[benlessard]

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.

[Cascade]

Isn't the easy way to see this that the problem is symmetrical in the defenders.

If either defender started with four hearts to the jack then he had no choice in his discards (only the order).

In addition if either defender held four hearts and the spade king then we would have seen either a heart discard or the spade king.

There is no information that suggests one rather than the other opponent had a choice in his plays.

[HighLow21]

No matter what you assume about discarding strategies, it is more likely that RHO has the long hearts than LHO, simply because RHO has more vacant spaces to hold JXX♥ after trick 7. I personally think that RHO is far more likely to have it than LHO, but it's still always bigger no matter how you think through the math.

Here's my explanation.

After trick 7 (1♥ 1♠ 2♣ and 3♦ have been played) there are 12 unseen cards: 2♦ 3♥ and 7♠.
The defenders randomly choose 6 of them; there are 12!/(6!x6!) = 924 unique ways to distribute them.

Of these, LHO will have no diamonds 210 times, 2 diamonds 210 times, and 1 diamond 504 times. (These are 6/12 x 5/11 x 924, 6/12 x 5/11 x 924, and 6/12 x 6/11 x 924 x 2, respectively.) By the end of trick 9, we're restricted to the 210 cases with LHO having started life with 5 diamonds.

Of these:
- LHO will have the JXX♥ without the K♠ 5 times. This is about 2% of the time.
- LHO will have the JXX♥ with the K♠ 2 times. (By the time the J♠ is pitched from dummy, these 2 cases can be eliminated, because LHO is squeezed.)
- LHO will have 2 hearts 63 times and 1 heart 105 times. This is 80% of the time.
- RHO will have the JHH♥ 35 times. This is about 16.7% of the time. Of these, LHO will usually have the K♠, but certainly not always. (I still haven't decided how to treat the inference that RHO does or does not pitch K♠ after this trick ends. I tend to think it's only marginally relevant.)

NOW: An intelligent defender who fully understands the situation after trick 7 will reason as follows: "The only relevant cards are the diamonds, the K♠, and the 3 ♥. I will ONLY pitch diamonds on the clubs if I hold 3 hearts or no hearts; otherwise I will pitch spades."

(Note that in this reasoning diamonds ARE relevant because of restricted choice, in a sense: each defender must have at least 2 spades, but neither defender must have another diamond.)

So back to the 924 cases after Trick 7 has ended.
1. 7 times, West will have all 5 remaining ♥/♦. Of these, 2 can be eliminated because 2 of them will include K♠.
2. 7 times, East will have all 5 remaining ♥/♦.
3. 42 times, West will have 3 ♥ and 1 ♦.
4. 42 times, East will have 3 ♥ and 1 ♦.
5. 35 times, West will have 3 ♥ and no ♦.
6. 35 times, East will have 3 ♥ and no ♦.

Under this strategy, West will only pitch two diamonds on the clubs and FAIL to pitch the K♠ while holding the ♥ stack 5 times; he will do so 35 times when East has the ♥ stack. Different reasoning, exact same result.

There are other lines of reasoning that get you different probabilities. For example, you could start at the beginning of trick 10 and say there are 4 relevant cards (K♠ and the 3 hearts); there are 70 ways to arrange them, and after the J♠ is discarded, 4 times West has all the hearts without the K♠ and 5 times, East does. In this case, the relative probability of East having the ♥ stack is lower, but it's still higher than the probability of West holding them.

[phil_20686]

mikeh, on 2012-January-31, 19:02, said:

While that statement is, I think, fair in general, there are types of falsecards that one should use all the time....so-called mandator falsecards.

For example, here is a well known one:

Dummy has AQ10x and declarer is known to hold 4 in the suit.....you hold J9xx in front of dummy.....declarer cashes the A and partner follows low. If you play low, declarer leads to his K and hooks your J. So you have to play the 9...now declarer can pick up Jxxx (assuming he has the 8) by cashing the Q....and you have created a stopper....this only works when you falsecard the 9....you have to do it all the time....you never gain when you don't.

I think this is in that category. I think there is no edge to ever not playing the K...the more frequently you falsecard, the closer to making declarer's choice a guess.

I got started on an attempt to show this mathematically and realized that I was far too ignorant of the math to even attempt. I suspect that I could have done this readily enough 40 years ago, but my engineering degree has sat unused for some 40 years, and my math skills, never great, have languished along with it.

We basically agree. But I should point out that in mandatory false cards represent situations where without the false card declarer has no losing options. In this case he has a losing option, he can finesse the other way. Attempting to look like I was squeezed is flat out deception and as such one should try to avoid doing it so often that declarer can draw any inference from your failure to attempt the deception.

[billw55]

Maybe slightly off topic, but I wonder what the actual best line of play is against a club lead. Is the line given in the OP best, or one of the squeezes, or something else?

[HighLow21]

I was wondering the same thing. Wouldn't it make sense to play the major suit aces first (discarding a blocking middle ♥ on the A♥), run the clubs, and then and only then play off the 3 top diamonds? The Q♠ can be discarded on the 3rd ♦, and the early ♣ plays force the defenders into discards when much less is known about the unseen hands. Expert defenders may not have any trouble, or they might; anyone below expert level probably will give something away. The Vienna coup play enforces an autosqueeze on anyone holding 4♥ and K♠, and communications still exist to play either opponent for 4♥ if they have them. I think this line makes it much more likely that declarer will get either (1) sufficient information to play hearts with certainty, or at least (2) enough information to make a strong inference one way or the other.

I realize the post was about which defender is more likely to hold the 4♥ strictly given the actual line of play, but it still is an interesting question to me. Does anyone see an improvement on the line I described?

[benlessard]

Another way to check your solution is to find at what point do you know East or West is favorite to be longer in H and why does it change at that point. When declarer cashed 3D+1H+1S+2C the position at this point is totally symmetric (so both defender have the same expected H lenght). After the 3rd club (defense 1st discard) are the defense significant discard (IMO no) ? Does it change something that west discarded a D when he could have discarded a S (no again) ? What about the 2nd discard ? (imo its still even no matter what the discard are since both players have idle cards) What if on the 3rd discard (on the last club) you see west a computer player discarding the K of S or you see a human player conceding or see the player go into thinking and discard a H ? How do you think the H split now ? IMO its clear that the odds are shifting exactly when the 5th club is cashed and west make is 3rd discard.

[benlessard]

In addition if either defender held four hearts and the spade king then we would have seen either a heart discard or the spade king.

The problem is that east can discard the K of S after the dummy. So he can discard the K of S from spade lenght randomly or to fool you into thinking he is long in H.

[HighLow21]

The 2 diamonds are the only significant discards on correct defense as my previous post explains. They reveal the number of spaces that are available for the only 4 relevant cards remaining (K♠ and Jxx♥) and the probability of them lying in one particular hand. I'm becoming more and more convinced that West's pitching them at the first opportunity was either a horrible blunder or a genius bit of reverse psychology if he were indeed 2542 without K♠.

Of course, a heart discard or the K♠ discard is also relevant, but I'm assuming we're not playing against relatives here.

[gszes]

fred, on 2012-January-31, 10:18, said:

---

You get a club lead against 7NT.

Rightly or wrongly, you play a second round of clubs (both follow) and then 3 rounds of diamonds (both follow) discarding the Queen of spades from your hand. Then you cash a top heart from your hand (both follow small) and the Ace of spades (both follow small) before finishing the clubs.

On the run of the clubs, LHO discards 2 diamonds and a small spade while RHO discards 3 small spades. You discard dummy's Jack of spades on your fifth club.

Which opponent do you play for Jxxx of hearts and why?

Fred Gitelman
Bridge Base Inc.
www.bridgebase.com

deleted garbage due to misread of trick 10

[Cascade]

gszes, on 2012-February-02, 20:13, said:

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]

Quote

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]

benlessard, on 2012-February-04, 05:57, said:

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]

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...

---

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

[A] seems more probable.

---

[WellSpyder]

JLOGIC, on 2012-February-01, 13:27, said:

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]

WellSpyder, on 2012-February-08, 07:15, said:

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]

gnasher, on 2012-February-08, 07:43, said:

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.