Strange passage in Reese on Play

[Stephen Tu]

 

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]

 

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]

 

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]

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]

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]

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:

 

[hv=d=n&v=n&n=sq432hk2dc&s=sakjh543dc]133|200|Scoring: IMP[/hv]

 

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...

[H_KARLUK]

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.

Dburn: I nominate your thread. My reason : You violated Isaac Newton's rule .“Tact is the art of making a point without making an enemy.” (English Mathematician and Physicist, "father of the modern science", 1642-1727)

 

But i see your default mood : “It's only arrogance if you're wrong” .

 

Now i expect your honest apologize to ArtK78. Well, if you think that you are a King or extracurricular one I doubt your abilities.

 

Are you really sure that you are here to discuss an idea politely or to name in an ugly way whenever you upset ? As you see it's not so hard.

[ArtK78]

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...

Thank you. I spent some time trying to come up with a position where the decision of whether to play for the 3-3 break or the 4-2 break might occur. Taking it one step further to figure out how that position would arise in real life is a bigger challenge. Clearly, there can be no outside entry to the hand with the Q432 (or the opponents knock out that entry by leading the suit on opening lead).

 

As for not being able to find out anything about the location of the opponents' high cards in the heart suit of the distribution in the spade suit (you substituted diamonds for spades in your discussion), there are hands where you do not find out any relevant information, or the information that you do find out does not point you in one direction or the other. Good opponents can make it difficult to find out any helpful information about the location of the unknown cards.

 

As for your example, it would work better if you reduced your heart holding to Qxx opposite Kx and changed the requirement to declarer needs 5 of the last 6 tricks.

[hackenbush]

Some of you may already have ready this from Pavlicek's site.

 

How Percentages Change

[EricK]

Can anybody answer this question:

 

You have a 3-3-3-4 distribution dummy is 3-3-4-3. You get a spade lead. You cash ♠AK, ♥AK ♦AK all following.

 

At this point, what are the probabilities for each club break?

 

Ignore the fact that the information so obtained might be useless and that we have almost certainly already misplayed the hand!

[dburn]

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.

I'm very sorry, Art. I had misread your post, thinking that it was a reference to the original problem (AK10x facing Qxx), not to the more recently discussed situation (AKJx facing Qxx).

 

In that situation, of course, the chances are as you say: when West follows to the third round of the suit, either he began with three cards (an initial 36% chance) or with four cards (an initial 24% chance). As to a practical application of the principle, one might consider:

 

[hv=d=s&v=b&n=s32hk32dq432c5432&s=saj4h654dakjcakqj]133|200|Scoring: IMP[/hv]

South, declarer in 3NT, wins the opening lead of ♠K (for fear of a heart shift). East indicates an even number, and declarer assumes correctly that spades are 4-4.

 

On four rounds of clubs, West follows twice and discards a spade and a heart; East follows three times and discards a spade. Both follow to ♦AK, and West follows to ♦J. West, who has fewer black cards than East, is a favourite to have ♥A, so should one retain the lead in South and play a heart towards the king, or should one overtake ♦J?

 

(Note: one should probably do neither of the above; instead, one might play three rounds of diamonds, overtaking in dummy, before playing the fourth round of clubs. If diamonds do not break, there is still the option on a heart to the king. But, especially in the beginners' lounge, the clubs might be played off early...)

[ArtK78]

Thank you, David, for your most recent post.

 

The hand that you post as a possible application of this principle contains the suit combinations noted in one of my earlier posts. Putting together a hand where this is the only way to deal with the problem is a bigger challenge. As you state in the hand that you posted, you could test the diamond suit (winning the third round in dummy by overtaking) to see if it broke 3-3 before cashing all of the clubs.

 

If you weaken the spade position to Axx opposite xx, then there is more of a problem. If you assume that the opponents' spades are breaking 4-4, the option of cashing the diamonds by leading the A, K and J and overtaking with the Q in dummy could establish a diamond as the 5th defensive trick when the ♥A is onside. So, in that case, it would make sense to cash all of the clubs first hoping for a useful discard.