[barmar]

whereagles, on Mar 13 2009, 06:51 PM, said:

I think the point is, the further you play on, if bad breaks were about, they would have shown already. So stuff rates to break ok.

That's how I interpreted it as well. In suit contracts, opponents are likely to lead their short suits in an attempt to get a ruff. At NT, they'll lead their long suits to try to set them up. So towards the end, you mainly have the even suits left.

[Echognome]

As to the original question and to a point made earlier, if there are 3 tricks left and neither defender has played the suit then it is a certainty that they split 3:3. If there are 4 tricks left then the odds are 4:3 in favor of a 3:3 split. The odds start out at against a 3:3 split. I leave it as an exercise to show this relationship is monotonic and to calculate the number of remaining tricks left when the odds become greater than "evens".

[H_KARLUK]

quiddity, on Mar 13 2009, 09:35 PM, said:

In "Reese on Play", in Chapter 1 ('The Simple Probabilities'), Reese writes:

"There is another factor to be considered.  The further the play has advanced, the more likely are the even divisions.  For example, if in the middle of the play a declarer with a combined seven cards in a suit plays two rounds, to which all follow, it is better than evens that the two outstanding cards will break 1-1; in other words a 3-3 break has become more likely than 4-2." 

How can this be right?

When opponents    Division   Probability
hold
      6 cards                  3-3            36%
      6 cards                  4-2            48%
      6 cards                  5-1            15%
      6 cards                  6-0              1%

A point worth remembering is that when th opponents have an odd number of cards in a suit th most probable division is th most even one. When opponents have an even number of cards, however, an uneven break is th most probable.
There is no need to commit this table to memory, but a knowledge of th main points can be useful in helping to determine th best line of play. It is immediately seen, for instance, that a simple finesse (50 %) is a better shot than relying on a 3-3 break (36%), but a worse bet than playing for a 3-2 break (68 %).

Finesse or drop
When opponents    Probability of an honor card being
have                        Singleton    Doubleton    Trebleton
    6 cards                  2.5%              16%              36%

When opponents have
2 cards               Play for th drop
3 or 4 cards        Finesse against th king but not against th queen or knave
5 or 6 cards        Finesse against th king or queen but not against th knave
7 cards               Finesse against th king, queen or knave

[Trumpace]

jdonn, on Mar 13 2009, 07:11 PM, said:

ArtK78 said:

HOWEVER, when you play the third round of the suit and the next player follows, things have changed. Now you can eliminate half of the possible 4-2 breaks. At this point, the odds of a 3-3 break are higher than the odds of the only remaining possible 4-2 break. The odds in favor of the 3-3 break are now 35.5 to 21.2, or in excess of 3-2 in favor of the 3-3 break.

Your third paragraph is the same argument by which beginners try to drop the queen doubleton offside with AKJx and xxxx. They cash the ace then lead toward the king, lho follows, they say "well if the suit is 3-2 onside, it's as likely RHO's last card is the missing x as it is the missing Q, so my play is a 50-50 guess."

Art's reasoning seems right to me... What am I missing?

[awm]

Echognome, on Mar 14 2009, 12:19 AM, said:

As to the original question and to a point made earlier, if there are 3 tricks left and neither defender has played the suit then it is a certainty that they split 3:3. If there are 4 tricks left then the odds are 4:3 in favor of a 3:3 split. The odds start out at against a 3:3 split. I leave it as an exercise to show this relationship is monotonic and to calculate the number of remaining tricks left when the odds become greater than "evens".

Not true.

The problem is that there is a difference between the following processes:

(1) Deal out eight cards to two people, of which six are clubs. Check the club break.

(2) Deal out twenty-six cards to two people, of which six are clubs. Ask each person to discard nine non-clubs if possible. Check the club break.

In the first case, the odds of a 3-3 club break are 4/7 and a 4-2 club break is 3/7, exactly the odds given by Echognome and 4:3 in favor of a 3-3 split.

But in the second case, the odds of a 3-3 club break before the discards were 35.5% and a 4-2 break was 48.5%. The discards of non-clubs don't change anything here (except eliminating the 5-1 and 6-0 break possibilities) so the odds remain the same, roughly 4:3 in favor of a 4-2 split!

[Stephen Tu]

Trumpace, on Mar 14 2009, 10:53 AM, said:

ArtK78 said:

HOWEVER, when you play the third round of the suit and the next player follows, things have changed. Now you can eliminate half of the possible 4-2 breaks. At this point, the odds of a 3-3 break are higher than the odds of the only remaining possible 4-2 break. The odds in favor of the 3-3 break are now 35.5 to 21.2, or in excess of 3-2 in favor of the 3-3 break.

Art's reasoning seems right to me... What am I missing?

I don't know where he is getting those numbers, he is way off. If you are playing a suit like AKTx vs. Qxx, cash A/Q then lead the third, and LHO follows, then at that point you have eliminated all but one specific 3-3 break and all but one specific 4-2 break. The odds of any one 3-3 break are 1.7764%, of one 4-2 are 1.6149%, so the drop is a 52.38% favorite. Nowhere near 3-2.

[Trumpace]

Stephen Tu, on Mar 14 2009, 02:57 PM, said:

Trumpace, on Mar 14 2009, 10:53 AM, said:

ArtK78 said:

HOWEVER, when you play the third round of the suit and the next player follows, things have changed. Now you can eliminate half of the possible 4-2 breaks. At this point, the odds of a 3-3 break are higher than the odds of the only remaining possible 4-2 break. The odds in favor of the 3-3 break are now 35.5 to 21.2, or in excess of 3-2 in favor of the 3-3 break.

Art's reasoning seems right to me... What am I missing?

I don't know where he is getting those numbers, he is way off. If you are playing a suit like AKTx vs. Qxx, cash A/Q then lead the third, and LHO follows, then at that point you have eliminated all but one specific 3-3 break and all but one specific 4-2 break. The odds of any one 3-3 break are 1.7764%, of one 4-2 are 1.6149%, so the drop is a 52.38% favorite. Nowhere near 3-2.

The numbers could be off, but the reasoning that 3-3 is better than 4-2 seems right. I was wondering about Jdonn's post about this reasoning being similar to beginner's reasoning in AKJx vs xxxx (and so incorrect).

[ArtK78]

Stephen Tu, on Mar 14 2009, 02:57 PM, said:

Trumpace, on Mar 14 2009, 10:53 AM, said:

ArtK78 said:

HOWEVER, when you play the third round of the suit and the next player follows, things have changed. Now you can eliminate half of the possible 4-2 breaks. At this point, the odds of a 3-3 break are higher than the odds of the only remaining possible 4-2 break. The odds in favor of the 3-3 break are now 35.5 to 21.2, or in excess of 3-2 in favor of the 3-3 break.

Art's reasoning seems right to me... What am I missing?

I don't know where he is getting those numbers, he is way off. If you are playing a suit like AKTx vs. Qxx, cash A/Q then lead the third, and LHO follows, then at that point you have eliminated all but one specific 3-3 break and all but one specific 4-2 break. The odds of any one 3-3 break are 1.7764%, of one 4-2 are 1.6149%, so the drop is a 52.38% favorite. Nowhere near 3-2.

I was not discussing SPECIFIC 3-3 and 4-2 breaks containing a particular card. I was discussing 3-3 breaks and 4-2 breaks in general.

I am hoping that I am not beating a dead horse, but suppose you have a 4-3 fit in a suit and you have played two rounds of the suit, with both opponents following. The relative chances of 4-2 breaks and 3-3 breaks have not changed (assuming that you have no information from the play of the other suits that have an impact on the likelihood of 4-2 and 3-3 breaks in this suit). So, the relative chances of the suit breaking 4-2 or 3-3 is still 4-3 in favor of the 4-2 break. But there are two different 4-2 breaks - one in which LHO has 4 and the other in which RHO has 4. So, when you play the third round in the suit and LHO follows, you have eliminated the chance that RHO has 4. Now the odds are roughly 3-2 in favor of a 3-3 break.

In the situation that you are discussing, there is an important card missing - the J. That changes the calculation significantly. If you play two rounds of the suit with both opponents following and the J has not appeared, and then you play the third round of the suit and LHO follows small, you are down to only two possibilties: Jxxx on your left or xxx on your left. The a priori possibility of these SPECIFIC two holdings (all other things being equal) is 16.149% vs. 17.764%, or approximately 52.38% in favor of the drop.

By the way, the calculation in my prior post was incorrect, in that it should have been 35.5 to 24.2, not 35.5 to 21.2. So the odds in favor of the 3-3 break after the opponents followed to 2 1/2 rounds of the suit were slightly less than 3-2 in favor, not slightly more than 3-2 in favor.

[ArtK78]

By the way, in the situation where you have a 4-3 fit with all of the high cards, I know of no practical application of the fact that the odds change in favor of a 3-3 break from 4-3 against to 3-2 in favor after you have played 2 rounds of the suit and you play the third round of the suit and the next opponent follows suit. Obviously, the situation involving the missing Jack is of far more practical significance. The only reason I brought it up was in the context of the topic being discussed in the OP.

[dburn]

ArtK78, on Mar 14 2009, 03:53 PM, said:

I am hoping that I am not beating a dead horse, but suppose you have a 4-3 fit in a suit and you have played two rounds of the suit, with both opponents following.  The relative chances of 4-2 breaks and 3-3 breaks have not changed (assuming that you have no information from the play of the other suits that have an impact on the likelihood of 4-2 and 3-3 breaks in this suit).  So, the relative chances of the suit breaking 4-2 or 3-3 is still 4-3 in favor of the 4-2 break.  But there are two different 4-2 breaks - one in which LHO has 4 and the other in which RHO has 4.  So, when you play the third round in the suit and LHO follows, you have eliminated the chance that RHO has 4.  Now the odds are roughly 3-2 in favor of a 3-3 break.

Is there an award for the most staggering lot of nonsense ever advanced on the BBO forums? Because if there is, the above "reasoning" would win not only this year's prize, but every prize for which it was entered from here to perpetuity.

[Stephen Tu]

OK, Art is right for the case where the opps have all low cards and will play them completely randomly. Of course I can't see how this would ever have any practical significance at the table.

[655321]

ArtK78, on Mar 14 2009, 03:53 PM, said:

I was not discussing SPECIFIC 3-3 and 4-2 breaks containing a particular card.  I was discussing 3-3 breaks and 4-2 breaks in general.

I am hoping that I am not beating a dead horse, but suppose you have a 4-3 fit in a suit and you have played two rounds of the suit, with both opponents following.  The relative chances of 4-2 breaks and 3-3 breaks have not changed (assuming that you have no information from the play of the other suits that have an impact on the likelihood of 4-2 and 3-3 breaks in this suit).  So, the relative chances of the suit breaking 4-2 or 3-3 is still 4-3 in favor of the 4-2 break.  But there are two different 4-2 breaks - one in which LHO has 4 and the other in which RHO has 4.  So, when you play the third round in the suit and LHO follows, you have eliminated the chance that RHO has 4.  Now the odds are roughly 3-2 in favor of a 3-3 break.

This is not correct. ArtK is assuming that all 4-2 breaks and all 3-3 breaks are still possible.

There is only one possible 3-3 break remaining (1.7764 a priori), and only one possible 4-2 break remaining (1.6149). So after 2 1/2 rounds have been played, using ArtK's method, we get a 52.4% chance of a 3-3 break.

Coincidentally (?!) this is the same probability we get using vacant places - i.e after 2 1/2 rounds, there are 10 spaces in the hand with 3 cards in the suit, and 11 spaces in the other hand.

[Stephen Tu]

No, actually Art is correct in the case of random low cards because of restricted choice principles.

Say we are missing the 234567. LHO has shown 457 in some order, RHO has shown 23 in some order.

The relevant possibilities remaining are 754-632 and 7654-23.
But we have to multiply the a priori odds by the odds that an opponent chose to conceal a particular card. Assuming random carding, the odds of the 3-3 break are reduced by factor of 3 (1.7764/3 = 0.5921), and the odds of the 4-2 splits are reduced by 4. (1.6149/4 = 0.4037)

Which works out to the same ratios Art is using.

Of course when the J is missing the numbers change since LHO won't randomly play J from Jxxx in the first 3 rounds, nor will RHO randomly play the J from Jxx the first couple rounds.

[655321]

OK, makes sense.

[hanp]

I also think Art is correct and I can't think of a single bridge context in which anybody would care.

[ArtK78]

hanp, on Mar 14 2009, 09:23 PM, said:

I also think Art is correct and I can't think of a single bridge context in which anybody would care.

I thought that I said that in my post. It is just a curiosity - it has no practical application.

And I don't appreciate the insults. It is beneath the usual dignity of the poster.

[hanp]

I didn't mean to insult you or anybody else in this thread. I did miss that you already said that there were no applications.

[ArtK78]

Han, I was not referring to you. You did state that I was right even if there was no practical application, although you did not phrase it exactly that way. I apologize for quoting you and then mentioning the insults - I did not intend to connect the two.

[ArtK78]

At the risk of provoking the ire of some posters, I believe I may have come up with a practical application of the situation in which you have a 4-3 fit and you need to decide whether to play for a 3-3 break or a 4-2 break.

Suppose you reach this end position, the contract being no trump, you need 4 tricks to make your contract and the possibility of making an overtrick is not important. Furthermore, all the cards in the other suits are gone and you have no information about the distribution in the remaining two suits or the location of the high cards in the remaining two suits, neither of these suits having been played to this point in the hand:

Scoring: IMP

---

You play the A, K and J of spades, RHO following to the A and K, and LHO following to the A, K and J. You need to decide whether to overtake your ♠J, hoping the suit breaks 3-3, or holding the lead with the ♠J so that you can lead towards the ♥K at trick 11.

If you have absolutely no information about the distribution of these two suits or the location of the high cards in hearts, then holding the lead with the ♠J to lead towards the ♥K is a 50-50 play. But, as has been demonstrated above, once RHO has followed to two rounds of spades and LHO has followed to 3 rounds of spades, overtaking the ♠J with the ♠Q is almost a 3-2 favorite to win 4 spade tricks.

[hanp]

First of all, well done finding a somewhat reasonable application.

Second, If all minor suit cards are out yet you know nothing of their distribution then you haven't been paying much attention

Thirdly, If LHO holds 4 spades and the QJ of hearts then winning the spade jack and taking the heart hook still works.

Continuing on your idea one could come up with a hand where the bidding is 2NT-3NT and dummy has ♦Qxxx and ♥Kx, with declarer holding ♥QJx and ♦AKJ. Dummy has no further entries. Declarer receives an unrevealing attitude lead in some side suit and to make her contrract she absolutely needs 4 diamond tricks. She plays AKJ of diamonds and all follow, should she overtake or hope the ace of hearts is onside...