do you open 1 Spade with this hand?

[JLOGIC]

Because you can show your feature

 

FYP

[gwnn]

No-one seems to have mentioned the old story of a (supposedly strong) player passing a hand like this and, as all four of them were putting their hands back in the board, asking his partner what he had held. "Oh, just three bare aces". I guess they deserved each other.

 

btw What is the Ogust response that shows a solid seven-card suit?

To answer your question, 3NT.

 

Additionally, I should like to point out that your partner will have 3+ card support for you 65.63 percent of the time. Accordingly opening 2♠ in no way rules out reaching 4♠ but also preserves the possibility of reaching 3NT.

The question was about Ogust, so maybe this is the first time in history that gwnn wins an argument against JLOGIC?

[VM1973]

65.63%

 

Either you are funnier than I think you are capable of being, or you are nuts.

It's simple math.

 

There are 39 unknown cards, 13 of which are in partner's hand.

The number of hands possible with the following number of spades in partner's hand are:

 

0) 8122425444

1) 48734552664

2) 1.21836E+11

3) 1.62449E+11

4) 1.21836E+11

5) 48734552664

6) 8122425444

 

As you can see, the probability of partner having 0 is the same as the probability of partner having all 6 remaining cards. Same with 1 or 5 as well as 2 or 4. Three is the most likely holding.

 

There are 5.19835E+11 possible hands your partner might have. Accordingly the chance of your partner having each holding is:

 

0 1.56%

1 9.38%

2 23.44%

3 31.25%

4 23.44%

5 9.38%

6 1.56%

 

Now anyone with a calculator can add up 31.25+23.44+9.38+1.56 and reach 65.63%

 

Would it also interest you to know that given this holding the chances of your partner having 5+ hearts is 80.62 percent? In fact your partner's most likely holding given this hand is 6 hearts some 22.56% of the time. So for those of you who looked at that hand and thought "the opponents almost certainly have a heart fit" ... think again.

[gwnn]

39 outstanding cards, 6 of which are spades.

 

if n is the number of spades partner has, then the probability of that is

 

p(n)= C(33,13-n) * C (6,n)/ C(39,13).

 

C(33,13-n) means that we need 13-n cards out of the 33 non-spades.

C(6,n) means that we need n cards out of the 6 spades.

The product of the two above number is what we are looking for.

 

C(39,13) means that we need 13 cards out of the 39 outstanding cards, i.e. all possible hands.

 

enter wolfram alpha:

http://www.wolframalpha.com/input/?i=Table%5B{100.0*C%5B33%2C13-n%5D*C%5B6%2Cn%5D%2FC%5B39%2C13%5D}%2C{n%2C0%2C6}%5D

 

n | p(n), %

0 | {7.05659}

1 | {26.2102}

2 | {35.7412}

3 | {22.7915}

4 | {7.12234}

5 | {1.02562}

6 | {0.0525957}

 

p(3)+p(4)+p(5)+p(6)~31%.

 

I am not sure what you missed but my maths looks correct to me and your result is clearly wrong. There are 6 outstanding spades and 3 opponents, so the expected value of their spade holdings is 2. My values conform to that (you can check it yourself by summing up n*p(n), it will give you 2).

[gwnn]

I am puzzled by your calculations, I see you are calling the number of hands with 6 spades and 0 spades each as 8122425444, but that is exactly C(39,13), i.e. the total number of hands partner might have. That is wrong.

 

Your number for the total number of hands seems to be just a sum of the above numbers you have, i.e. 64*C[39,13]

[VM1973]

39 outstanding cards, 6 of which are spades.

 

if n is the number of spades partner has, then the probability of that is

 

p(n)= C(33,13-n) * C (6,n)/ C(39,13).

 

C(33,13-n) means that we need 13-n cards out of the 33 non-spades.

C(6,n) means that we need n cards out of the 6 spades.

The product of the two above number is what we are looking for.

 

C(39,13) means that we need 13 cards out of the 39 outstanding cards, i.e. all possible hands.

 

enter wolfram alpha:

http://www.wolframalpha.com/input/?i=Table%5B{100.0*C%5B33%2C13-n%5D*C%5B6%2Cn%5D%2FC%5B39%2C13%5D}%2C{n%2C0%2C6}%5D

 

n | p(n), %

0 | {7.05659}

1 | {26.2102}

2 | {35.7412}

3 | {22.7915}

4 | {7.12234}

5 | {1.02562}

6 | {0.0525957}

 

p(3)+p(4)+p(5)+p(6)~31%.

 

I am not sure what you missed but my maths looks correct to me and your result is clearly wrong. There are 6 outstanding spades and 3 opponents, so the expected value of their spade holdings is 2. My values conform to that (you can check it yourself by summing up n*p(n), it will give you 2).

Well it's not impossible that I'm doing something wrong. I'm trying to put English formulae into a Spanish-speaking computer and half the time it gives me the ¿#Nombre? error which means I'm misspelling the name of the formula. If you're doing this on an Excel spreadsheet, I'd be much obliged if you'd email it to me so I can see the Spanish-language translation of the formulae in question.

[Bbradley62]

Well it's not impossible that I'm doing something wrong. I'm trying to put English formulae into a Spanish-speaking computer and half the time it gives me the ¿#Nombre? error which means I'm misspelling the name of the formula. If you're doing this on an Excel spreadsheet, I'd be much obliged if you'd email it to me so I can see the Spanish-language translation of the formulae in question.

Here's the simple version: There are 6 outstanding spades, and you are claiming that there's a 60+% chance that your partner has at least half of them. Shouldn't it be equally likely that LHO has at least half of them? And also equally likely that RHO has at least half of them? And isn't there some possibility that none of the three has at least half of them? Clearly, this adds up to too much.

 

Edit: Yes, the probabilities overlap, as 3-3-0 is possible, but still...

Edited August 6, 2011 by Bbradley62

[Bbradley62]

It's simple math...

 

As you can see, the probability of partner having 0 is the same as the probability of partner having all 6 remaining cards...

Clearly, this is incorrect. Partner, LHO and RHO should all be equally likely to hold all 6... Partner has 0 anytime LHO has all 6, plus anytime RHO has all 6, plus some other times when the 6 are split between LHO and RHO.

[VM1973]

Here's the simple version: There are 6 outstanding spades, and you are claiming that there's a 60+% chance that your partner has at leasat half of them. Shouldn't it be equally likely that LHO has at least half of them? And also equally likely that RHO has at least half of them? And isn't there some possibility that none of the three has at least half of them? Clearly, this adds up to too much.

 

Edit: Yes, the probabilities overlap, as 3-3-0 is possible, but still...

Yes, you're right. That's clearly impossible.

 

I would like to say, though, that I ran gwnn's numbers through the anticipated calculations of how many fast defensive tricks AKQxxxx is likely to take at any given suit contract and still came up with 1.23 fast defensive tricks, on average.

[Finch]

39 outstanding cards, 6 of which are spades.if n is the number of spades partner has, then the probability of that isp(n)= C(33,13-n) * C (6,n)/ C(39,13).C(33,13-n) means that we need 13-n cards out of the 33 non-spades.C(6,n) means that we need n cards out of the 6 spades.The product of the two above number is what we are looking for.C(39,13) means that we need 13 cards out of the 39 outstanding cards, i.e. all possible hands.

 

 

Well it's not impossible that I'm doing something wrong. I'm trying to put English formulae into a Spanish-speaking computer and half the time it gives me the ¿#Nombre? error which means I'm misspelling the name of the formula. If you're doing this on an Excel spreadsheet, I'd be much obliged if you'd email it to me so I can see the Spanish-language translation of the formulae in question.

 

If you name a cell containing the number of cards "n" then use gwnn's formula above - C(a,b) in Excel in English is COMBIN(a,b) so that formula is

=COMBIN(33,13-n)*COMBIN(6,n)/COMBIN(39,13)

 

But I think BBradley's reply is more to the point: a bit of common sense should show you that 65% can't be right. Indeed, you mention that you have the same result for 0 spades in partner's hand as 6 spades. There's only one way he can have 6 spades, so you only have 7 remaining cards to play around with, out of the 33 total non-spades. If he has 0 spades then you have 13 cards out of 33 to select. We can look at the formula to prove that this second number is a lot larger, but thinking about it should also make that clear. (You get the most possibilities at half the total i.e. at 16 or 17 out of 33, and the least at the extremes i.e. 1 or 32 out of 33, and then it's symmetric)

[Trumpace]

I am puzzled by your calculations, I see you are calling the number of hands with 6 spades and 0 spades each as 8122425444, but that is exactly C(39,13), i.e. the total number of hands partner might have. That is wrong.

 

Your number for the total number of hands seems to be just a sum of the above numbers you have, i.e. 64*C[39,13]

 

I guess the mistake is this: *C[6%2Cn]}%2C{n%2C0%2C6}]"]Wolfram Alpha Table

[Bbradley62]

I'm having trouble emailing you an Excel spreadsheet, but it matches gwnn's numbers and looks like this:

 

P(0)=(26*25*24*23*22*21)/(39*38*37*36*35*34)

P(1)=(26*25*24*23*22*13)*(6)/(39*38*37*36*35*34)

P(2)=(26*25*24*23*12*13)*(6*5/2)/(39*38*37*36*35*34)

P(3)=(26*25*24*11*12*13)*(6*5*4)/(3*2)/(39*38*37*36*35*34)

P(4)=(26*25*10*11*12*13)*(6*5/2)/(39*38*37*36*35*34)

P(5)=(26*9*10*11*12*13)*(6)/(39*38*37*36*35*34)

P(6)=(8*9*10*11*12*13)/(39*38*37*36*35*34)

[VM1973]

Trying again, I have these numbers for the chance of partner having the following numbers of hearts:

 

0) 0.25%

1) 2.57%

2) 10.59%

3) 22.85%

4) 28.56%

5) 21.65%

6) 10.10%

7) 2.89%

8) 0.49%

9) 0.05%

10) 0.002%

11) 0.00005%

12) 0.0000003%

 

For the total chance of partner having 4+ hearts as: 63.74%

 

Agree or disagree?

[gwnn]

I agree, those numbers are correct, but who said opponents were almost sure to have a heart fit? Edited August 6, 2011 by gwnn

[VM1973]

I agree, those numbers are correct, but who said opponents were almost sure to have a heart fit?

As we have only one heart, our partner having 4+ does not automatically preclude the opponents from having an 8 card fit.

 

Hypothesis:

 

4♠ is a bad bid under the following circumstances:

 

A) The hand belongs to our side AND 4♠ does not make.

OR

B) The hand belongs to our side AND 3NT takes the same number of tricks as 4♠

OR

C) The hand does not belong to our side, but 4♥ does not make.

OR

D) Our partner has 3+ spades.

 

Agree or disagree?

[gwnn]

Oh, this part of the discussion is not really interesting to me. I like mathematics and making obvious bridge bids. I regard 4♠ as obvious in this case. I don't think your points are especially productive as long as you cannot ever estimate their respective likelihoods. But even without exact probabilities, I would say:

 

a) this seems very unlikely - the hand belongs to a partnership, but they can't make game in their 7 card major?

b) as long as our partnership knows for certain that 3NT is a good spot from the auction - I can't see this happening with any confidence. I think the probability of such a deal is maybe 5% and the probability of us confidently identifying is 1%, so in total it is .05%

c) definitely disagree - maybe it belongs to them but they defend undoubled. maybe it belongs to them, 4♥ doesn't make but 7♦ makes!

d) I don't understand this at all - if partner has 3+ spades, we are more likely to have game on and they are very likely of having a big fit somewhere, hence preempting will work out better than if partner has 0-2 spades. If you want to say that 3♠ will work just as well because partner will raise us anyway, I emphatically disagree. we are letting in LHO much cheaper and our opps will gain a lot of info.

 

so I disagree.

[VM1973]

Oh, this part of the discussion is not really interesting to me. I like mathematics and making obvious bridge bids. I regard 4♠ as obvious in this case. I don't think your points are especially productive as long as you cannot ever estimate their respective likelihoods. But even without exact probabilities, I would say:

 

a) this seems very unlikely - the hand belongs to a partnership, but they can't make game in their 7 card major?

b) as long as our partnership knows for certain that 3NT is a good spot from the auction - I can't see this happening with any confidence. I think the probability of such a deal is maybe 5% and the probability of us confidently identifying is 1%, so in total it is .05%

c) definitely disagree - maybe it belongs to them but they defend undoubled. maybe it belongs to them, 4♥ doesn't make but 7♦ makes!

d) I don't understand this at all - if partner has 3+ spades, we are more likely to have game on and they are very likely of having a big fit somewhere, hence preempting will work out better than if partner has 0-2 spades. If you want to say that 3♠ will work just as well because partner will raise us anyway, I emphatically disagree. we are letting in LHO much cheaper and our opps will gain a lot of info.

 

so I disagree.

 

A) It's not a question of the likelihood of that because it's something we would calculate. Your partner have, for example, ♥KQJx or ♦KQJx which would be less than ideal for offense.

B) Considering that the honors that are likely to help you in hearts and diamonds are the Aces (for example partner holding 3 aces or ♥AK and ♦A, or ♥AQ ♦A with a heart lead giving a trick) the hand is likely to make the same number of tricks in NT as in the suit.

C) I was going to say that the strongest defensive possibility is to find your partner with ♦A and one or more spade honors cashing to beat 4♥ (or partner with a favorable trump holding) - under those circumstances it seems quite unlikely that 4♥ will fail but 7♦ rolls home. Additionally, if their spot is diamonds the same logic applies to beating that... cashing ♠A, ♥A, heart ruff and then looking around for more.

D) Not just because those who bid 3♠ will get to the same contract but also because your partner may bid on unwisely to 5♠ when looking at spade support.

[gwnn]

yes, I'm done with this. please don't consider me rude for doing this, but I am simply not interested. I did not want to participate in this argument at all, but I wanted to correct the maths calculations. good luck with passing, or bidding 3♠, or what you think is right.

[aguahombre]

IMHO, posts 2,4,7,8,11 and 14 were appropriate and sufficient to cover this B/I topic.

[Lurpoa]

no

 

♥♥♥ WHY ? ♥♥♥

[nige1]

My partner opened 3♠ with this hand and I passed with: ♠ 8 2 ♥ K 7 5 2 ♦ A K Q 10 6 ♣ J 5

so I guess the next question is: if partner opens 3♠ do you raise to 4 with this hand (same conditions, MP's)

no

♥♥ WHY ? ♥♥

No and it isn't close. KQJ109xx xx x xxx Is there a bridge player with a pulse who wouldn't open 3♠ at favourable even in 2nd seat? Mps is about plus scores: if I held this hand and one of my partners opened 3♠ at favourable I would be far more concerned that we are going -50 against air than that we have just missed a game. Even at imps, I'd pass...now red v white, 2nd seat, imps...I'd bid...ony because a 40% chance of making makes it a good call,

Lurpoa, we all ♥ love ♥ you :) too but please don't ask why when there is a well-reasoned explanation.

[awm]

I've done a pretty thorough bridge browser study on these sorts of hands and 4♠ is a significantly winning bid at the table.

 

What VM1973's analysis misses is that opponents will often make incorrect competitive decisions over such a high-level opening. For example:

 

(1) We make 4♠, but opponents have a good sacrifice at the five-level or even a double-game swing. If we open 4♠ they will often not find it. If we open 1♠ (or 3♠) they often will.

 

(2) We cannot make 4♠ on best-defense, but opponents make the wrong lead and we grab some quick pitches. Again, a slower auction might help opponents get in a lead directional call or hear what responder's long suit is.

 

(3) Opponents mistakenly double a making 4♠, or fail to double a 4♠ that doesn't make (again when a slower auction would help them).

 

(4) Opponents take a phantom sacrifice over a non-making 4♠ and go for a number.

 

All of these sorts of things are difficult to evaluate by generating random hands and running simulations, but you can see them by exploring a large database of hands from real play.

 

Finally, it's important to remember that opening 1♠ does not solve all problems. Yes, it is possible to stay out of game if that is right, or to play in 3NT. But you will not always get these decisions correct; there can easily be hands where you miss a making game (if you plan to rebid 2♠ over 1NT for example and partner passes with i.e. ♠x ♥Axxx ♦xxxx ♣KQxx). Opening 4♠ puts a lot of pressure on the opponents to make a correct competitive decision (pass, double, or compete) and then get the subsequent defense right. In the long run this kind of action will be successful, even though it is easy to give examples where it doesn't work out, or even to argue statistically that it won't work against opponents who are (impossibly) double-dummy perfect in bidding and defense.

[VM1973]

I've done a pretty thorough bridge browser study on these sorts of hands and 4♠ is a significantly winning bid at the table.

 

What VM1973's analysis misses is that opponents will often make incorrect competitive decisions over such a high-level opening. For example:

 

(1) We make 4♠, but opponents have a good sacrifice at the five-level or even a double-game swing. If we open 4♠ they will often not find it. If we open 1♠ (or 3♠) they often will.

 

(2) We cannot make 4♠ on best-defense, but opponents make the wrong lead and we grab some quick pitches. Again, a slower auction might help opponents get in a lead directional call or hear what responder's long suit is.

 

(3) Opponents mistakenly double a making 4♠, or fail to double a 4♠ that doesn't make (again when a slower auction would help them).

 

(4) Opponents take a phantom sacrifice over a non-making 4♠ and go for a number.

 

All of these sorts of things are difficult to evaluate by generating random hands and running simulations, but you can see them by exploring a large database of hands from real play.

 

Finally, it's important to remember that opening 1♠ does not solve all problems. Yes, it is possible to stay out of game if that is right, or to play in 3NT. But you will not always get these decisions correct; there can easily be hands where you miss a making game (if you plan to rebid 2♠ over 1NT for example and partner passes with i.e. ♠x ♥Axxx ♦xxxx ♣KQxx). Opening 4♠ puts a lot of pressure on the opponents to make a correct competitive decision (pass, double, or compete) and then get the subsequent defense right. In the long run this kind of action will be successful, even though it is easy to give examples where it doesn't work out, or even to argue statistically that it won't work against opponents who are (impossibly) double-dummy perfect in bidding and defense.

What you haven't said is if you specifically took the seat into account. I completely agree that a 4♠ opening is likely to work in 1st and 3rd seat. You are, however, in 2nd seat and the opener has passed.

 

Using Bridge Baron I dealt 10 hands to analyze them. In all of the hands I held:

♠AKQ10987

♥2

♦2

♣5432

 

White vs. Red, 2nd seat, matchpoints, opener passes. This is, in my opinion, superior than culling a large database of people of all seats and drawing generalizations about what will happen. These were the results:

 

Hand 1:

4S and 1S lead to +450

3S and Pass lead to +200.

 

Hand 2:

4S and 1S lead to +420

3S leads to +100 (opponents bid 5♦)

Pass leads to 4S doubled making.

 

Hand 3:

The most interesting hand, dummy hits with:

♠Jx

♥AKQxx

♦AKxxx

♣x

 

After a low heart lead I won on the board, ruffed a small heart in hand and ran the spades hoping for some kind of squeeze. In the end hearts broke 4-3 and +510 for everyone - flat.

 

Hand 4:

4S and 1S lead to +450

3S +200

Pass leads to 4S doubled +5

 

Hand 5:

4S and 1S lead to +450

3S and Pass lead to +200

 

Hand 6:

4S and 1S lead to +420

3S +170

Pass -> 4S*+4

 

Hand 7:

Regardless what you open you reach 4S*+5. Flat

 

Hand 8:

4S, 3S, and Pass lead to 4S*-1

1S +140

 

Hand 9:

Another very interesting hand.

4S gets cracked right off the bat and dummy hits with ♠Jxxxx (!) Routine play leads to 12 tricks.

3S, 1S, Pass lead to +480

 

Hand 10:

RHO has ♠Jxxxx of spades and 4S never has a prayer, but doesn't get doubled, either.

4S-2

3S-1

1S+100 (4H-1)

Pass+100 (4H-1) Although in all fairness, had I not previously played the hand 3 times, I might have taken an action after P-P-1H-P-1S-???

 

End Result:

3S +9

4S +16.5

Pass +17

1S +17.5

 

My conclusion: 3S doesn't work out, but the other 3 options are too close to call.

[wyman]

10 hands is a laughable sample size from which to draw any conclusion, and you should know that.

[gwnn]

Hand 2:

4S and 1S lead to +420

3S leads to +100 (opponents bid 5♦)

Pass leads to 4S doubled making.

 

May I ask how this happened? Opps find 5♦ after 3♠, but not after 1♠ or pass? Indeed, after pass they make the worst decision?