[cherdano]

Scoring: BAM

4N (P) 5D AP

---

LHO leads ♥A, RHO following with the 6.
How do you play the trumps on a heart continuation? Pretend that LHO is an expert.

[jdonn]

I can add to the problem as an opponent. RHO played the 8 not the 6, then on the second round LHO continued with the T and when dummy ruffed RHO played the 6. Opponents play UDCA but needless to say it's up to you to draw whatever conclusions you want about what their cards mean in this unusual situation.

[helene_t]

I would just play ♦J and play low if RHO does not cover. Obviously I must be missing something.

[dburn]

Relevant cases appear to be: West began with seven hearts and a singleton diamond honour; West began with a diamond void. In the first case I should play ace and another diamond; in the second I should take a first-round finesse in diamonds.

Since the first case has an a priori probability of around 0.5% and the second an a priori probability of around 11%, I will take a first-round finesse in diamonds.

[dicklont]

With 3 diamonds with West I'm already down. If East has them all I must finesse.
With diamonds 2-1 either way the finesse is ok too. In diamonds I see no problem.

With diamonds 1-2 and clubs 5-0 West can give East the ruff he could already have given him.
Clubs 5-0 is 4%, diamonds 3-0 is 22%. Finesse East is much better.

When the expert has ♦K and 5 clubs and sees that giving the ruff immediately would gain nothing
since South would probably drop the remaining diamonds I congratulate him.

[cherdano]

dburn, on Dec 15 2007, 04:34 AM, said:

Relevant cases appear to be: West began with seven hearts and a singleton diamond honour; West began with a diamond void. In the first case I should play ace and another diamond; in the second I should take a first-round finesse in diamonds.

Since the first case has an a priori probability of around 0.5% and the second an a priori probability of around 11%, I will take a first-round finesse in diamonds.

I think your odds are wrong, the first case is around 1.3%.

How do you play if LHO shifts to a club at trick two?

[dake50]

Am I to be outsmarted or out-dumbed? Only H7-2 or C5-0 with RHO DHx loses a ruff.
I can be dumb and lose to DKQx RHO, trying to score 'smart' vs LHO clever.

[neilkaz]

dake50, on Dec 15 2007, 10:11 AM, said:

Am I to be outsmarted or out-dumbed? Only H7-2 or C5-0 with RHO DHx loses a ruff.
I can be dumb and lose to DKQx RHO, trying to score 'smart' vs LHO clever.

And note that LHO really can't be sure if we have only 3♦ in which case he certainly wants to give PD a ♣ ruff.

I also finesse here.

[ArtK78]

Sure seems like passing the ♦J is the percentage move.

If LHO has a singleton diamond honor and 7 hearts and continues hearts, that is life.

IF LHO shifts to a club at trick two, it is more likely that LHO has the singleton club rather than 4 clubs. Nevertheless, it does change the problem. I am sure that passing the diamond J is still the right play, but I am more worried than I would have been if LHO had continued hearts.

[dburn]

cherdano, on Dec 15 2007, 11:02 AM, said:

dburn, on Dec 15 2007, 04:34 AM, said:

Relevant cases appear to be: West began with seven hearts and a singleton diamond honour; West began with a diamond void. In the first case I should play ace and another diamond; in the second I should take a first-round finesse in diamonds.

Since the first case has an a priori probability of around 0.5% and the second an a priori probability of around 11%, I will take a first-round finesse in diamonds.

I think your odds are wrong, the first case is around 1.3%.

How do you play if LHO shifts to a club at trick two?

I agree with you; I was calculating in my head and forgot to factor in the extent to which West's holding seven hearts to East's two affected the probability that he would hold a singleton diamond honour. I think I also divided by two once more often than necessary. But it was early in the morning, or if it wasn't, it felt as though it was.

To illustrate what I ought to have done:

The chance that West has seven hearts and East two is about 4.28%.

The chance that West has a singleton diamond honour given that he has seven hearts is about 32.35%.

The chance that West has both danger holdings is thus, as cherdano rightly says, the product of those two numbers - about 1.39%.

The chance that West has a diamond void, regardless of what he has in hearts, is exactly 11%. On ace and another heart, I would still be inclined to take the safety play in diamonds.

For the case where West cashes ♥A and shifts to a club, we need the chance that West began with four clubs - about 14.13% - multiplied by the chance that he has a singleton diamond honour given that he has four clubs - about 29.77%.

The product of these two numbers is about 4.21%; still significantly less than the chance that West has no diamonds, and the safety play in diamonds is still better in a vacuum.

Of course, as always we are not operating in a vacuum but at a bridge table. An interesting question is this: how should West (having led ♥A) defend with five clubs and the singleton king of diamonds?

[Halo]

Very difficult hand.

People talk about 'extended restricted choice'. Here we have a heart versus a club continuation.

I'm finessing, because I don't believe any of my opponents are that good, including the ones I've played in dburn's class. On this layout I don't feel very confident and would tip my cap to opponent in the next room who 'got it right' if he did.

[ArtK78]

dburn approaches the statistical analysis in the usual manner, but it is actually not as simple as presented.

Multiplying the probabilities of two events to determine the probability of both events occurring is correct if the events are totally independent. However, the fact that hearts are 7-2 and diamonds are 1-2 are not independent events. It is more likely that diamonds are 1-2 when hearts are 7-2 than it would be if hearts were 6-3, 5-4, 4-5, and so on.

Nevertheless, the probabilities calculated by dburn are close to the actual probabilites.

It has been quite some time since I studied statistics and probability in college, so I am not prepared to determine what the actual odds are. And I suspect that it would be a waste of time to do so, since the odds are close to what you get by treating the events as independent events.

[dburn]

ArtK78, on Dec 15 2007, 04:30 PM, said:

dburn approaches the statistical analysis in the usual manner, but it is actually not as simple as presented.

Multiplying the probabilities of two events to determine the probability of both events occurring is correct if the events are totally independent.  However, the fact that hearts are 7-2 and diamonds are 1-2 are not independent events.  It is more likely that diamonds are 1-2 when hearts are 7-2 than it would be if hearts were 6-3, 5-4, 4-5, and so on.

Nevertheless, the probabilities calculated by dburn are close to the actual probabilities.

It has been quite some time since I studied statistics and probability in college, so I am not prepared to determine what the actual odds are.  And I suspect that it would be a waste of time to do so, since the odds are close to what you get by treating the events as independent events.

They aren't "close to the actual probabilities". They are the actual probabilities. This ought to have been apparent from my use of the phrase "the chance that West has a singleton diamond honour given that he has seven hearts".

The a priori probability that a given West hand about which nothing else is known has a singleton diamond honour, for example, is exactly 26% (when North-South have ten diamonds). West will have one diamond exactly 39% of the time; it will be an honour in exactly two thirds of those cases. But the probability that the West hand has a singleton diamond honour given that it has seven of East-West's nine hearts rises to 32.35%, which is the figure I quoted in my post.

As I have already explained to Hannie, I am not a mathematician. But - or therefore - I know how to count.

[jdonn]

ArtK78, on Dec 15 2007, 04:30 PM, said:

dburn approaches the statistical analysis in the usual manner, but it is actually not as simple as presented.

Multiplying the probabilities of two events to determine the probability of both events occurring is correct if the events are totally independent.  However, the fact that hearts are 7-2 and diamonds are 1-2 are not independent events.  It is more likely that diamonds are 1-2 when hearts are 7-2 than it would be if hearts were 6-3, 5-4, 4-5, and so on.

Nevertheless, the probabilities calculated by dburn are close to the actual probabilites.

It has been quite some time since I studied statistics and probability in college, so I am not prepared to determine what the actual odds are.  And I suspect that it would be a waste of time to do so, since the odds are close to what you get by treating the events as independent events.

Not that he needs me to defend him, but you are completely wrong. I mean you are right about what he has to do, but wrong when you say he didn't do it. His calculation is accurate to the degree it was rounded. You don't need a complex statistical formula to solve this, a basic understanding is plenty good enough.

If hearts are 7-2 then west has 6 empty spaces (13-7), east has 11.

The odds of west having singleton king are (6/17)*(11/16)*(10/15), in other words (empty spaces in west / total EW empty spaces) * (east spaces / total remaining spaces) * (east spaces / total remaining spaces). This equals about 16.18%.

You can follow this process to find, given that west has 7 hearts and east has 2, the odds of west having a diamond holding of...
void = (11/17)*(10/16)*(9/15) = 24.26%
5 = (6/17)*(11/16)*(10/15) = 16.18%
Q = (6/17)*(11/16)*(10/15) = 16.18%
K = (6/17)*(11/16)*(10/15) = 16.18%
Q5 = (6/17)*(5/16)*(11/15) = 8.09%
K5 = (6/17)*(5/16)*(11/15) = 8.09%
KQ = (6/17)*(5/16)*(11/15) = 8.09%
KQ5 = (6/17)*(5/16)*(4/15) = 2.94%
As a check the total is ..... 100%! Ok, 100.01%, stupid rounding error.

So as I suspected all along, David was completely correct. The odds of west having a singleton diamond honor, GIVEN that west has exactly 7 hearts, are 16.18*2 = 32.35% (more rounding error). Your criticism of his calculation is entirely unfounded.

More relevant may be waiting to see what east follows with when you lead off dummy. If the assumption is still west has 7 hearts, and east follows with the 5 of diamonds, the odds he has KQ5 are 24.26/(24.26+16.18+16.18+8.09) = 37.5%, whereas the odds he has either K5 or Q5 are 50%. So if you knew hearts were 7-2 (which of course you don't) then you should go up ace by a 4 to 3 margin. This, combined with all appropriate bridge inferences (which are truly the interesting part of this problem IMO, and are probably David's greatest strength over the rest of us here) is how the problem should be solved.

[dburn]

jdonn, on Dec 15 2007, 05:14 PM, said:

ArtK78, on Dec 15 2007, 04:30 PM, said:

dburn approaches the statistical analysis in the usual manner, but it is actually not as simple as presented.

Multiplying the probabilities of two events to determine the probability of both events occurring is correct if the events are totally independent.  However, the fact that hearts are 7-2 and diamonds are 1-2 are not independent events.  It is more likely that diamonds are 1-2 when hearts are 7-2 than it would be if hearts were 6-3, 5-4, 4-5, and so on.

Nevertheless, the probabilities calculated by dburn are close to the actual probabilites.

It has been quite some time since I studied statistics and probability in college, so I am not prepared to determine what the actual odds are.  And I suspect that it would be a waste of time to do so, since the odds are close to what you get by treating the events as independent events.

Not that he needs me to defend him, but you are completely wrong. I mean you are right about what he has to do, but wrong when you say he didn't do it. His calculation is accurate to the degree it was rounded. You don't need a complex statistical formula to solve this, a basic understanding is plenty good enough.

If hearts are 7-2 then west has 6 empty spaces (13-7), east has 11.

The odds of west having singleton king are (6/17)*(11/16)*(10/15), in other words (empty spaces in west / total EW empty spaces) * (east spaces / total remaining spaces) * (east spaces / total remaining spaces). This equals about 16.18%.

You can follow this process to find, given that west has 7 hearts and east has 2, the odds of west having a diamond holding of...
void = (11/17)*(10/16)*(9/15) = 24.26%
5 = (6/17)*(11/16)*(10/15) = 16.18%
Q = (6/17)*(11/16)*(10/15) = 16.18%
K = (6/17)*(11/16)*(10/15) = 16.18%
Q5 = (6/17)*(5/16)*(11/15) = 8.09%
K5 = (6/17)*(5/16)*(11/15) = 8.09%
KQ = (6/17)*(5/16)*(11/15) = 8.09%
KQ5 = (6/17)*(5/16)*(4/15) = 2.94%
As a check the total is ..... 100%! Ok, 100.01%, stupid rounding error.

So as I suspected all along, David was completely correct. The odds of west having a singleton diamond honor, GIVEN that west has exactly 7 hearts, are 16.18*2 = 32.35% (more rounding error). Your criticism of his calculation is entirely unfounded.

More relevant may be waiting to see what east follows with when you lead off dummy. If the assumption is still west has 7 hearts, and east follows with the 5 of diamonds, the odds he has KQ5 are 24.26/(24.26+16.18+16.18+8.09) = 37.5%, whereas the odds he has either K5 or Q5 are 50%. So if you knew hearts were 7-2 (which of course you don't) then you should go up ace by a 4 to 3 margin. This, combined with all appropriate bridge inferences (which are truly the interesting part of this problem IMO, and are probably David's greatest strength over the rest of us here) is how the problem should be solved.

I understood most of that, because it agrees with what I think, so I was a heavy favourite to understand it. But I am not sure about the last part - in fact, I don't understand it at all.

If East follows with a diamond other than the five when you lead from dummy (or does not follow with any diamond) you can forget about computing probabilities and concentrate on not revoking.

But if East does follow with ♦5, you need to compute the probabilities as we (or at any rate some of us, cherdano being primus inter pares) have already done. The only relevant cases are those in which East has three diamonds (you should finesse) or West has one diamond honour and seven hearts (you should play the ace). Since it is about nine times more likely that the first case holds than the second, you should finesse by a 9 to 1 margin, not go up ace by any margin at all, let alone 4 to 3.

It is worth remarking that East, with three or more hearts and ♦KQx, might follow upwards in hearts (to pretend to have a doubleton playing udca). I gather that in fact he followed downwards, to represent three or more. Now, should I play him to have been false-carding? The original problem invited me to pretend that West was an expert, but who is this East guy?

[han]

dburn, on Dec 15 2007, 05:14 PM, said:

As I have already explained to Hannie, I am not a mathematician. But - or therefore - I know how to count.

Ah, therefore. So not only do mathematicians not know how to count, now you claim that all non-mathematicians know how to count? That's shouldn't be hard to check, let me call my mother.

I agreed with your analysis and Arend's play at the table. I think that he should finesse in diamonds no matter how the opponents defend.

[jdonn]

dburn, on Dec 15 2007, 05:36 PM, said:

But if East does follow with ♦5, you need to compute the probabilities as we (or at any rate some of us, cherdano being primus inter pares) have already done. The only relevant cases are those in which East has three diamonds (you should finesse) or West has one diamond honour and seven hearts (you should play the ace). Since it is about nine times more likely that the first case holds than the second, you should finesse by a 9 to 1 margin, not go up ace by any margin at all, let alone 4 to 3.

I didn't mean to confuse the situation. What I tried to state was:

IF we somehow know west has 7 hearts, THEN if east follows the the 5 of diamonds on a lead off dummy it is exactly 4 to 3 odds to go up with the ace.

This was perhaps an unnecessary diversion and I realize that is not a solution to the actual problem, since of course we don't know with any certainty how hearts are distributed. It is, however, a valid thought process, and does not contradict your admittedly more relevent statement about what one should actualy do in that position (although it's not 9 times more likely to work, it's about 7.9 times more likely.)

What I said could, as an aside, solve the problem of how to play if west had opened 4♥. I suspect in that case we would all play the ace of diamonds at trick 3 anyway, but I for one am rather shocked to find that it is nearly as good (about 43% to 57%) to finesse in that scenario.

I agreed with the finesse also. West actually had JTx AQJT9xx K xx. I told Arend to post just to make sure we weren't missing some inference, but I guess to think that would just be resulting.

The only thing you said that I don't agree with regards east's plays. I'm not sure why you believe they would be count at all, I would take it as suit preference (do I have the ace of clubs / club void / club singleton or not?)

[nick_s]

For anyone wanting to know more about calculating the odds using vacant spaces, I heartily recommend 'Bridge Odds for Practical Players' by Hiugh Kelsey and Michael Glauert.

[ceeb]

Win in HAND and play a low diamond. This has a small chance to lose when diamonds are 2-1 and a significant chance to gain when LHO has all three, given the uninformative bidding and the fact of BAM scoring.

[Echognome]

ceeb, on Dec 15 2007, 04:13 PM, said:

Win in HAND and play a low diamond. This has a small chance to lose when diamonds are 2-1 and a significant chance to gain when LHO has all three, given the uninformative bidding and the fact of BAM scoring.

I wouldn't worry about the BAM scoring. As it happens the board was won (or lost, depending on how you look at it) on the lead. Just play to make your contract.

Oh, and I agree with Arend's play as well.