?.? (AKQJT98xx) 9/26 unused

H's: AQJTxxx..?.? (K98xxx) 6/26 unused

D's: Tx.AQJ98xx.?.? (Kxxx) 4/26 unused

C's: Axx.KJ9.?.? (QT8xxxx) 7/26 unused

 

Now let's say we want to create the most likely shapes for a 3rd hand given the above.

A simple procedure to do so would be to iteratively "pull" a card from the most likely suit until we have taken 13 cards.

So the 1st card would be a S (8+6+4+7= 25 left).

So would the 2nd. (7+6+4+7= 24 left)

Then a S+C (6+6+4+6= 22 left)

Then a S+H+C (5+5+4+5= 19 left)

and again S+H+C (4+4+4+4= 16 left)

We have only 3 spaces left and 4 equally probable suits they can come from, so

the last 3 cards can be any of (S+H+D, S+H+C, S+D+C, H+D+C)

 

A shorthand for describing the most likely "pulls" of cards for GOP is

9, 8, 7|7, 7, 6|6|6, 6|6, 6, 5|5|5, 5|5, 5, 4|4|4|4, 4|4|4, 4|4

ssscshcshc + (hdc =5314, or shd =6313 or sdc =6214 or shc =6304)

 

Once we know the most likely shapes, straight combinatorics gives us the likelihood of a specific pattern of honors and "x" 's in each suit.

 

This leads to the single most likely hand pattern for GOP to have being

HHxxx.xxx.x.Qxxx or HHxxx.Kxx.x.Qxxx (with equal weight of 60*10*3*20)

 

As I said before "HH" is just as likely to be "AK" as "QJ" here from the set (A,K,Q,J)

 

I'm certainly fallible, but I believe the logic as to the conditional probabilities is correct here.

My calculation is posted above, yours is nonsense.

[foo]

mikeh,

 

answer to p1=

The most likely H holdings in GOPs hand are any of xxx, Kxx, or xx

If the rest of the hand is suitable for a balance, especially with 5+S, would you not always balance?

 

answer to p2=

Do you raise all your partner's balances? Do you expect your partner to raise all of yours? I'll bet not in either case. Why should this situation be any different?

 

answer to p3 and p4=

The nice thing about doing things based on what cards are likely to be dealt where rather than by attempting to simulate the circumstance of a specific question is that I don't have to explicitly ask any question in order to figure out what is most likely to happen. I just have to be careful and honest about putting the cards where they are most likely to go.

[foo]

Hannie, I don't think you can legitimately do what you are doing since you have to give no more and no less than 13 cards to each player.

 

...and that is dependent on how many cards of =each= suit are available, not simply a two of them.

 

...and trust me C(n,m) was used extensively when it came time to start making hand patterns with honor holdings and sort them according to likelihood! :)

[cherdano]

Hannie, I don't think you can legitimately do what you are doing since you have to give no more and no less than 13 cards to each player.

 

...and that is dependent on how many cards of =each= suit are available, not simply a two of them.

Wrong again.

 

Edit: If you did use binomial coefficients, you should learn how to use them. I hear the University of Wisconsin has a great class for this, "topics in finite mathematics".

[han]

I'm going to do you a favor and explain why your argument is wrong. This is not easy material if you think about it for the first time and it is tough to explain it using this medium but I will try.

 

Suppose you have a bag with 6 balls, 4 green and 2 blue. You draw two of those balls. What is more likely, two green balls or a green and a blue ball? Using your argument two green balls would be most likely (the first is most likely to be green and the second is also more likely to be green because of the 5 balls left 3 are green). This is not correct.

 

There are 6 possible ways to draw 2 green balls. An easy way to see this is to mark them A, B, C and D. The possible drawings are AB, AC, AD, BC, BD and CD. Let's say that the blue balls are marked X and Y. Then there are 8 ways to draw a blue and a green ball: 4 choices for the green ball and 2 for the blue ball. The drawings are AX, BX, CX, DX, AY, BY, CY and DY. So it is more likely that you get a blue and a green ball than that you get two green balls.

 

If you couldn't follow this then you can check it with an easy experiment. Really make a bag and put in 6 balls, 4 marked green and 2 marked blue. Draw two balls from the bag and write down your results. I suggest you do this 150 times. You should get roughly 60 times two green balls, and 80 times a green and a blue ball. I'll let you guess what you will get the other 10 times.

 

If you followed this, look at your computation again and maybe you can see where you went wrong. Then you can also look at the short computation that I wrote earlier.

[han]

Hannie, I don't think you can legitimately do what you are doing since you have to give no more and no less than 13 cards to each player.

 

...and that is dependent on how many cards of =each= suit are available, not simply a two of them.

Wrong again.

 

Edit: If you did use binomial coefficients, you should learn how to use them. I hear the University of Wisconsin has a great class for this, "topics in finite mathematics".

Yeah Yeah, very funny Arend. But binomial coefficients is a good term to google.

[Finch]

To Hannie & foo

 

Maybe I'm wrong, but I think foo was working out partner's potential distributions given both our hand and the 4H opener's actual hand.

 

Hannie was working them out given only our hand. Hence the difference.

 

I really don't see the point of the former calculation: if we could see RHO's hand before bidding every time we'd all have much better results!

[Ant590]

Hannie, I don't think you can legitimately do what you are doing since you have to give no more and no less than 13 cards to each player.

 

...and that is dependent on how many cards of =each= suit are available, not simply a two of them.

Wrong again.

 

Edit: If you did use binomial coefficients, you should learn how to use them. I hear the University of Wisconsin has a great class for this, "topics in finite mathematics".

Too subtle perhaps - see http://www.math.wisc.edu/~peters/ :(

[han]

To Hannie & foo

 

Maybe I'm wrong, but I think foo was working out partner's potential distributions given both our hand and the 4H opener's actual hand.

 

Hannie was working them out given only our hand. Hence the difference.

 

I really don't see the point of the former calculation: if we could see RHO's hand before bidding every time we'd all have much better results!

No you are wrong. I was doing what foo proposed: working out the probabilities given the TWO hands. I agree that this is not useful but I couldn't stand it.

 

foo was attempting the same but it was just nonsense.

[han]

I'm still cranking the math, but using the actual 4H hand and this hand, I can tell you that GOPs most likely shapes are

=5314, =6313, =6214, =6304 in that order.

Let's see once again. There are 9 spades left, 6 hearts, 4 diamonds and 7 clubs. You think 5314 is the most likely distribution for partner to have? How about LHO, does LHO also most likely have a 5314 distribution? How likely is it that there are 2 diamonds lying on the floor somewhere?

 

I haven't done the calculations but I'm willing to bet that the two most likely distributions for partner are 4-3-2-4 and 5-3-2-3 (you can easily see that they are exactly equally likely). Any takers?

[Finch]

I still can't see the point of doing calculations based on knowing both our hand and LHO's.

 

If you know your hand and give LHO 7 hearts, there are obviously many symmetries (7 hearts and 7 diamonds known, you are exactly 3-3 in the blacks) the most likely distributions for partner are, in order,

 

4,2,3,4 3.81%

4,3,2,4 3.81%

5,2,2,4 3.43%

4,2,2,5 3.43%

3,3,3,4 2.90%

4,3,3,3 2.90%

3,2,3,5 2.61%

5,2,3,3 2.61%

5,3,2,3 2.61%

3,3,2,5 2.61%

5,1,3,4 1.83%

4,1,3,5 1.83%

5,3,1,4 1.83%

4,3,1,5 1.83%

5,1,2,5 1.65%

5,2,1,5 1.65%

4,2,4,3 1.63%

3,2,4,4 1.63%

4,4,2,3 1.63%

3,4,2,4 1.63%

6,2,2,3 1.63%

3,2,2,6 1.63%

2,3,3,5 1.31%

5,3,3,2 1.31%

3,3,4,3 1.24%

3,4,3,3 1.24%

6,1,2,4 1.14%

4,1,2,6 1.14%

4,1,4,4 1.14%

6,2,1,4 1.14%

4,2,1,6 1.14%

4,4,1,4 1.14%

3,1,3,6 0.87%

6,1,3,3 0.87%

6,3,1,3 0.87%

3,3,1,6 0.87%

4,3,4,2 0.82%

2,3,4,4 0.82%

4,4,3,2 0.82%

2,4,3,4 0.82%

2,2,3,6 0.82%

6,2,3,2 0.82%

6,3,2,2 0.82%

2,3,2,6 0.82%

3,1,4,5 0.78%

5,1,4,3 0.78%

5,4,1,3 0.78%

3,4,1,5 0.78%

2,2,4,5 0.73%

5,2,4,2 0.73%

5,4,2,2 0.73%

2,4,2,5 0.73%

[Finch]

And if you can be bothered to do the sums knowing that RHO is 1723, then partner's most likely distributions are (as Hannie points out the symmetries are obvious):

 

4,3,2,4 5.09%

5,3,2,3 5.09%

5,2,2,4 3.82%

4,4,2,3 3.82%

4,3,3,3 3.39%

5,3,1,4 3.39%

4,2,3,4 2.54%

5,2,3,3 2.54%

4,4,1,4 2.54%

5,4,1,3 2.54%

3,4,2,4 2.54%

6,2,2,3 2.54%

4,2,2,5 2.29%

5,4,2,2 2.29%

3,3,3,4 2.26%

6,3,1,3 2.26%

4,3,1,5 2.04%

5,3,3,2 2.04%

3,3,2,5 2.04%

6,3,2,2 2.04%

3,4,3,3 1.70%

6,2,1,4 1.70%

5,2,1,5 1.53%

4,4,3,2 1.53%

5,1,3,4 1.02%

4,5,1,3 1.02%

3,2,3,5 1.02%

6,1,2,4 1.02%

6,2,3,2 1.02%

3,4,1,5 1.02%

3,5,2,3 1.02%

6,4,1,2 1.02%

5,1,2,5 0.92%

4,5,2,2 0.92%

2,4,3,4 0.73%

7,2,1,3 0.73%

[david_c]

I still can't see the point of doing calculations based on knowing both our hand and LHO's.

 

If you know your hand and give LHO 7 hearts, there are obviously many symmetries (7 hearts and 7 diamonds known, you are exactly 3-3 in the blacks) the most likely distributions for partner are, in order,

I think that for once Frances may have got this wrong. Looking at our hand and giving RHO seven hearts, the two hands have exactly seven hearts between them, but at least seven diamonds. So the other two hands should not be symmetric between the red suits, but will on average have more hearts than diamonds.

 

[Edited: I suppose you may mean to give RHO at least seven hearts, rather than exactly seven hearts. But that still doesn't work because he's much more likely to have additional diamonds than additional hearts.]

[jtfanclub]

No, he's saying using the ACTUAL two hands, not the potential ones. In other words, if your RHO's bid showed this exact shape, what would your partner be most likely to have?

[jtfanclub]

I'll agree with the many 5♦ bidders.

 

One point people seem to be neglecting is that the 5♦ bid doesn't have to be right for it to work. After I bid 5♦, opponents will also be put to a guess.

Suppose that you asked your LHO (in a face to face game) what the bid showed, and he replied 'a solid suit'. You then asked for more details, and received "Probably a 7 card suit missing at most one honor, and some side value, an ace or king or maybe a QJ combination". So not only do you know what opener has, but you know that his partner knows.

 

Would you still bid 5♦? Or does that move you back the other way?

[Finch]

I did a simulation on this.

As Justin says, it's quite possible my results are biased so make of this what you will.

If anyone really cares I'll do another one (I've lost the first one now) and post all the hands.

 

I gave us our hand, and RHO 7 or 8 hearts, not a 7222 unless the hearts were solid, 12 HCP or fewer. I dealt 100 hands and then eliminated about 25% of them as not being 4H openings for anyone vaguely sane. The remainder could be split between 4H openers the world would open 4H, and ones that are warped in one way or another (i.e. would be passed or opened either 1H or 3H by a significant percentage of people).

 

The summary of my findings is:

 

i) It's definitely right to bid 5D about 35% of the time (if you don't you miss game or slam, or you miss a 1-off-on-best-defence save against 4H).

 

ii) It's definitely right to pass about 20% of the time. The main danger is that you go for a nasty penalty, not that you miss a spade contract.

 

iii) It's unclear about 25% of the time. That is, double-dummy it's marginally wrong to bid (-500 against -420 or -100/200 against +50/+100) but sometimes they will bid 5H anyway, sometimes they will bid 5H and go off, sometimes they won't double, sometimes they will lead the wrong thing.

 

iv) It doesn't matter about 20% of the time (either they are going to play in 5/6H whatever you do, or partner is going to protect with a double and, I assume, you bid 6D).

 

A few additional comments:

- On rather limited evidence, if you pass and partner bids 4S, pass it.

- The sounder RHO's pre-empting style is, the more it is right to bid. The more RHO has solid hearts, the fewer heart honours partner has, and the greater than chance that they are making 4H (which improves your equity from bidding)

- Your heart void is worth a huge amount. 2-1-7-3 is often two tricks worse in offence (sometimes because RHO can win the opening lead and play a club through, but mainly because LHO has to lead the right thing at trick 1: he can't cash a heart and decide which black suit to switch to - would you find a spade lead against 5D (doubled or not) from something like AQxx Ax xx Axxxx?)

- At matchpoints it's much closer whether to pass or bid, because -500 against -420/-450 is a bit more pricey, and LHO is more likely to find a light double.

 

- At any other vulnerability it's not even close whether to bid or not. But that probably isn't a surprise.

[foo]

It is double dummy obvious that NS should have been in 4S or 5S, not 5D, on this board. Before the analysis, this wasn't worth much mention because it would've been Resulting.

 

Now it is mathematically provable that the odds are so good that we belong in a Spade contract that we should not make the unilateral bid of a direct 5D.

 

While there are plenty of hands with the expected 5+S where GOP is highly likely to balance, apparently far more than the direct 5D bidders thought, there are definitely some where even if the odds are in your favor you have to have "big brass ones" to balance w/ them ATT, so I sympathize with the problem the direct 5D bidders have w/ passing.

 

I'm seriously considering whether X followed by a suit at this level should be redefined as not showing extras, simply not liking Advancer's response.

Sort of an extension to the Equal Level Conversion idea.

 

Then

=our 7+ loser hands pass. We need too much from GOP to assume it.

=our 6 loser hands can X if they have support for a 4 level contract (correcting Advancer if they have to),

OTOH, 6 loser hands that can only consider 5m pass (4 cover cards is too much to assume in GOPs hand; and if he's got them he'll very likely balance)

=our 5- loser hand can do whatever they darn want since they can afford to.

 

Thus in this sort of situation holding =3073, 6 losers, and decent controls (Bridge is the game of A's and K's after all), you X intending to sit for GOPs bid.

OTOH, had you had the same hand but =2173, you should probably pass in this situation. (I haven't analyzed this yet.)

 

Nothings perfect, preempts =do= work after all, but it appears the above might be better than Standard.

 

 

To Hannie: You want to have a discussion as to whether what I did was mathematically sound, let's do it privately. Especially, if you want to be insulting. Better yet, publish a proof as to why what I did was wrong. Just because I approached it differently than you would does not mean the approach is wrong.

I'm perfectly aware of standard combinatorics, thanks.

 

 

However, it appears the real lesson of this thread is that many of the regular posters don't care what anyone else thinks or what logic or math that differs from their own prejudices say.

 

Inference taken, Lesson learned. Bye.

[jdonn]

Good post Frances. My only real criticism is the opener could easily have 9 hearts also (or even more?) which seems to make bidding an even better option.

 

Hannie's math is quite correct. Foo's is meaningless gibberish. But can I ask a silly question? What is the point about doing a calculation based on knowing 4♥ opener's hand, which you don't know about at the time you have to bid? It would obviously change the bidding problem to know he had the minimum heart length and maximum strength in his range, but sadly at the table we are not privy to this information.

[foo]

Clarification as to methods:

 

I =started= by using the actual two hands. That was Part I.

 

In Part II, I changed the 4H bidder's hand from a specific instance to a class where the constraints were:

a= a single suited hand containing 7-8 H's

b= 3-S, usually 2-S

c= 1-4 Honors

d= Since the bidder was at White vulnerability, 7-6 losers.

e= Not so many HCPs as to be in serious danger of missing a slam

(Yes that means 11 counts are a tiny bit heavy here, but since it happened ATT...)

 

I then published =preliminary= results. Which I'm not going to bother finishing given the "pig pile" I got sacked under (USA football reference) for bothering.

 

My results are more or less in line with Frances' simulation, except that I could not find =any= 12 counts, and hardly any 11 counts, that seemed sane 4H openings. Frankly, the given 4H hand is too heavy.

[mikeh]

I couldn't stand it any longer, plus I was intrigued by Frances' results. So I ran it under DealMaster Pro, specifying for opener 7-9 ♥s, 5-12 hcp and no other constraints. As I suspected, the main problems were trying to figure out which hands qualified for a 4♥ opening bid and what the other two bidders would do.

 

I lacked the time required to deal with 4th chair balancing issues, altho a handful were clear balances had the 5♦ hand passed, including a laydown ♠ slam unbiddable after 5♦ (imo).

 

My results, probably not as well-analyzed as Frances', were a little different.

 

I ran 90 hands, but ended up rejecting 35 of them as not a 4♥ bid (note, my constraints were not quite Frances'). Of the remaining 55, on 26 of them 5♦ worked better than pass... in a significant minority, this was because LHO took the push to 5♥ down 1. On 20 of them, pass was clearly best... usually because we were going down, either turning a plus (on defence) into a minus or by going down 500... more than 500 was uncommon. On a few hands, passing would lead to a better ♠ spot, but I didn't analyze these to see whether this required, on average, that the 5♦ hand pass or raise... I accept that Frances' results suggest pass of 4♠, but I defy a real world bridge player to pass 4♥ and then pass 4♠ by partner!

 

On 9 of the 55, I really couldn't tell what was best due to uncertainty about the defence or about what would happen in the bidding either over 5♦ or pass.

 

So my results suggest that bidding is slightly superior to passing, but my sample was both limited and probably flawed in that I am sure that others would have included some of the hands I rejected, rejected some of the hands I included, and disagreed with my view of how the other hands would bid/defend, etc.

 

Oh well, at least I reinforced my own preferences :)

[jtfanclub]

in a significant minority, this was because LHO took the push to 5♥ down 1.

I don't think you can do that.

 

Bidding 5♥ here is a very stylistic thing. Your LHO is going to take the push based on a lot of things we can't know...how long you took to bid 5♦, their tendencies for opening at the 4 level, state of the match, etc. I think there are very few hands where, after 4♥ 5♦, the entire field would bid 5♥. If it's strong enough to bid 5♥ to make by the guys whose partners tend to have a 5 count, it's strong enough to bid 6 by the guys whose partners tend to have an 11 count. Bidding 5♥ as a sacrifice is even more tricky to predict.

 

I think what you've shown is if YOU open 4♥, and I hold this hand, I should bid 5♦.

 

Which, come to think of it, is a good thing to know. :)

[Finch]

My results, probably not as well-analyzed as Frances', were a little different.

 

Of the remaining 55, on 26 of them 5♦ worked better than pass... in a significant minority, this was because LHO took the push to 5♥ down 1. On 20 of them, pass was clearly best... usually because we were going down, either turning a plus (on defence) into a minus or by going down 500... more than 500 was uncommon.

Actually, your results aren't that different.

 

You had a ratio of 26:20 in favour of bid over pass, + a small amount of uncertainty on other hands

 

I had a ratio of 35:20 in favour of a bid + a rather larger amount of uncertainty on other hands.

 

Given the small number of actual hands we ended up with, I don't think these results are outside the range of experimental error: bidding is better, but not by an enormous margin.

 

The one conclusion I've seen elsewhere that I now disagree srongly with is that 4S is much more likely to be right than 5D. I learnt from the simulation that this is very unlikely to be the case.