Declarer play problem

[Cascade]

J

Q7 (hopefully we unblocked the T on the heart A)

void

void

 

void

K98

void

void

 

This is not the position as the ♠J has been discarded.

[gszes]

The chosen LOP gave away too much information

making the general overall discarding by the

opps too easy thus no useful information can

be gleaned from their carding.

 

the end postions are either

 

Kxx void void void vs void jxx void void

void Jxx void void vs Kxx void void void

 

for position 1 to be reached lho would

have had to start with 5152

and rho with 4432

 

for position 2 to be reached lho would have

had to start with 2452

and rho 7132

 

the odds of positon 1 are around 21.5%

the odds of positon 2 are around 6%

 

% of all (including 3/2 hearts) possible holdings

which means positon 1 is a 3 1/2 to 1 favorite of

relevant positions.

 

The sad part is rho can do nothing to dissuade

us from playing for postion 1. It is up to lho

to weave a believeable tale of woe due to

forced discards that they have 4 hearts.

[diejowae]

Not wrong, now assuming the position of how the card,LHO 5152 RHO 4432 OR LHO 2452 RHO 7132. I select the first position, a small play Heart to Q.

[benlessard]

for position 1 to be reached lho would

have had to start with 5152

and rho with 4432

 

the odds of positon 1 are around 21.5%

the odds of positon 2 are around 6%

 

Most of the 5152 hes going to discard all the S and keep the D. So with perfect defense only part of the 5152 are there not the full 21.5%

[benlessard]

Interesting side question you are west and you hold the 5152, you are playing against a perfect computer who remember all the hands that you have played against him. What is your discard strategy ? How often are you going to discard 2D & 1S ?

[ron_ron]

Interesting side question you are west and you hold the 5152, you are playing against a perfect computer who remember all the hands that you have played against him. What is your discard strategy ? How often are you going to discard 2D & 1S ?

 

First consider the simpler situation that Fred described, where the defenders have all the top spades except for the ace. In this situation, if you were playing an infinite number of hands against this perfect computer, with 5152 you must discard 2D & 1S 1/5 of the time. There are a few ways to see this. One way is as follows: the only relevant layout in which LHO might discard 2D & 1S are 5152 and 2452. On the given information, the former is 5 times more likely, because there are 7*1 ways to put one of the remaining seven spades and all of the remaining three hearts in LHO's hand and 35*1 ways to put four of the remaining seven spades and all of the remaining three hearts in LHO's hand. If you discard 2D & 1S more or less frequently than 1/5 of the time with 5152, declarer will have better than a 50-50 guess about which of the two layouts exists.

 

The question is more complicated for the original problem. The play in diamonds is more flexible because the SK has a greater effect on declarer's strategy (and secondarily, because RHO can vary the frequency with which he plays the SK when he has the heart shortness, based on the defenders' diamond strategy). The short answer (if I haven't made a mistake) is that for any possible optimal defensive strategy, LHO must discard 2D & 1S from 5152 between 6/35 to 34/35 of the time.

 

Long answer that most people will probably not want to read:

 

There are 9 relevant holdings for LHO:

 

S H D probability*140

 

x Jxx xx 6

xxxx -- xx 15

Kxxx -- xx 20

 

xx Jxx x 30

xxxxx -- x 12

Kxxxx -- x 30

 

xxx Jxx -- 20

xxxxxx -- -- 1

Kxxxxx -- -- 6

 

If the defense discards small spimonds randomly and RHO does not play the SK unduly often, declarer can do no better than play RHO for Jxxx, picking up 84/140 = 3/5 of the layouts. If LHO discards diamonds more or less frequently than random, RHO may or may not be able to make up for this with his SK strategy. If LHO discards 2D & 1S from a holding including short H less than 6 times every 140 relevant hands (only on 35 hands is this possible), then declarer will gain by playing LHO for the long H when he sees 2D & 1S. Likewise LHO needs to discard 1D & 2S from short hearts at least 30 times every 140 relevant hands and 0D & 3S from short hearts at least 20 times every 140 relevant hands. These three conditions are also sufficient to ensure that RHO has an optimal SK strategy. In each of the three cases (n = 6, 30, 20), short-hearted RHO just needs to divide out his n "last discards" between SK and non-SK in such a way that LHO is more likely to have short hearts in each case (for example RHO could play SK exactly balancing RHO's forced SK's, and non-SK slightly less than RHO's forced non-SK's).

 

Since LHO needs to discard fewer than 2D (from a short H holding) on at least 50 out of 140 relevant hands, but there are only 49 such hands on which LHO actually has fewer than 2D, he must discard 2D on at most 34 of the 35 relevant hands with 2D and 1H.

 

I assumed RHO discards randomly from small spimonds and that this doesn't affect the result. Maybe that's not true, but this is more than anyone wanted to read anyway :) I was just curious what the answer was.

[nige1]

[hv=pc=n&s=sahakt9dakqcakqjt&w=shdc&n=st98765432hq432dc&e=shdc]399|300|

 

Analogous problem, South declares 7N against good opponents. Both opponents follow to ♠A (with ♠Q and ♠J), ♥A, three ♦ and four ♣. On the fifth ♣, declarer discards dummy's last ♠, leaving...[/hv]

[hv=pc=n&s=shkt9dc&w=shdc&n=shq43dc&e=shdc]399|300|

 

The ♠K has not appeared. Hence neither opponent can have been dealt ♠K and four ♥. Thus, the critical cases are:

[A] LHO has ♠K and RHO has ♥Jxx OR

LHO has ♥Jxx and RHO has ♠K (having chosen not to discard it).

 

[A] seems more probable.

[/hv]

[WellSpyder]

Very smart people in this thread. Great read, enjoyed it.

Not entirely sure what to read into this, but would it be fair to say that in your experience the ability to solve a problem like this isn't actually relevant to whether or not one can get to the BB final? If so, I would find this slightly reassuring, since I find that when the discussion turns to the importance of volunteered vs non-volunteered information the discussion can quickly get very confusing....

 

I have absolutely no confidence in my "reasoning" standing up to scrutiny on this sort of hand, but I think I would approach it much more simply than many posters here by asking when the two different lines would work. Playing LHO for 4 hearts at the critical point means the line will work whenever hearts are 3-2 and on the half of the 4-1 breaks when they are with leftie. Playing RHO for 4 hearts will work whenever hearts are 3-2, the half of the 4-1 breaks when they are with rightie, but also half of the other 4-1 breaks when LHO has SK as well as 4 hearts. That surely makes it a better line. I know at the point the problem is presented we actually know we haven't squeezed LHO out of SK and 4 hearts, but I'm not convinced this actually changes the answer - wouldn't allowing it to be like saying in a restricted choice situation missing QJxx that when an opponent drops J you know they didn't start with a singleton Q so that possibility is no longer relevant to the odds?

[gnasher]

I have absolutely no confidence in my "reasoning" standing up to scrutiny on this sort of hand, but I think I would approach it much more simply than many posters here by asking when the two different lines would work. Playing LHO for 4 hearts at the critical point means the line will work whenever hearts are 3-2 and on the half of the 4-1 breaks when they are with leftie. Playing RHO for 4 hearts will work whenever hearts are 3-2, the half of the 4-1 breaks when they are with rightie, but also half of the other 4-1 breaks when LHO has SK as well as 4 hearts. That surely makes it a better line. I know at the point the problem is presented we actually know we haven't squeezed LHO out of SK and 4 hearts, but I'm not convinced this actually changes the answer - wouldn't allowing it to be like saying in a restricted choice situation missing QJxx that when an opponent drops J you know they didn't start with a singleton Q so that possibility is no longer relevant to the odds?

I think that's an excellent way of looking at it. Another, equivalent, approach is to consider the situation after we have cashed the last club. Of the four layouts:

(1) LHO 4 hearts, RHO ♠K

(2) LHO ♠K, RHO 4 hearts

(3) LHO 4 hearts + ♠K

(4) RHO 4 hearts + ♠K

1 and 2 are equally likely; 3 is known to be impossible; 4 is still possible. Thefeore we should play RHO for four hearts.

 

What's difficult about this is getting to the point where we realise that the answer is simple. To do that, we have to understand that:

(a) The opponents' small spades and small diamonds are all equivalent, so we should ignore which ones they throw

(b) From RHO's point of view in the 3-card ending, ♠K is just another irrelevant small card, so we don't care whether he's thrown it or not.

[phil_20686]

I think that's an excellent way of looking at it. Another, equivalent, approach is to consider the situation after we have cashed the last club. Of the four layouts:

(1) LHO 4 hearts, RHO ♠K

(2) LHO ♠K, RHO 4 hearts

(3) LHO 4 hearts + ♠K

(4) RHO 4 hearts + ♠K

1 and 2 are equally likely; 3 is known to be impossible; 4 is still possible. Thefeore we should play RHO for four hearts.

 

What's difficult about this is getting to the point where we realise that the answer is simple. To do that, we have to understand that:

(a) The opponents' small spades and small diamonds are all equivalent, so we should ignore which ones they throw

(b) From RHO's point of view in the 3-card ending, ♠K is just another irrelevant small card, so we don't care whether he's thrown it or not.

 

Are one and two equally likely? I think they are not. for one thing the minors are divided 7-5, and the spades being divided 2-7 is less likely. Somehow we also need to fold this distributional probability in.

 

For example, if lho has 4 hearts its 7/9 for the spade K to be on the right, but if rho has four hearts the spade is only 5/4 to be on the left. Equivalently, while three is impossible, it is must the least likely of the four scenarios anyway, since lho is less likely to have four hearts anyway due to the vacant spaces in the minors, and also because if he does the spade K is very likely to be with RHO.

[phil_20686]

It a simpler but equivalent solution to what i was thinking before. Before the run of the clubs West has 7 know cards and 6 vacant spaces & so do East. This IMO is the important starting point. We know that West has at least 3 idles cards (he cannot have Ks and the 3 remaining hearts) and that west has at least 2 idle cards (for him the K of S is an idle card since he can discard it easily after dummy discard his J of S). So the H distribution should be ---5 vacants spaces for West and 6 vacants spaces for East. West is going the have Jxx in hearts slightly more than 6% and East has twice the odds (more than 12%) to have the 3 remaining H.

 

As I understand vacant spaces, you should only include suits where the suit break is known completely, so diamonds cannot be included until the thirteenth is discarded on the run of the clubs.

[gnasher]

What's difficult about this is getting to the point where we realise that the answer is simple.

See what I mean?

[mikeh]

I think the main idea that you can take out of this is that (against optimal defenders) the discards of the opponents (for example the fact that LHO pitched two diamonds) are completely irrelevant, but the fact that LHO cannot have 4 hearts and the spade king is relevant. This makes it more likely that RHO has 4 hearts.

 

For Bluecalm the following argument might be convincing. There are these possibilities:

 

A) hearts split 3-2.

 

B) LHO has the king of spades and Jxxx of hearts.

 

C) LHO has the king of spades and RHO has Jxxx of hearts.

 

D) RHO has the king of spades and Jxxx of hearts.

 

E) RHO has the king of spades and LHO has Jxxx of hearts.

 

Scenarios A and B are irrelevant since you always make it in these cases.

 

In the remaining 3 scenarios RHO has Jxxx of hearts twice (C and D) and LHO has Jxxx of hearts once (E). Since C and E are exactly equally likely, we should play RHO for the heart length.

 

To make this argument correct we have to convince ourselves that the discards are really irrelevant. See my previous post for how you can do that, it takes some work but it is not hard.

At the risk of merely emphasizing my ignorance in these matters, I take what you are saying to be that we have 3 scenarios that we need to consider: C, D and E. Since C and E cancel each other out, the fact that the 'un-cancelled' D gives the Jxx to RHO means we play him for it.

 

But isn't D known, on the original post, not to be present? By the time we make the decision, RHO is reduced to 3 cards, and he didn't pitch the spade K, therefore he didn't have D.

 

How then, when comparing which of C and E existed (the only two scenarios remaining as possible) do we give any weight to D?

 

Isn't what we are doing simply trying to assess the a priori chances that LHO was dealt a hand that resulted in C compared to a hand that resulted in D?

 

And that entails figuring out whether 5=1=5=2 is more or less probable than 2=4=5=2 as of the time of the actual deal. Thus, before anyone looked at their cards, LHO was probably some 4432 and almost as likely to be some 4333 and very unlikely to be either 2=4=5=2 or 5=1=5=2, but we've eliminated all but the last two shapes from consideration, and the ratio of their probabilities remains unchanged, assuming uninformative discards.

 

I suspect something is wrong with this, since I have earlier suggested that RHO should throw the spade K when he has it in order to make it look as if D existed.

 

if we assume that he will sometimes throw the spade K when he holds no hearts, the fact that in the actual case he didn't, makes E less likely than C. If he always throws the spade K when he has it then at the point of decision, RHO would have pitched the spade K and thus D would be back into the equation. Conversely, if RHO always pitches the spade K when he has it, the lack of the pitch takes E out. A blended strategy appears to be indicated.

 

My head hurts.

 

(ducking now in anticipation of a hopely polite explanation of my undoubted error)

[WellSpyder]

My head hurts.

I'm not surprised! How about this for a simple pragmatic reason for playing RHO for 4 hearts?

 

a) half the replies/analysis seems to suggest it makes no difference who you play for 4 hearts. In that case, you lose nothing by playing for it to be RHO

b) the other half argue that it is better to play RHO for 4 hearts.

c) as far as I can see, no-one thinks it is better to play LHO for 4 hearts.

 

So whatever (non-zero) odds you put on a) or b) being right you are better off playing RHO for 4 hearts! Simple, and you can rest your head because you don't even have to decide who is right :)

[fred]

It seems like some people are still having trouble (as I did) with the basic point of this problem: that the information regarding the diamond layout is not relevant. What follows is what I think is a good way to understand this point.

 

I am going to assume that the spades are xx opposite Ax in order to simplify things.

 

Consider the strategy that a defender (either one) with 5152 might adopt (and assume that declarer knows what that strategy is).

 

If a defender with 5152 never discards two diamonds then declarer will guess correctly 100% of the time that he sees two diamond discards (because he will know that the two diamond discards were forced plays from a defender with 2452).

 

If a defender with 5152 always discards two diamonds then declarer will guess correctly well over 50% of the time that he sees two diamond discards by playing the defender that discarded diamonds for 5152 (because that shape is considerably more likely than 2452).

 

So in the never case declarer can do better than a 50-50 guess by playing the diamond discarder for 2452 and in the always case declarer can do better than a 50-50 guess by playing the diamond discarder for 5152. I am not sure if I could prove it formally, but to me the implication is that there exists a certain amount of sometimes (which obviously falls between never and always) that a 5152 hand should discard diamonds that will leave declarer with a complete guess as to if he should play the diamond discarder for 5152 or 2452.

 

If the defenders adopt this strategy (discarding diamonds from 5152 the appropriate amount of time) then diamond discards are not relevant - declarer has a pure guess regardless of what the discards are.

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[mikeh]

If a defender with 5152 always discards two diamonds then declarer will guess correctly well over 50% of the time that he sees two diamond discards by playing the defender that discarded diamonds for 5152 (because that shape is considerably more likely than 2452).

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

I would have thought, and a brief search online seems to confirm this, that a priori a 5422 shape is far more probable than a 5521 shape. Admittedly, these are with respect to any 5422 and any 5521, and we are now reduced to more specific possibilities, but is Fred right on his assertion that 5152 is 'considerably more likely than 2452?.

[omarsh10]

The D discards for LHO are the must. The holder of the K of spades would have tried to inform his p about it unless he has both 4-hearts and K of spades. LHO failure to discard spades first or second time points to RHO as likely K-of-s holder so I would play LHO for having 4 hearts

[MrAce]

LHO failure to discard spades first or second time points to RHO as likely K-of-s holder so I would play LHO for having 4 hearts

 

LHO can not possibly discard ♠K because dummy's ♠ J is not discarded yet.

[fred]

I would have thought, and a brief search online seems to confirm this, that a priori a 5422 shape is far more probable than a 5521 shape. Admittedly, these are with respect to any 5422 and any 5521, and we are now reduced to more specific possibilities, but is Fred right on his assertion that 5152 is 'considerably more likely than 2452?.

One easy way to see this: if 2452 then spades are 2-7, but if 5152 then spades are 5-4.

 

Or maybe this works better for you: who is more likely to have 4 hearts? The hand with 5 diamonds (hence 2452) or the hand with 3 diamonds (hence 5152)?

 

A priori doesn't really matter since we know how diamonds and clubs break and since we are assuming hearts are 4-1.

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

[HighLow21]

I think that's an excellent way of looking at it. Another, equivalent, approach is to consider the situation after we have cashed the last club. Of the four layouts:

(1) LHO 4 hearts, RHO ♠K

(2) LHO ♠K, RHO 4 hearts

(3) LHO 4 hearts + ♠K

(4) RHO 4 hearts + ♠K

1 and 2 are equally likely; 3 is known to be impossible; 4 is still possible. Thefeore we should play RHO for four hearts.

 

What's difficult about this is getting to the point where we realise that the answer is simple. To do that, we have to understand that:

(a) The opponents' small spades and small diamonds are all equivalent, so we should ignore which ones they throw

(b) From RHO's point of view in the 3-card ending, ♠K is just another irrelevant small card, so we don't care whether he's thrown it or not.

 

(1) and (2) are unequivocally NOT equally likely. I know A/E players have developed a disdain for the concept of Vacant Spaces, but it is absolutely germane in this case. We KNOW LHO started with 2 more diamonds than RHO, so RHO is much more likely to hold the remaining heart stack. Whether Lefty gave us that info accidentally or on purpose, we have it. 5-to-1 is the calculation I explained earlier, and it has been corroborated by the analyses of others.

 

I realize that LHO might have been playing a deceptive game from 5152 shape to try to fool us, or from 2452 to try to fool us, but his cleverness (or lack thereof, depending on the true shape) cannot change the fact that he's shown us the 2 missing diamonds and therefore is much less likely to have also been longer in ♥.

 

Put simply, he started with 6 major suit cards originally, and his partner, 8. The original division of 2♠4♥ to the left and 7♠1♥ to the right is far less likely than 5♠1♥ at left and 4♠4♥ at right.

 

And by the way-->if he was anything other than 2452, his two diamond discards were a mistake, because a smart declarer who appreciates the importance of the diamond discards will then play ♥ correctly, much to his partner's chagrin. (And if Lefty WAS 2452, you should congratulate him on his brilliant defense!)

 

Yes, if he was 2452 originally, his diamond discards were forced, lest he give up the hand entirely, but that is far less likely than 5152, the only other relevant case.

[mikeh]

One easy way to see this: if 2452 then spades are 2-7, but if 5152 then spades are 5-4.

 

Or maybe this works better for you: who is more likely to have 4 hearts? The hand with 5 diamonds (hence 2452) or the hand with 3 diamonds (hence 5152)?

 

A priori doesn't really matter since we know how diamonds and clubs break and since we are assuming hearts are 4-1.

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

I have come up with another way of looking at it (I can hear the groans already).

 

My earlier reference to a priori was mistaken. We should instead look at the odds of either hand shape not on the deal (true a priori) but after we have looked at the hands held by declarer and responder.

 

The cards dealt to our opps are far more constrained than on the original deal.

 

When we are assigning 52 cards equally between 4 hands, 2452 is more likely than 5152.

 

But we know we're not doing that. We are assigning 9=5=8=4 cards equally between 2 hands.

 

We do that, and surely 5=1=5=2/4=4=3=2 is more likely than 2=4=5=2/7=1=3=2?

 

Each option contains one balanced/semi-balanced hand and one shapely hand.

 

if we compare 4432 to 5422, we'd expect 4432 to be more common. When we compare the unbalanced companion hands, we'd expect 5521 to be more common than 7321.

 

So when we compare the combination of 5152/4432 to 7132/2452, we find that both elements on the first are more common than their counterpart on the second.

 

So given the constraints arising from our hands being known, and given the elimination of all other shapes, it seems to me to be clear to hook rho.

 

I'm actually quite proud of this....even though it took me a long time (and a lot of help from others here) to get to this point.....so I really hope I don't shortly learn that I've screwed it up again.....like I said, my head hurts.

[benlessard]

If a defender with 5152 never discards two diamonds then declarer will guess correctly 100% of the time that he sees two diamond discards (because he will know that the two diamond discards were forced plays from a defender with 2452).

 

If a defender with 5152 always discards two diamonds then declarer will guess correctly well over 50% of the time that he sees two diamond discards by playing the defender that discarded diamonds for 5152 (because that shape is considerably more likely than 2452).

 

So in the never case declarer can do better than a 50-50 guess by playing the diamond discarder for 2452 and in the always case declarer can do better than a 50-50 guess by playing the diamond discarder for 5152. I am not sure if I could prove it formally, but to me the implication is that there exists a certain amount of sometimes (which obviously falls between never and always) that a 5152 hand should discard diamonds that will leave declarer with a complete guess as to if he should play the diamond discarder for 5152 or 2452.

 

If the defenders adopt this strategy (discarding diamonds from 5152 the appropriate amount of time) then diamond discards are not relevant - declarer has a pure guess regardless of what the discards are.

 

Fred Gitelman

Bridge Base Inc.

www.bridgebase.com

The theoretical anwser is that West should discard randomly from a pool of S and D. (and of course never discard a H or the K of S).

 

he has 4S/2D at this point and he has to make 3 discards. One of them is always a S. So he should discard SDD about 10%,SSS about 30% and SSD 60%. So this is why this problem is counter intuitive. After west discard 2 D we know that Spades are more likely to break a priori 54 than 27. However only 10% of the 5152 really count (because 90% of the times when west is 5152 hes NOT going to discard 2 diamonds). So that why if dummy had Ax in S and you play against a perfect defender you would have no clues on who to finesse because 5152--4432x10% equal the frequency of 2452-7132x100% (West with a 2452 is always going to discard 2D+1S)

[phil_20686]

The theoretical anwser is that West should discard randomly from a pool of S and D. (and of course never discard a H or the K of S).

 

he has 4S/2D at this point and he has to make 3 discards. One of them is always a S. So he should discard SDD about 10%,SSS about 30% and SSD 60%. So this is why this problem is counter intuitive. After west discard 2 D we know that Spades are more likely to break a priori 54 than 27. However only 10% of the 5152 really count (because 90% of the times when west is 5152 hes NOT going to discard 2 diamonds). So that why if dummy had Ax in S and you play against a perfect defender you would have no clues on who to finesse because 5152--4432x10% equal the frequency of 2452-7132x100% (West with a 2452 is always going to discard 2D+1S)

 

While there are cases where a mixed strategy by the defence lets them hide all info, surely this is not one of them, as the discards of the person who has 4 hearts is mandated.

 

My head hurts, going to bed.

[han]

I recommend searching back to a short post by gnasher which contains a 1-line explanation for why you should play RHO for the heart length, followed by some thinking about why this single line is correct.

 

(to be more clear, I recommend the searching followed by the thinking, I did not mean to suggest that gnasher's post contains thinking)

[phil_20686]

Suppose that the hands were 4432-5152 LHO has to discard three spades. if 3442 then he gets two spades and one diamonds, and if 2452 then two diamonds and one spade. So if LHO is any of the 5152 6142 or 7132, then on the third he must discard three spades, if the middle or the first I have some choices.

 

However, I need to use my mixed strategy not only to give equal probability to 5152 and 2452, but also to balance 4432 and 6142 and it is not clear to me that these aims are simultaneously acheivable. For example, 4432 could be enough more common than any other layout that the best you can for your overall strategy is to always try to look like 4432 whenever you do not have four hearts. In that case the best strategy overall would never involve discarding two diamonds from 5152.

 

Gnashers argument of treating them all like one suit would be true if all the distributions were close to equally likely, because in that case you can definitely achieve an optimal mixed strategy. I have no idea if a perfect mixed strategy is acheivable here. If it is then the odds are for RHO because lho was not squeezed, because you will be able to balance the relative likelyhood of diamonds being 5-3 in discards on other hands, so its irrelevant. If you cannot acheive a mixed strategy, or if your best mixed strategy involves mostly trying to look like diamonds are 4-4 then it would almost never involve discarding two diamonds, and so I should include that information.